Sobolev spaces and linear elliptic partial differential equations

Project Abstract

Sobolev spaces are function spaces equipped with a norm that incorporates information about both the function and its derivatives. These spaces play a fundamental role in the study of partial differential equations (PDEs), particularly in the context of linear elliptic PDEs. Linear elliptic PDEs are a class of PDEs that arise in various scientific and engineering applications, including heat conduction, fluid dynamics, and electromagnetism. In the study of linear elliptic PDEs, Sobolev spaces provide a natural framework for understanding the regularity properties of solutions. The Sobolev embedding theorems, which establish the inclusion relationships between different Sobolev spaces, are crucial tools in this context. These theorems provide insights into the smoothness of solutions to elliptic PDEs and help characterize the behavior of solutions near singular points. One of the key advantages of Sobolev spaces is their ability to handle functions with weak derivatives, making them well-suited for dealing with PDEs involving non-smooth coefficients or domains. This flexibility is particularly valuable in the study of elliptic PDEs, where the coefficients may exhibit irregular behavior that cannot be captured by classical function spaces. Additionally, Sobolev spaces offer a natural setting for defining weak solutions to elliptic PDEs. Weak solutions provide a generalized notion of solutions that allows for a broader class of functions to be considered. By working in Sobolev spaces, researchers can establish the existence and uniqueness of weak solutions to elliptic PDEs, providing a powerful method for solving these equations in a variational sense. Moreover, the theory of Sobolev spaces allows for the development of appropriate function spaces for the formulation and analysis of boundary value problems associated with elliptic PDEs. The trace theorems in Sobolev spaces, which relate the behavior of functions on the boundary of a domain to their properties in the interior, are essential for studying PDEs in bounded domains. In conclusion, Sobolev spaces play a central role in the study of linear elliptic PDEs, providing a rich mathematical framework for understanding the behavior of solutions to these equations. By leveraging the tools and techniques offered by Sobolev spaces, researchers can make significant progress in analyzing the regularity, existence, and uniqueness of solutions to a wide range of elliptic PDEs, contributing to advancements in various fields of science and engineering.

Project Overview

<p> </p><p>Spaces of Functions<br>In the following,<br>is a nonempty open subset of RN with the Lebesgue measure<br>dx.<br>1.1 Lp-spaces and some of its properties<br>1.1.1 Basic Integration Results<br>Theorem 1.1.1.1 (Monotone Convergence Theorem) Let ffng be a nondecreasing<br>sequence of integrable functions such that:<br>sup<br>Z</p><p>fn dx &lt; 1:<br>Then ffng converges pointwise to some function f. Futhermore f is integrable<br>and<br>lim<br>n!+1<br>Z<br>jfn ô€€€ fj dx = 0:<br>Theorem 1.1.1.2 (Lebesgue Dominated Convergence Theorem) Let ffng be<br>a sequence of integrable functions such that:<br>(i) fn(x) ! f(x) a.e on<br>,<br>(ii) there exists a function g, integrable and jfn(x)j g(x) a.e on<br>.<br>Then f is integrable and<br>lim<br>n!+1<br>Z<br>jfn ô€€€ fj dx = 0:<br>1<br>Theorem 1.1.1.3 (Fatou Lemma) Let ffng be a sequence of integrable functions<br>such that:<br>(i) 8 n; fn(x) 0 a.e on<br>,<br>(ii) sup<br>R<br>fn dx &lt; 1.<br>For x 2<br>, set f(x) = lim infn fn(x). Then f is integrable and<br>Z<br>f dx lim inf<br>n<br>Z<br>fn dx:<br>1.1.2 Definition and basic properties<br>Definition Let 1 p &lt; 1. We define:<br>(i) Lp(<br>) as the set of measurable functions f :<br>! R such that:<br>Z</p><p>jf(x)jp dx &lt; +1<br>and<br>(ii) L1(<br>) as the set of measurable functions f :<br>! R such that:<br>ess sup jfj &lt; +1<br>where<br>ess sup jfj = inf fK 0; jf(x)j K; a.e x 2<br>g<br>Definition We say that two functions f and g are equivalent if f = g almost<br>everywhere. Then we define Lp(<br>) spaces as the equivalent classes for this relation.<br>Remark 1.1.2.1 The space Lp(<br>) can be seen as a space of functions. We do<br>however, need to be careful sometimes. For example, saying that f 2 Lp(<br>) is<br>continuous means that f is equivalent to a continuous function. Now for f 2 Lp(<br>),<br>we define:<br>kfkp =<br>Z</p><p>jf(x)jp dx<br>1<br>p<br>; 1 p &lt; +1 (1.1)<br>kfk1 = ess sup jfj: (1.2)<br>Theorem 1.1.2.1 (Holder’s Inequality) . Let 1 p &lt; +1, we define p0 by<br>1=p + 1=p0 = 1: If f 2 Lp(<br>) and g 2 Lp0(<br>), then fg 2 L1(<br>) and<br>kfgk1 kfkpkgkp0 : (1.3)<br>2<br>Proof. The cases p = 1 and p0 = +1 are easy to prove. Now assume 1 &lt; p &lt;<br>+1. We use the following Young’s inequality: Let 1 &lt; p &lt; +1, a; b 0 then<br>ab<br>ap<br>p<br>+<br>bp0<br>p0 :<br>Assume that kfkp 6= 0 and kgkp0 6= 0 otherwise, nothing to do. Using Young’s<br>inequality, we have<br>jfj<br>kfkp</p><p>jgj<br>kgkp0</p><p>1<br>p<br>jfjp<br>kgkpp<br>+<br>1<br>p0<br>jgjp0<br>kfkp0<br>p<br>:<br>Thus<br>Z</p><p>jfj<br>kfkp</p><p>jgj<br>kgkp0<br>dx<br>1<br>p<br>Z</p><p>jfjp<br>kfkpp<br>dx +<br>1<br>p0<br>Z</p><p>jgjp0<br>kgkp0<br>p<br>dx =<br>1<br>p<br>+<br>1<br>p0 = 1:<br>Hence Z</p><p>jfj jgj dx kfkp kgkp0 :<br>Theorem 1.1.2.2 (Minkowski’s Inequality) . If 1 p +1 and f; g 2 Lp(<br>)<br>then<br>kf + gkp kfkp + kgkp: (1.4)<br>Proof. If f + g = 0 a:e, then the statement is trivial. Assume that f + g 6= 0<br>and p &gt; 1 (the case p = 1 is easy to check). We evaluate as follows:<br>jf + gjp = jf + gjjf + gjpô€€€1 (jfj + jgj)jf + gjpô€€€1:<br>Integrating over<br>, we get<br>Z</p><p>jf + gjp dx<br>Z</p><p>(jfj + jgj)jf + gjpô€€€1 dx<br>=<br>Z</p><p>jfjjf + gjpô€€€1 dx +<br>Z</p><p>jgjjf + gjpô€€€1 dx<br>Using Holder’s inequality in the right hand side, we obtain<br>Z</p><p>jf + gjp dx (kfkp + kgkp)kf + gkp=q<br>p ;<br>from which it follows<br>kf + gkp kfkp + kgkp:<br>3<br>1.1.3 The Main properties of Lp(<br>)<br>Lp-Spaces are Banach<br>Theorem 1.1.3.1 The Lp-spaces are Banach for 1 p +1.<br>Proof.<br>Case1. Assume that p = 1. Let ffng be a Cauchy sequence in L1. Let<br>k 1, there exists Nk such that<br>kfm ô€€€ fnkp<br>1<br>k<br>8 n;m Nk:<br>There exists a set of measure zero Ak such that<br>jfm(x) ô€€€ fn(x)jp<br>1<br>k<br>8 x 2<br>ô€€€ Ak; 8 n;m Nk: (1.5)<br>Let A = [Ak (A is of measure zero) and forall x 2<br>ô€€€ A the sequence ffn(x)g<br>is Cauchy in R. Let fn(x) = limn fn(x) forall x 2<br>ô€€€ A. Letting m goes to +1<br>in (1.5), we obtain<br>jfn(x) ô€€€ f(x)jp<br>1<br>k<br>8 x 2<br>ô€€€ Ak; 8 n Nk:<br>Thus f 2 L1 and kfn ô€€€ fkp 1=k; 8 n Nk: So kfn ô€€€ fkp ! 0:<br>Case2. Assume that 1 p &lt; +1. Let (fn)n1 be a Cauchy sequence in<br>Lp(<br>), then there exists a subsequence (fnk)k1 of (fn) such that:<br>kfnk+1 ô€€€ fnkkp<br>1<br>2k ; 8 k 1: (1.6)<br>To simplify the notations, let us replace fnk by fk so that:<br>kfk+1 ô€€€ fkkp<br>1<br>2k ; 8 k 1: (1.7)<br>Now set:<br>gn(x) =<br>Xn<br>k=1<br>jfk+1(x) ô€€€ fk(x)j:<br>It follows that:<br>kgnkp 1; 8 n 1:<br>Thus, from the monotone convergence theorem, gn(x) converge pointwise to some<br>g(x) almost every where and g 2 Lp: On the other hand we have: for all n;m 2<br>jfm(x) ô€€€ fn(x)j jfm(x) ô€€€ fmô€€€1(x)j + + jfn+1(x) ô€€€ fn(x)j g(x) ô€€€ gnô€€€1(x):<br>4<br>It follows that (fn(x)) is Cauchy in R and converges to some f(x) a.e. Letting<br>m goes to +1 leads to:<br>jf(x) ô€€€ fn(x)j g(x); 8 n 2:<br>Therefore f 2 Lp and by using dominate convergence theorem we have<br>kfn ô€€€ fkp ! 0:<br>We complete the proof by applying the following lemma<br>Lemma 1.1.3.1 Let E be a metric space and (xn) be a cauchy sequence in E. If<br>(xn) has a convergence subsequence, then it converges to the same limit.<br>The preceding proof contains a result which is interesting enough to be stated<br>separetely:<br>Theorem 1.1.3.2 (Convergence criteria for Lp functions) Let 1 p &lt; +1.<br>Let (fn) and f in Lp(<br>) such that (fn) converges to f in Lp(<br>). Then there exists<br>a subsequence (fnk) of (fn) and h 2 Lp(<br>) such that fnk(x) ! f(x) for a.e,<br>x 2<br>and fnk(x) h(x), a.e x 2<br>.<br>Remark 1.1.3.1 It is in general not true that the entire sequence itself converge<br>pointwise to the limit f, without some futher conditions holding.<br>Example 1.1.3.1 Let X = [0; 1], and consider the subintervals<br>h<br>0;<br>1<br>2<br>i<br>;<br>h1<br>2<br>; 1<br>i<br>;<br>h<br>0;<br>1<br>3<br>i<br>;<br>h1<br>3<br>;<br>2<br>3<br>i<br>;<br>h2<br>3<br>; 1<br>i<br>;<br>h<br>0;<br>1<br>4<br>i<br>;<br>h1<br>4<br>;<br>2<br>4<br>i<br>;<br>h2<br>4<br>;<br>3<br>4<br>i<br>;<br>h3<br>4<br>; 1<br>i<br>;<br>h<br>1;<br>1<br>5<br>i<br>;<br>Let fn denote the indicator function of the nth interval of the above sequence.<br>Then kfnkp ! 0, but fn(x) does not converge for any x 2 [0; 1].<br>Example 1.1.3.2 Let<br>= R, and for n 2 N, set fn = X[n; n + 1]. Then<br>fn(x) ! 0 as n ! 1; but kfnkp = 1 for p 2 [0;1). Thus fn converge pointwise<br>but not in norm.<br>Theorem 1.1.3.3 Let 1 p &lt; 1. Let ffng be a sequence in Lp such that<br>fn(x) ! f(x) a.e. If<br>lim<br>n<br>kfnk = kfk<br>then ffng converges to f in norm.<br>Theorem 1.1.3.4 The Lp spaces are reflexive for 1 &lt; p &lt; 1.<br>5<br>Proof. For 2 p &lt; 1. We have the follwing first Clarkson inequality:</p><p>f + g<br>2</p><p>p<br>p<br>+</p><p>f ô€€€ g<br>2</p><p>p<br>p</p><p>1<br>2</p><p>kfkpp<br>+ kgkpp</p><p>; 8 f; g 2 Lp:<br>For 1 &lt; p 2, we have the second Clarkson inequality:</p><p>f + g<br>2</p><p>p0<br>p<br>+</p><p>f ô€€€ g<br>2</p><p>p0<br>p</p><p>h1<br>2<br>kfkpp<br>+<br>1<br>2<br>kgkpp<br>i1=(pô€€€1)<br>; 8 f; g 2 Lp:<br>Using the Clarkson inequalities, we prove that Lp is uniformly convex for 1 &lt;<br>p &lt; 1. So it is reflexive by Milman-Pettis Theorem<br>Theorem 1.1.3.5 Let 1 p &lt; 1. Then Lp is separable.<br>Proof. Let (i)i2I be the family of N-cubes of RN of the form =<br>YN<br>k=1<br>]ak; bk[<br>where ak; bk 2 Q and<br>. Let E be the Q-vector space spanned by the<br>functions Xi .<br>Claim: E is a countable dense subspace of Lp.<br>Remark 1.1.3.2 L1 is not separable. To establish this, we need the following:<br>Lemma 1.1.3.2 Let E be a banach space. We assume that there exists a familly<br>(Oi)i2I such that:<br>(i) For all i 2 I Oi is a nonempty open subset of E;<br>(ii) Oi Oj = ; if i 6= j;<br>(iii) I is uncountable.<br>Then E is not separable.<br>Now we apply this lemma for L1 as follows:<br>For all a 2<br>, let ra such that 0 &lt; ra &lt; d(a;<br>c). Set fa = XB(a;ra) and<br>Oa = ff 2 L1 j kf ô€€€ fak1 &lt;<br>1<br>2<br>g:<br>One can check that the family (Oa)a2<br>satisfies (i), (ii) and (iii).<br>6<br>1.1.4 Dual Space<br>Theorem 1.1.4.1 (Riesz representation theorem.) Let 1 &lt; p &lt; +1 and let<br>2 (Lp)0. Then there exists a unique g 2 (Lp)0 such that:<br>h; fi =<br>Z</p><p>g f dx; 8 f 2 Lp(<br>):<br>Futhermore<br>kk(L1)0 = kgk1:<br>Proof. Let 1 &lt; p &lt; +1 and let p0 such that 1=p + 1=p0 = 1. For g 2 Lp0(<br>), we<br>define<br>Tg : Lp(<br>) ! R; hTg; fi =<br>Z</p><p>f g dx:<br>Using Holder’s inequality, we observe that Tg is well defined, linear and<br>jhTg; fij kgkp0kfkp:<br>Thus<br>kTgk(Lp)0 kgkp0 :<br>In fact we have kTgk(Lp)0 = kgkp0 . This follows by choosing f = jgjp0ô€€€2g.<br>Now we define the map<br>T : Lp0<br>! (Lp)0; by T(g) = Tg 8 g 2 Lp0<br>:<br>We have to prove that T is onto. For this, let E = T(Lp0). We have to show that<br>E is closed and dense in (Lp). E is closed by using the fact that kTgk = kgkp0<br>and Lp0 is Banach. For density we will show that if L 2 (Lp)00 and L = 0 on E<br>then L = 0 on (Lp)0. Since Lp is reflexive, we identify (Lp)00 to Lp through the<br>canonical embeding. Thus there exists f 2 Lp such that hL; i = h; fi, for all<br>2 (Lp)0. So L = 0 on E leads hTg; fi = 0 for all g 2 Lp0 and this implys that<br>f = 0 so L is.<br>Theorem 1.1.4.2 (Dual space of L1). Let 2 (L1)0, then there exists a unique<br>g 2 L1 such that<br>h; fi =<br>Z</p><p>g f dx; 8 f 2 Lp(<br>):<br>and<br>kk(L1)0 = kgk1:<br>Remark 1.1.4.1 The spaces L1(<br>) and L1(<br>) are not reflexive.<br>7<br>Indeed assume that L1 is reflexive and let<br>open such that assume that 0 2<br>.<br>Let fn = nXB(0;1=n), where n =</p><p>B(0; 1=n)</p><p>ô€€€1<br>so that kfnk1 = 1: For n large<br>enough, we have B(0; 1=n)<br>. By reflexivity, ffng has a weakly convergence<br>subsequence fnk to some function f in L1(<br>). Thus<br>Z</p><p>fnk’ dx !<br>Z</p><p>f’ dx; 8 ‘ 2 L1(<br>): (1.8)<br>So for ‘ 2 Cc(<br>ô€€€ f0g), we have<br>Z</p><p>fnk’ dx = 0 for k large enough. By (1.8) it<br>follows that Z</p><p>f’ dx = 0; 8′ 2 Cc(<br>ô€€€ f0g):<br>Thus Z f = 0 a.e on<br>. On the other hand, taking ‘ 1 in (1.8) leads to</p><p>f dx = 1. Contradiction. So L1(<br>) is not reflexive.<br>Since a Banach space is reflexive if and only if its dual E0 is reflexive, then<br>L1(<br>) is not reflexive.<br>Remark 1.1.4.2 Since (L1)0 = L1, then from Banach-Alaogulu theorem any bounded<br>sequence in L1(<br>) has a w-convergence subsequence.<br>Proposition 1.1.4.1 There exists a linear continuous forms on L1(<br>) such that<br>there is no g 2 L1(<br>) such that<br>hT; fi =<br>Z</p><p>g f dx; 8 f 2 L1(<br>):<br>Proof. Let<br>an open subset of Rn such that 0 2<br>. Let<br>0 : Cc(<br>) ! R; h0; ‘i = ‘(0):<br>0 is a linear continuous form on (Cc(<br>); k k1). So by the Hann-Banach<br>extension theorem, 0 can be extended to a continuous linear form on L1(<br>),<br>say . We summarize the main properties of the Lp spaces as follows:<br>Completeness Reflexivity Separability Dual Space<br>Lp; 1 &lt; p &lt; 1 yes yes yes Lp0 ; 1=p + 1=p0 = 1<br>L1 yes no yes L1<br>L1 yes no no Contains strctly L1<br>8<br>1.1.5 Convolutions and Mollifiers<br>Two usefull theorems<br>Let<br>1 RN,<br>2 RN open subsets of RN and F :<br>1<br>2 ! R be a measurable<br>function.<br>Theorem 1.1.5.1 (Tonelli) Assume that<br>Z</p><p>2<br>jF(x; y)j dy &lt; 1 a:e x 2<br>1<br>and Z</p><p>1<br>Z</p><p>2<br>jF(x; y)j dy</p><p>dx &lt; 1:<br>Then F 2 L1(<br>1<br>2).<br>Theorem 1.1.5.2 (Fubini) Assume that F 2 L1(<br>1<br>2).<br>Then for a.e x 2<br>1<br>F(x; ) 2 L1(<br>2) and<br>Z</p><p>2<br>F(; y) dy 2 L1(<br>1):<br>Similarly, for a.e y 2<br>2<br>F(; y) 2 L1(<br>1) and<br>Z</p><p>1<br>F(x; ) dx 2 L1(<br>2):<br>Futhermore, we have<br>Z</p><p>1<br>Z</p><p>2<br>F(x; y) dxdy =<br>Z</p><p>2<br>Z</p><p>1<br>F(x; y) dx</p><p>dy =<br>Z</p><p>1<br>Z</p><p>2<br>F(x; y) dy</p><p>dx:<br>Definition Let f and g be measurable functions on RN. We define the convolution<br>product f g of f and g by:<br>f g(x) =<br>Z<br>RN<br>f(x ô€€€ y)g(y) dy<br>for those x, if any, for which the integral converges.<br>Theorem 1.1.5.3 (Minkowski’s Inequality) . Let 1 p &lt; +1 and let<br>(X;A; dx) and (Y; B; dy) be -finite measure spaces. Let F be a measurable function<br>on the product space X Y . Then<br>Z<br>X</p><p>Z<br>Y<br>F(x; y) dy</p><p>p<br>dx<br>1<br>p</p><p>Z<br>Y<br>Z<br>X<br>jF(x; y)jp dx<br>1<br>p<br>dy;<br>9<br>in the sense that the integral on the left hand side exists if the one on the right<br>hand side is finite, and in this case the inequality holds. Note that the inequality<br>may also be writen as:</p><p>Z<br>Y<br>F(; y) dy</p><p>p</p><p>Z<br>Y<br>kF(; y)kp dy:<br>Theorem 1.1.5.4 Let 1 p +1. If f 2 L1(RN) and g 2 Lp(RN) then<br>f g(x) =<br>Z<br>RN<br>f(x ô€€€ y)g(y) dy<br>exists for almost all x and defines a function f g 2 Lp(RN). Moreover<br>kf gkp kfk1kgkp:<br>Proof.<br>Case1. If p = +1, we have<br>Z<br>RN<br>jf(x ô€€€ y)g(y)j dy kgk1<br>Z<br>RN<br>jf(x ô€€€ y)j dy = kgk1kfk1;<br>by invariance of Lebesgue’s measure under translation. Thus f g(x) exists a.e<br>and<br>jf g(x)j kgk1kfk1; a.e x 2 RN:<br>So f g 2 L1(<br>) and<br>kf gk1 kfk1kgk1:<br>Case2. For p = 1, let<br>F(x; y) = f(x ô€€€ y)g(y):<br>For almost every y 2 RN, we have<br>Z<br>RN<br>jF(x; y)j dx = jg(y)j<br>Z<br>RN<br>jf(x ô€€€ y)j dx = kfk1jg(y)j &lt; 1<br>and Z<br>RN<br>Z<br>RN<br>jF(x; y)j dx</p><p>dy = kfk1kgk1 &lt; 1:<br>Using Tonelli’s Theorem, we have F 2 L1(RN RN). By Fubini’s Theorem, we<br>obtain Z<br>RN<br>jF(x; y)j dy &lt; 1 a.e x 2 RN<br>10<br>and Z<br>RN<br>Z<br>RN<br>jF(x; y)j dy</p><p>dx kfk1kgk1:<br>So<br>kf gk1 kfk1kgk1:<br>Case3. For 1 &lt; p &lt; +1, let q be the conjugate exponent of p. From Case2., we<br>know that for a.e x 2 RN fixed, y 7! jf(xô€€€y)jjg(y)jp is integrable or equivalently<br>y 7! jf(xô€€€y)j1=pjg(y)j is in Lp(RN). Since y 7! jf(xô€€€y)jq is in Lq(RN), we have<br>from Holder’s inequality that<br>jf(x ô€€€ y)jjg(y)j = jf(x ô€€€ y)jq jf(x ô€€€ y)j1=pjg(y)j 2 L1(RN)<br>and<br>jf(x ô€€€ y)jjg(y)j<br>Z<br>RN<br>jf(x ô€€€ y)jjg(y)jp dy<br>1=p<br>kfk1=q<br>1<br>i.e<br>jf g(x)jp (jfj jgjp)(x) kfkp=q<br>1 :<br>Using again case2. we have<br>f g 2 Lp(<br>) and kf gkp kfk1kgkp:<br>Definition Let 2 L1(RN) such that<br>Z<br>RN<br>(x) dx = 1. Let (x) =<br>1<br>N (<br>x</p><p>).<br>The family Z of functions ; &gt; 0, is called a mollifier with kernel . Note that<br>RN<br>dx = 1.<br>Definition If f is a function on RN and a 2 RN, we define the translation of f<br>by a, af as follow:<br>af(x) = f(x ô€€€ a)<br>Proposition 1.1.5.1 Let be a mollifier, 1 p &lt; +1 and f 2 Lp(RN). Then<br>for each &gt; 0<br>kf ô€€€ fkp<br>Z<br>RN<br>kyf ô€€€ fkpj(y)j dy: (1.9)<br>Proof. Since<br>Z<br>RN<br>(x) dx = 1 we have<br>f (x) ô€€€ f(x) =<br>Z<br>RN<br>[f(x ô€€€ y) ô€€€ f(x)](y) dy:<br>11<br>by Minkowski’s inequality (1.1.5.3)<br>kf ô€€€ fkp =<br>Z<br>RN</p><p>Z<br>RN<br>[f(x ô€€€ y) ô€€€ f(x)](y) dy</p><p>p<br>dx<br>1<br>p</p><p>Z<br>RN<br>Z<br>RN<br>jf(x ô€€€ y) ô€€€ f(x)jpj(y)j dx<br>1<br>p<br>dy<br>=<br>Z<br>RN<br>kyf ô€€€ fkpj(y)j dy:<br>Corollary 1.1.5.1 If is such that<br>Z<br>RN<br>(x) dx = 0 then<br>kf kp<br>Z<br>RN<br>kyf ô€€€ fkpj(y)j dy:<br>Theorem 1.1.5.5 Assume that 0. Let f be a bounded continuous function<br>on RN. Then f is continous on RN for each &gt; 0 and for each x 2 RN we<br>have<br>lim<br>!0+<br>f (x) = f(x):<br>Proof. Let &gt; 0, we have<br>f (x) =<br>Z<br>RN<br>f(x ô€€€ y)(y) dy =<br>Z<br>RN<br>f(x ô€€€ y)(y) dy:<br>Let M be the bound on the absolute value of f. Then jf(x ô€€€ y)(y)j M(y)<br>a.e. Since 2 L1(RN) and the function x ! f(x ô€€€ y)(y) is continuous a.e<br>y 2 RN then by Lebesgue dominated convergence theorem f is continuous.<br>Now, fix x 2 RN. Since<br>Z<br>(y) dy = 1 we have:<br>f (x) ô€€€ f(x) =<br>Z<br>RN<br>[f(x ô€€€ y) ô€€€ f(x)](y) dy:<br>Let &gt; 0. By the continuity of f at x, there is &gt; 0, such that<br>jf(x ô€€€ y) ô€€€ f(x)j</p><p>2<br>; for jyj &lt; :<br>Since Z<br>jyj<br>(y) dy =<br>Z<br>jyj</p><p>(y) dy ! 0; as ! 0<br>then there exists 0 &gt; 0 such that<br>Z<br>jyj<br>(y) dy &lt;</p><p>4M<br>; for &lt; 0<br>12<br>It follows that for all such &gt; 0, we can write the integral as a sum over jyj &lt;<br>and jyj and get<br>jf (x) ô€€€ f(x)j</p><p>2<br>+</p><p>2<br>= :<br>1.1.6 Density of Cc(<br>) in Lp(<br>)<br>Proposition 1.1.6.1 Let<br>be an open subset of RN. Let (Uj)j2J be a collection<br>of open subsets of<br>with union U. Let E U. If E Uj is a set of Lebesgue<br>measure 0 for each j 2 J then E has measure 0.<br>Proof. Let Q be the countable set consisting of all open balls in RN with rational<br>radius and rational center coordinates. Then for each j 2 J<br>Uj =<br>[<br>fB j B 2 Q; B Ujg<br>so E is a countable union of sets of measure 0 of the form E B.<br>Note that it is important that be Uj to be open.<br>Now let f 2 L1(<br>). Then by the proposition above there exists a largest open<br>subset U of<br>on which f is 0 almost everywhere, just take the union of open<br>sets on which f vanishes.<br>Definition The complement of U is called the support of f in<br>and is denoted<br>by supp(f).<br>Proposition 1.1.6.2 If f :<br>! R is continuous then the support of f in<br>is<br>the closure of<br>fx 2<br>j f(x) 6= 0g<br>Definition If<br>is an open subset of RN, we denote by Cc(<br>) the set of continuous<br>functions on RN with compact support in<br>. We denote by D(<br>) the set of<br>infinitely continuously differentiable functions with compact support in</p><p>Let : RN ! R defined by<br>(x) =<br>8&gt;&lt;<br>&gt;:<br>c(1 ô€€€ kxk) if kxk 1;<br>0 if kxk &gt; 1 :<br>(1.10)<br>where the constant c is chosen so that<br>Z<br>RN<br>(x) dx = 1. Then is a continuous<br>mollifier and moreover supp () is the -Ball B0(0; ).<br>13<br>Lemma 1.1.6.1 (Uryshon.) Let<br>be an open subset of RN and K<br>be a<br>compact set. Then there exists 2 Cc(<br>) such that 0 1 and = 1 on<br>some neighborhood of K.<br>Proof. Let be a continuous mollifier as above and let L be the closed –<br>neighborhood of K, that is<br>L = fx 2 RN; j dist(x;K) g<br>where =<br>1<br>3<br>dist(K; @<br>). Let<br>(x) = XL (x) =<br>Z<br>RN<br>XL(x ô€€€ y)(y) dy =<br>Z<br>L<br>(x ô€€€ y) dy<br>For 0 &lt; &lt; , we have 2 C(<br>), has it support in the closed 2-neighborhood<br>of K and so has compact support in<br>, 0 1 and = 1 on the ( ô€€€ )-<br>neighborhood of K.<br>Theorem 1.1.6.1 (Density of Cc(<br>) in Lp(<br>) ) . Let<br>be an open subset of<br>RN and let 1 p &lt; +1. Then Cc(<br>) is dense in Lp(<br>).<br>Proof. We denote the Lebesgue measure of measurable set B by m(B). Since<br>the simple functions are dense in Lp(<br>) for finite p, it suffices to show that we can<br>approximate the characteristic function XA of a measurable set A of finite measure<br>by function in Cc(<br>). Let &gt; 0. By the regularity of Lebesgue measure there<br>exits a compact set K A and an open set U, A U such that m(U ô€€€K) &lt; p.<br>From Uryshon’s Lemma, there is 2 Cc(U) such that 0 1 and = 1 on<br>K. We have jXA ô€€€ j XU ô€€€ XK and so<br>kXA ô€€€ kp m(U ô€€€ K)<br>1<br>p &lt; :<br>Remark 1.1.6.1 If 1 p &lt; 1, Theorem 1.1.6.1 says that Cc(<br>) is dense in Lp(<br>),<br>and Theorem 1.1.3.1 shows that Lp(<br>) is complete. Thus Lp(<br>) is the completion<br>of the metric space which is obtained by endowing C0(<br>) with the Lp-metric.<br>Of course, every metric space S has a completion S whose elements may be viewed<br>abstractly as equivalent classes of Cauchy sequence in S. The important point in<br>the present situation is that the various Lp-completion of Cc(<br>) again turn out to be<br>spaces of functions on<br>.<br>The case p = +1 differs from the cases p &lt; 1. The L1-completion of Cc(<br>)<br>is not L1(<br>), but is C0(<br>), the spaces of all continuous functions on<br>which vanish<br>at infinity.<br>14<br>Definition A function f :<br>! R is said to vanish at infinity if for every &gt; 0,<br>there exists a compact set K<br>such that jf(x)j &lt; for all x not in K.<br>We denote by C0(<br>),the class of all continuous functions on<br>which vanish<br>at infinity. It is clear that Cc(<br>) C0(<br>).<br>Theorem 1.1.6.2 C0(<br>) is the completion of Cc(<br>), relative to the metric defined<br>by the supremum norm:<br>kfk1 = sup<br>x2</p><p>jf(x)j:<br>Proof. An elementary verification shows that C0(<br>) satisfies the axioms of a<br>metric space if the distance between f and g is taken to be kf ô€€€ gk1. We have<br>to show that (i) Cc(<br>) is dense in C0(<br>) and (ii) C0 is complete.<br>To prove (i), let f 2 C0(<br>) and &gt; 0, there exists a compact set K<br>such<br>that jf(x)j &lt; outside K. Uryshon’s lemma gives us that there exists a function<br>‘ 2 C0(<br>) such that 0 ‘ 1 and ‘(x) = 1 on K. Put h = ‘f. Then<br>h 2 Cc(<br>) and kf ô€€€ hk1 &lt; .<br>To prove (ii), let ffng be a Cauchy sequence in C0(<br>). Using the definition<br>of Cauchy sequence and supremum norm, we can assume that ffng converges<br>uniformly. Then its pointwise limit function f is continuous. Given &gt; 0, there<br>exists an N so that kfN ô€€€ fk1 &lt; =2 and there exists a compact set K so that<br>jfN(x)j &lt; =2 outside K. Hence jf(x)j &lt; outside K, and we have proved that<br>f vanishes at infinity. Thus C0(<br>) is complete.<br>Proposition 1.1.6.3 (Continuity of Translation in Lp(<br>)) . Let 1 p &lt;<br>+1 and f 2 Lp(RN). Let : RN ! Lp(RN) be the map defined by<br>(y) = yf; 8 y 2 RN:<br>Then is uniformly continuous on RN.<br>Proof. Let &gt; 0. By density choose g 2 Cc such that kf ô€€€ gkp &lt;</p><p>3<br>. Let<br>y; z 2 RN and v = y ô€€€ z, then<br>k(y) ô€€€ (z)kp = kyf ô€€€ zfkp kyf ô€€€ ygkp + kyg ô€€€ zgkp + kzg ô€€€ zfkp</p><p>2<br>3<br>+ kyg ô€€€ zgkp</p><p>2<br>3<br>+ kvg ô€€€ gkp<br>by translation invariance of Lebesgue measure. Since g has compact support,<br>then the support of vg stays in a fixed compact set K for kvk 1. Since g is<br>bounded we have<br>jvgj M XK:<br>15<br>It follows that<br>jvg ô€€€ gjp (2M)pXK 2 L1(RN); for kvk 1<br>since g is continuous we have vg ! g as v ! 0 pointwise. By the dominate<br>convergence theorem<br>Z<br>RN<br>jvg ô€€€ gjp dx ! 0 as v ! 0. Thus there exists &gt; 0<br>such that 0 &lt; &lt; 1 and<br>kvg ô€€€ gkp &lt;<br>1<br>3<br>; if kvk &lt; :<br>Hence, the uniform continuity of follows.<br>Theorem 1.1.6.3 Let 2 L1(RNZ ), 1 p &lt; +1 and let f 2 Lp(RN). If<br>RN<br>dx = 1 then f ! f in Lp(RN) as ! 0. If<br>Z<br>RN<br>dx = 0 then<br>f ! 0 in Lp(RN) as ! 0<br>Proof. for the first case, we know that:<br>kf ô€€€ fkp<br>Z<br>RN<br>kyf ô€€€ fkpj(y)j dy:<br>The integrand is bounded by 2kfkpjj 2 L1(RN) and goes to 0 as ! 0 by<br>continuity of the translation. Thus the Lebesgue dominated convergence theorem<br>yields the desired result.<br>Corollary 1.1.6.1 Let 1 &lt; p &lt; +1 and q 1 such that 1=p + 1=q = 1. If<br>f 2 Lp(RN) and g 2 Lq(RN) then f g is uniformly continuous.<br>Proof. We have<br>fg(x)ô€€€fg(z) =<br>Z<br>RN<br>(f(x ô€€€ y) ô€€€ f(z ô€€€ y)) g(y)dy =<br>Z<br>RN<br>(ô€€€xf(ô€€€y) ô€€€ ô€€€zf(ô€€€y)) g(y) dy<br>therefore by using Holder’s inequality we have<br>jf g(x) ô€€€ f g(z)j kô€€€xf ô€€€ ô€€€zfkpkgkq:<br>We conclude by using the fact that the translation is uniformly continuous.<br>1.1.7 Density of D(<br>) in Lp(<br>).<br>One important application of the convolution product is regularization of functions,<br>that is, the approximation of functions by smooth functions. Let<br>u(t) =<br>(<br>eô€€€1=t if t &gt; 0;<br>0 if t 0 :<br>(1.11)<br>16<br>Since for any integer k, lim<br>t!0<br>1<br>tk eô€€€1<br>t = 0, then u 2 C1(R). Let (x) = cu(1 ô€€€<br>kxk2); x 2 RN. Then 2 C1(RN) and (x) = 0 if kxk 1. Moreover, for a<br>suitable choice of the constant c we have (x) 0 and<br>Z<br>RN<br>(x) dx = 1. Let<br>(x) =<br>1<br>N (<br>x</p><p>):<br>Then<br>1. 2 C1(RN),<br>2. supp() = B0(0; ) = fx 2 RN j kxk g,<br>3. (x) 0;<br>4.<br>Z<br>RN<br>(x) dx = 1:<br>Any family () satisfying these four properties is called Friedrichs’s mollifier.<br>Theorem 1.1.7.1 Let be a Friedrichs’s mollifier. If f 2 L1(RN; loc) the<br>convolution<br>f (x) =<br>Z<br>RN<br>f(x ô€€€ y)(y) dy =<br>Z<br>RN<br>f(x ô€€€ y)(y) dy<br>exists for each x 2 RN. Moreover<br>1. f 2 C1(RN),<br>2. supp(f ) supp(f) + B0(0; ),<br>3. if 1 p &lt; +1 and f 2 Lp(RN), then f ! f in Lp(RN), as ! 0. In<br>fact we have kf ô€€€ fkp sup<br>kyk<br>kyf ô€€€ fkp<br>4. If K, the set of continuity points of f is compact, then f ! f uniformly<br>on K as ! 0<br>Proof. The convolution exists for each x because the mollifier has compact<br>support. Note that f (x) =<br>Z<br>RN<br>(x ô€€€ y)f(y) dy implies f 2 C1(RN) by<br>standard results on differentiating under the integral sign (since has compact<br>support). The second is obvious and the third follows from<br>kf ô€€€ fkp<br>Z<br>RN<br>kfyf ô€€€ fkp(y) dy supkykkyf ô€€€ fkp<br>17<br>Assume that K the set of continuity points of f is compact. Then f is uniformly<br>continuous on K and this shows a little bit more: let &gt; 0, then there exists &gt; 0<br>such that if x 2 K, z 2 RN and kxô€€€zk &lt; then it follows that jf(x)ô€€€f(z)j &lt; .<br>Note that we do not require z to be in K. Now<br>f (x) ô€€€ f(x) =<br>Z<br>kyk<br>(f(x ô€€€ y) ô€€€ f(x)) (y) dy:<br>Hence if 0 &lt; &lt; then<br>jf (x) ô€€€ f(x)j<br>Z<br>kyk<br>jf(x ô€€€ y) ô€€€ f(x)j (y) dy</p><p>Z<br>RN<br>(y) dy =<br>for each x 2 K:<br>Corollary 1.1.7.1 Let<br>be an open subset of RN and K is a compact subset of<br>RN then there exits 2 D(<br>) with 0 1 and = 1 on K.<br>Proof. Let =<br>1<br>3<br>dist(K; @<br>). Let L be the closed -neighbborhood of K, that<br>is:<br>L := fx 2 RN j dist(x;K) g<br>Let f be the characteristic function of L and let 0 &lt; &lt; . then = f 2<br>C1(RN) has its support in the closed 2-neighborhood of K and so has compact<br>support in<br>. We have that 0 1 and = 1 on the ( ô€€€)-neighborhood of<br>K.<br>Theorem 1.1.7.2 (Density of D(<br>) in Lp(<br>)) . Let<br>be an open subset of<br>RN and let 1 p &lt; +1. Then D(<br>) is dense in Lp(<br>)<br>Proof. Let f 2 Lp(<br>) and let &gt; 0. By theorem (1.1.6.1) there exits g 2 Cc(<br>)<br>such that kf ô€€€ gkp &lt;</p><p>2<br>. Let &gt; 0 and define g = g . Then g 2 C1(<br>) and<br>g ! g in Lp(<br>).<br>Moreover<br>supp(g) supp(g) + B0(0; )<br>Since g ! g in Lp(<br>) as ! 0, there exists &gt; 0 such that kg ô€€€ gkp &lt;</p><p>2<br>for<br>&lt; . Now let &lt; min(; dist(supp(g); @<br>)). Then g 2 D(<br>) and kf ô€€€gkp &lt;<br>Theorem 1.1.7.3 (Partition of unity) . Let<br>be an open subset of RN and<br>let (Uj)j2J be a locally finite open cover of<br>such that each Uj has compact closure<br>in<br>. Then there exsits j such that<br>j 2 D(Uj); j 0 and<br>X<br>j2J<br>j(x) = 1; 8 x 2</p><p>18<br>Proof. There exists an open covering (wj) of<br>such that wj Uj U<br>j for all<br>j 2 Uj . Choose j 2 D(Uj) such that 0 j 1 and j = 1 on wj . The sum<br>(x) =<br>X<br>j2J<br>j(x)<br>is locally finite and bounded below by 1. Thus 2 C1(<br>) and 1 . take<br>j =<br>j</p><p>The j are called a smooth partion of unity subordinate to the locally finite<br>open cover (Uj).<br>Theorem 1.1.7.4 (Finite partition of unity) . Let K be a compact subset of<br>RN and let (Uj)j=1; ;N be a finite open cover of K. Then there exists functions<br>j 2 D(Uj) such that j 0 and<br>XN<br>j=1<br>j = 1<br>in a neighborhood of K.<br>Proof. For x 2 K, let Vx be an open neighborhood of x such that Vx is a compact<br>subset of Uj with x 2 Uj . Since K is compact there exsit a finite set x1; ; xm<br>in K such that<br>K<br>m[<br>k=1<br>Vxk :<br>For each j let Kj be the union of those Vxk which are contained in Uj . Then Kj<br>is compact, Kj Uj and<br>K K1 [ [ KN<br>By corollary1.1.7.1, we may choose j 2 D(Uj), 0 j 1 in a neighborhood<br>of Kj and j = 1 on Kj . Finally let<br>1 = 1<br>2 = (1 ô€€€ 1) 2<br>3 = (1 ô€€€ 1)(1 ô€€€ 2) 3</p><p>N = (1 ô€€€ 1)(1 ô€€€ 2) (1 ô€€€ Nô€€€1) N<br>We have, j 0 and<br>1 + 2 + + N = 1 ô€€€ (1 ô€€€ 1)(1 ô€€€ 2) (1 ô€€€ N):<br>For each x 2 K there is j so that j(x) = 1. Thus 1 + 2 + + N = 1 on K<br>To obtain the equality on a neighborhood of K, we would enlarge K a bit.<br>19<br>Theorem 1.1.7.5 (Dubois-Reymond) . Let<br>be an open subset of RN. If<br>f 2 L1(<br>; loc) and<br>Z</p><p>f(x)(x) = 0 for each 2 D(<br>) then f = 0 a.e in<br>.<br>20<br>1.2 Distribution Theory<br>1.2.1 Test Functions<br>Let<br>be a nonempty open subset of RN.<br>Notations<br>If m 2 N, Cm(<br>) denotes the space of real-valued functions on<br>of class Cm and<br>C1(<br>) the space of those of class C1. By convention, C0(<br>) = C(<br>) the space<br>of continuous functions on<br>.<br>An element 2 Nn is called a multiindex. If = (1; ; n) is a multiindex,<br>we define the length of to be the sum jj = 1 + + n, and we put<br>! = 1! n!. We give Nn the product order: if ; 2 Nn , we write if<br>1 1; ; n n:<br>If 1 i n, we often use Di to denote @<br>@xi<br>. Then if is a multiindex, we write<br>D = D1<br>1 Dn<br>n =<br>Djj<br>@x1<br>1 @xn<br>n<br>:<br>The differential operator D is also denoted by @jj<br>x or @<br>x . By convention, D0<br>(the differential of order 0 with espect to any index) is the identity map. We see<br>that each operator D, where 2 Nn, acts on the space Cm(<br>), for jj m.<br>We recall the following classical result:<br>Proposition 1.2.1.1 (Leibniz’ formula) . Let u; v 2 Cm(<br>). For each multiindex<br>such that jj m,<br>D(uv) =<br>X</p><p>C<br>Dô€€€uDv;<br>where<br>C<br>=<br>Yn<br>i=1<br>i!<br>i!(i ô€€€ i)!<br>=<br>!<br>!( ô€€€ )!<br>:<br>+ We denote by Dm(<br>) the space of functions of class Cm having compact<br>support in<br>. In particular, D0(<br>) = Cc(<br>). Clearly, m0 m implies Dm0(<br>)<br>Dm(<br>). Now we set<br>D(<br>) =</p><p>m2N<br>Dm(<br>);<br>Thus D(<br>) is the space of functions of class C1(<br>) having compact support in</p><p>; such functions are called test functions on<br>.<br>21<br>Finally, if K is a compact subset of<br>, we denote by DK(<br>) the space of functions<br>of class C1 having support contained in K.<br>DK(<br>) =</p><p>m2N<br>Dm<br>K(<br>)<br>thus<br>D(<br>) =<br>[<br>K2K(<br>)<br>DK(<br>)<br>where K(<br>) is the set of compact subsets of<br>.<br>Clearly, a function in Dm(<br>) or D(<br>), when extended with the 0 value outside</p><p>becomes an element of Dm(RN) or D(RN), respectively. Thus Dm(<br>) and D(<br>)<br>can be considered as subspaces of Dm(RN) and D(RN). We will often make this<br>identification. Conversely, an element ‘ 2 Dm(RN) or D(RN) belong to all the<br>spaces Dm(<br>) or D(<br>) such that supp(‘)<br>.<br>1.2.2 Convergence in Function Spaces<br>Convergence in Dm<br>K(<br>) and DK(<br>). Let K be a compact subset of<br>. We say<br>that a sequence (‘n) in Dm<br>K(<br>) converges to ‘ 2 Dm<br>K(<br>), if for every multiindex<br>such that jj m, the sequence (D’n) converges uniformly to D’. An<br>analogous definition applies with the replacement of Dm<br>K(<br>) by DK(<br>) where<br>now there is no restriction on the multiindex 2 Nn.<br>The convergence thus defined on Dm<br>K(<br>) clearly corresponds to the convergence<br>in the norm k k(m) defined on Dm<br>K(<br>) by:<br>k’k(m) =<br>X<br>jjm<br>kD ‘k1;<br>where k k1 denote the uniform norm. In contrast, no norm on DK(<br>) yields the<br>notion of convergence we have defined in that space.<br>Convergence in Dm(<br>) and D(<br>). We say that a sequence (‘n) in Dm(<br>)<br>converges to ‘ in Dm(<br>) if the followings are satisfied:<br>(i) there exists a compact subset K of<br>such that<br>supp(‘) K and supp(‘n) K for all n;<br>(ii) the sequence (‘n) converge to ‘ in Dm<br>K(<br>).<br>An analogous definition applies with the replacement of Dm(<br>) and Dm<br>K(<br>) by<br>D(<br>) and DK(<br>).<br>22<br>Convergence in Cm(<br>) and C1(<br>). We say that a sequence (fn) in Cm(<br>)<br>converge to f 2 Cm(<br>), if for every multiindex such that jj m and for every<br>compact K in<br>, the sequence (Dfn) converges to (Df) uniformly on K. An<br>analogous definition applies with the replacement of Cm(<br>) by C1(<br>), where<br>now there is no restriction on the multiindex . For m = 0, the convergence in<br>C0(<br>) = C(<br>) thus defined coincides with the uniform convergence on compact<br>subsets.<br>Remark 1.2.2.1 The definitions of convergence of sequences just made extend immediately<br>to families (‘), where runs over a subset in R and ! 0; 0 2<br>[ô€€€1;+1].<br>We will see that it is possible to give the spaces DK(<br>), Cm(<br>) and C1(<br>)<br>complete metric structures for which convergence of sequences coincides with the<br>notions just defined. In contrast, one can show that the convergence we have<br>defined in Dm(<br>) and D(<br>) cannot come from a metric structure.<br>In fact, the only topological notions that we will use in connection with these<br>function spaces are continuity and denseness, and these notions, in the case of<br>metric spaces, can always be expressed in terms of sequences.<br>1.2.3 Continuity and Denseness on Dm(<br>) and D(<br>)<br>A subset C of Dm(<br>) or D(<br>) will be called dense in Dm(<br>) or D(<br>), if for<br>every ‘ in Dm(<br>) or D(<br>), there exists a sequence (‘n) in C converging to ‘ in<br>Dm(<br>) or D(<br>).<br>A function F on Dm(<br>) or D(<br>) and taking values in a metric space or in<br>one of the spaces just introduced will be called continuous, if for every sequence<br>(‘n) in Dm(<br>) or D(<br>) that converges to ‘ in Dm(<br>) or D(<br>), the sequence<br>(F(‘n)) converges to F(‘) in the space considered.<br>For example, the Canonical Injection from Dm(<br>) to Cm(<br>) is continuous. This<br>means simply that every sequence in Dm(<br>) that converges in Dm(<br>) also converges<br>in Cm(<br>) to the same limit.<br>Proposition 1.2.3.1 For every m 2 N, the space D(<br>) is dense in Dm(<br>). In<br>particular, D(<br>) is dense in Cc(<br>).<br>Lemma 1.2.3.1 Let<br>be an open subet of RN. For n 2 N, define<br>Kn = fx 2 RN j kxk n and d(x;<br>c)<br>1<br>n<br>g<br>where d is the usual distance in RN. Then<br>1. Each Kn is a compact subset of<br>and Kn Kn+1.<br>23<br>2.<br>=<br>1[<br>n=1<br>Kn =<br>1[<br>n=2<br>Kn.<br>3. For all compact K in<br>, there exists N 1 such that K KN.<br>Proposition 1.2.3.2 The space D(<br>) is dense in C1(<br>) and in Cm(<br>) for every<br>m 2 N<br>Proof. Let Kn be a sequence of compact subsets of<br>exhausting<br>(as above).<br>Then there exists for each n, an element ‘n 2 D(<br>) such that<br>0 ‘n 1; ‘n = 1 on Kn; supp(‘n) Kn+1:<br>Now let f 2 C1(<br>), we have (f’n) 2 D(<br>) for every n 2 N. If K is a compact<br>subset of<br>, there is an integer N such that K KN. Thus for every n N<br>and for every 2 Nn, we have D(f’n) = Df on K. By the definition of<br>convergence in C1(<br>), we deduce that (f’n) converge to f in C1(<br>).<br>1.2.4 Distributions<br>Definition A distribution on<br>is a continuous linear mapping T from D(<br>)<br>into R. The set of all distributions is denoted by D0(<br>).<br>Remark 1.2.4.1 By Linearity, to show that T is continuous, it is enough to show<br>that, if ‘n ! 0 in D(<br>) then (T; ‘n) ! 0 in R.<br>Theorem 1.2.4.1 Let T be a linear mapping from D(<br>) into R. Then T is a<br>distribution if and only if, for any compact set K in<br>, there exists an integer<br>nK 2 N and a positive constant CK such that:<br>j(T; ‘)j CK<br>X<br>jjnK<br>sup<br>K<br>jD'(x)j; 8 ‘ 2 DK(<br>):<br>Definition If nK can be chosen independent of K, then the smallest n with this<br>property is called the order of the distribution T.<br>Example 1.2.4.1 (Distribution given by a locally integrable function) Let<br>f 2 L1(<br>; loc), then f gives a distribution Tf defined by:<br>(Tf ; ‘) =<br>Z</p><p>f(x) ‘(x) dx; 8 ‘ 2 D(<br>):<br>24<br>The linearity of Tf follows from the linearity of integral. Now let K be a compact<br>subset of<br>and let ‘ 2 D(<br>) with supp(‘) K then we have<br>j(Tf ; ‘)j<br>Z<br>K<br>jf(x)j dx</p><p>sup<br>K<br>j'(x)j</p><p>;<br>so T is a distribution of order 0.<br>We define the maping f ! Tf . It is linear and one to one. In fact let f 2 L1(<br>; loc)<br>such that Tf = 0, Then by using Dubois-Reymond’s lemma, we show that f = 0 a.e<br>in<br>. From now, we can identify L1(<br>; loc) as a subset of D0(<br>).<br>Example 1.2.4.2 (Dirac distribution) .<br>Let x0 2<br>. We denote by x0 the linear form defined on D(<br>) by<br>(x0 ; ‘) = ‘(x0); 8 ‘ 2 D(<br>):<br>Let K be a compact subset of<br>. Since j(x0 ; ‘)j sup<br>K<br>j'(x)j for all ‘ 2 D(<br>)<br>with supp(‘) K, then x0 is a distribution.<br>Example 1.2.4.3 (The distribution Principal Value of 1=x) .<br>Consider the function x 7! 1=x from R to R. This function is clearly not locally<br>integrable on R but it is on R. We will see how we can extend to R the distribution<br>defined by this function on R.<br>Proposition 1.2.4.1 For every ‘ 2 D(R), the limit<br>(pv(1=x); ‘) = lim<br>!0+<br>Z<br>jxj<br>‘(x)<br>x<br>dx<br>exists. The linear form pv(1=x) thus defined a distribution of order 1 on R, and<br>is an extension to R of the distribution [1=x] 2 D0(R).<br>We call pv(1=x) the principal value of 1=x.<br>Example 1.2.4.4 (The distribution Finite part of 1=x2. ) Let ‘ 2 D(<br>),<br>we call the distribution Finite part, the distribution denoted by fp(1=x2) and defined<br>by:<br>(fp(1=x2); ‘) = lim<br>!0+<br>Z<br>jxj<br>‘(x)<br>x2 dx ô€€€ 2<br>‘(0)</p><p>8 ‘ 2 D(R):<br>Proposition 1.2.4.2 Let T be a distribution on<br>such that every point x 2</p><p>has an open neighborhood Vx such that (T; ‘) = 0 for all ‘ 2 D(Vx). Then T = 0.<br>25<br>Proof. Let ‘ 2 D(<br>), we will show that (T; ‘) = 0. Let K = supp(‘) and let<br>x 2 K then by hypothesis, there exists an open neighborhood Vx of x such that<br>(T; ‘) = 0 for all ‘ 2 D(Vx). Since K is compact, there exists x1; ; xN 2 K<br>such that<br>K<br>N[<br>i=1<br>Vxi<br>Let 1: ; N be a partition of unity associated to this open cover of K. Then<br>‘ =<br>XN<br>i=1<br>i’<br>Since supp(i’) Vxi then (T; i’) = 0 and so is (T; ‘).<br>1.2.5 The Support of a Distribution<br>Definition Let T be a distribution on<br>, an open of nullity of T is an open<br>subset U of<br>such that (T; ‘) = 0 for all ‘ 2 D(U).<br>Proposition 1.2.5.1 Any distribution T has a largest open of nullity<br>0. Its<br>complement is called the support of T and denoted by supp(T).<br>Proof. Let U be the collection of opens of nullity of T, and let<br>0 =<br>S<br>U2U U<br>be there union. It suffices to show that<br>0 is it self an open of nullity of T.<br>Take ‘ 2 D(<br>0). By compactness of the support of ‘, their exists a finite<br>collection of opens sets U1; ; UN whose union contains the support of ‘. Let<br>i; i = 1; ;N be a partion of unity associated to this open cover of supp(‘).<br>It follows that<br>‘ =<br>XN<br>i=1<br>‘ i:<br>Since each ‘ i is supported in the open of nullity Ui, this implies that<br>(T; ‘) =<br>XN<br>i=1<br>(T; ‘ ) = 0:<br>This proves that<br>0 is indeed an open of nullity of T, and by construction it is<br>the largest of such open sets.<br>Somes consequences of the definition:<br>1. x0 =2 supp(T) if and only if there exists an open neighborhood Vx0 of x0<br>such that<br>(T; ‘) = 0; 8 ‘ 2 D(Vx0);<br>26<br>2. x0 2 supp(T) if and only for all open neighborhood Vx0 of x0, there exists<br>‘ 2 D(Vx0) such that (T; ‘) 6= 0.<br>Proposition 1.2.5.2 Let T be a distribution on<br>and ‘ 2 D(<br>) such that<br>supp (T) supp(‘) = ;:<br>Then (T; ‘) = 0.<br>1.2.6 Distributions with Compact Support<br>Theorem 1.2.6.1 Let T be a distribution on<br>. A necessary and sufficient condition<br>for the support of T to be compact is that T has an extension to a continuous<br>linear form on C1(<br>). The extension is then unique.<br>Proof. Suppose first that the support of T is compact. Then there exists a<br>compact K in<br>whose interior contains the support of T. It follows from corollary<br>1.1.7.1 that there exists 2 D(<br>) such that 0 1 and (x) = 1 on K. We<br>then set, for f 2 C1(<br>),<br>( T; f) = (T; f): (1.12)<br>It is clear that this does define a linear form T on C1(<br>). On the other hand, if<br>‘ 2 D(<br>), we have<br>supp (‘ ô€€€ ‘)<br>ô€€€ K<br>ô€€€ supp (T);<br>this implies that<br>supp (‘ ô€€€ ‘) supp (T) = ;:<br>So by proposition 1.2.5.2, it follows that<br>( T; ‘) = (T; ‘):<br>Thus T is an extension of T to C1(<br>).<br>Finally, if (fn) is a sequence in C1(<br>) that converges to 0 in C1(<br>), then from<br>the definitions and Leibniz’s formula the sequence (fn) converges to 0 in D(<br>),<br>so that<br>lim<br>n!+1<br>( T; fn) = lim<br>n!+1<br>(T; fn) = 0<br>This proves that T is continuous on C1(<br>). Since D(<br>) is dense in C1(<br>), the<br>extension is unique.<br>For the converse, assume that T can be extended to a continuous linear form<br>T on C1(<br>). Let (Kn) be an exhausting sequence of compact subsets of<br>. If<br>the support of T is not compact, then there exists for each n 2 N, an element<br>‘n 2 D(<br>) such that<br>supp(‘n)<br>ô€€€ Kn and (T; ‘n) 6= 0:<br>27<br>Put<br>n =<br>‘n<br>(T; ‘n)<br>;<br>so we have<br>(T; n) = 1; 8 n 2 N:<br>Now we will show that the series<br>Xn<br>n=1<br>n converges in C1(<br>). To this end, let<br>K be a compact subset of<br>, then there exists N 2 N such that K KN. But<br>for n &gt; N, we have n = 0 on KN, and so on K, the sum<br>X1<br>n=0<br>n reduces to a<br>finite sum on K, and this holds for every compact subset K on<br>. So the sum<br>converges in C1(<br>). By the continuity of T, it follows that the series<br>X1<br>n=0<br>(T; n)<br>converges, contradicting the fact that (T; n) = 1.<br>Remark 1.2.6.1 The restriction to D(<br>) of a continuous linear form on C1(<br>) is a<br>distribution on<br>(since a sequence in D(<br>) that converges in D(<br>) also converges in<br>C1(<br>)), and by the preceding theorem this distribution has compact support. Thus<br>we can identify the space of distributions having compact support with the space of<br>continuous linear forms on C1(<br>) denoted by C1(<br>)0.<br>Proposition 1.2.6.1 Every distribution T with compact support in<br>has finite<br>order. More precisely, there exists an integer m 2 N and a constant C0 0 such<br>that<br>j(T; ‘)j C0k’k(m); 8 ‘ 2 D(<br>):<br>Proof. Let K be the support of T and let K1;K2 be compact subsets of<br>such<br>that<br>K K1 K1 K2 K2<br>:<br>Then by theorem 1.2.4.1, there exists an integer m 2 N and a constant C 0<br>such that<br>j(T; ‘)j Ck’k(m); 8 ‘ 2 DK2(<br>):<br>By corollary 1.1.7.1, there exists 2 D(<br>) such that 0 1, = 1 on K1<br>and supp( ) K2. If ‘ 2 D(<br>) then ‘ 2 DK2(<br>) and<br>supp (‘ ô€€€ ‘ )<br>ô€€€ K1<br>ô€€€ K:<br>Since K is the support of T, then<br>(T; ‘ ô€€€ ‘ ) = 0;<br>so there exists a positive constant C depending only on C0, m and such that<br>j(T; ‘)j = j(T; ‘ )j C0k’ k(m) Ck’k(m):<br>The last inequality being a consequence of Leibniz’ formula.<br>28<br>Remark 1.2.6.2 One can deduce from the preceding result that if, T is a distribution<br>with compact support, there exists an integer m 2 N such that T extends to a<br>continuous linear form on Cm(<br>) and this extension is unique.<br>1.2.7 Convergence of Distributions<br>We assume that<br>is an open subset of RN.<br>Definition Let (Tn)n2N be a sequence of distributions in<br>. We say that (Tn)<br>converges to the distribution T if<br>lim<br>n!+1<br>(Tn; ‘) = (T; ‘); for all ‘ 2 D(<br>):<br>Theorem 1.2.7.1<br>1. Let 1 p +1. If fn; f 2 Lp(<br>) with fn ! f in Lp(<br>) then fn ! f in<br>D0(<br>).<br>2. The pointwise convergence does not imply the convergence in D0(<br>).<br>Proof.<br>1. Let q such that 1=p + 1=q = 1. Then by Holder’s inequality we have:<br>j(fn; ‘) ô€€€ (f; ‘)j = j(fn ô€€€ f; ‘)j</p><p>Z</p><p>jfn ô€€€ fj j’j dx kfn ô€€€ fkpk’kq ! 0:<br>2. Let (fn) be the sequence of functions defined by<br>fn(x) =<br>p<br>neô€€€nx2<br>; x 6= 0;<br>then fn(x) ! 0 for all x 6= 0 but fn !<br>p<br>0 in D0(<br>). In fact let ‘ 2 D(<br>), by<br>Lebesgue dominated convergence therorem we have<br>(fn; ‘) =<br>p<br>n<br>Z<br>R<br>eô€€€nx2<br>‘(x) dx =<br>Z<br>R<br>eô€€€y2<br>‘(<br>y<br>p<br>n<br>) dy !<br>p<br>‘(0) =<br>p<br>(0; ‘)<br>Examples<br>Example 1.2.7.1 Let (Tn)n2N, be a sequence of distributions on R defined by:<br>Tn(x) = sin(nx):<br>Let ‘ 2 D(R), we have<br>(Tn; ‘) =<br>Z<br>R<br>sin(nx)'(x) dx =<br>1<br>n<br>Z<br>R<br>cos(nx)’0(x) dx ! 0:<br>So Tn converge to 0 in D0(R).<br>29<br>Example 1.2.7.2 For : &gt; 0 define<br>v(x) =<br>8&gt;&gt;&lt;<br>&gt;&gt;:<br>1</p><p>if x 2 [0; ];<br>0 if x =2 [0; ] :<br>and w(x) =<br>8&gt;&gt;&lt;<br>&gt;&gt;:<br>1<br>2<br>if x 2 [ô€€€; ];<br>0 if x =2 [ô€€€; ] :<br>Then we have:<br>v ! 0 in D0(R) as ! 0 and w ! 0 in D0(R) as ! 0:<br>1.2.8 Multiplication of Distributions<br>Now, we define the product of distribution by a smooth function. The definition<br>arises from the following lemma.<br>Lemma 1.2.8.1 Let 2 C1(<br>). The map ‘ ! ‘ from D(<br>) to D(<br>) is linear<br>and continuous. In other words if (‘n) is a sequence in D(<br>) converging to ‘ in<br>D(<br>) then the sequence (‘n) converges to ‘ in D(<br>).<br>Definition If T 2 D0(<br>) and 2 C1(<br>), the product distribution T on<br>is<br>defined by setting:<br>(T; ‘) = (T; ‘); 8 ‘ 2 D(<br>):<br>The fact that T defines a distribution follows from the preceding lemma.<br>The definition immediately implies that if 2 C1(<br>), the linear map T ! T<br>from D0(<br>) to D0(<br>) is continuous in the sense that, if (Tn) converge to T in<br>D0(<br>) then (Tn) converge to (T) in D0(<br>).<br>Proposition 1.2.8.1 Let T 2 D0(<br>) and 2 C1(<br>), then we have<br>supp(T) supp() supp(T):<br>Proof. Let ‘ 2 D(<br>). If supp(‘)<br>-supp(), then ‘ = 0, so (T; ‘) = 0. It<br>follows that<br>-supp() is contained in<br>-supp(T), so supp(T) supp().<br>Now if If supp(‘)<br>-supp(T), then supp(‘) supp(‘)<br>ô€€€supp(T), which<br>implies that (T; ‘) = 0. Therefore<br>ô€€€supp(T) is contained in<br>ô€€€supp(T) so<br>supp(T) supp(T).<br>The inclusion in the proposition may be strict. For example, if T = is the<br>dirac distribution in RN, and 2 C1(RN) is such that (0) = 0 and 0 2 sup()<br>(say (x) = x), then T = (0) = 0 and the support of T is empty, whereas<br>supp()supp(T) = f0g.<br>30<br>Proposition</p> <br><p></p>