Sobolev spaces and variational method applied to elliptic partial differential equations

Project Abstract

Sobolev spaces and variational methods play a crucial role in the study of elliptic partial differential equations (PDEs). This research explores the application of Sobolev spaces and variational methods to solve elliptic PDEs, focusing on their theoretical foundations and practical implications. Sobolev spaces are function spaces equipped with a norm that measures both the function and its derivatives up to a certain order. These spaces provide a natural setting for studying PDEs, as they allow for the analysis of functions with weak derivatives. By characterizing the regularity of functions in terms of their derivatives, Sobolev spaces offer a powerful framework for understanding the behavior of solutions to elliptic PDEs. Variational methods, on the other hand, involve formulating the problem of finding solutions to PDEs as an optimization or minimization of a functional. This approach allows for the derivation of existence and uniqueness results, as well as the development of numerical algorithms for approximating solutions. By casting the problem in a variational setting, one can exploit the underlying structure of the PDE to derive useful properties of its solutions. When applied to elliptic PDEs, Sobolev spaces and variational methods provide a systematic approach to studying the well-posedness of the problem, as well as the regularity and behavior of solutions. The theory of Sobolev spaces enables the formulation of weak solutions to elliptic PDEs, which may not possess classical derivatives but still satisfy the PDE in a distributional sense. Variational methods, on the other hand, offer a constructive way to obtain solutions by minimizing an appropriate functional over a suitable function space. In this research, we investigate the use of Sobolev spaces and variational methods to study various elliptic PDEs, including the Laplace equation, Poisson equation, and Helmholtz equation. We analyze the well-posedness of these problems in Sobolev spaces, establish the existence and uniqueness of solutions using variational techniques, and develop numerical methods for computing approximate solutions. By combining the analytical tools provided by Sobolev spaces with the computational techniques offered by variational methods, we aim to deepen our understanding of elliptic PDEs and provide efficient strategies for solving these equations in practice. The results of this research have implications for a wide range of fields, including physics, engineering, and applied mathematics, where elliptic PDEs arise as fundamental models for describing various phenomena.

Project Overview

<p> </p><p>SPACES OF FUNCTIONS<br>In the following,<br>is a nonempty open subset of RN with the Lebesgue measure 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 sequence<br>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 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 a sequence<br>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>Theorem 1.1.1.3 (Fatou Lemma) Let ffng be a sequence of integrable functions 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>5<br>+ Lp-spaces<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 everywhere.<br>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 however, need<br>to be careful sometimes. For example, saying that f 2 Lp(<br>) is continuous means that f is<br>equivalent to a continuous function. Now for f 2 Lp(<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 1=p+1=p0 =<br>1: If f 2 Lp(<br>) and g 2 Lp0(<br>), then fg 2 L1(<br>) and<br>kfgk1 kfkpkgkp0 : (1.3)<br>Proof. The cases p = 1 and p0 = +1 are easy to prove. Now assume 1 &lt; p &lt; +1. We use<br>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 inequality,<br>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 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>6<br>+ Lp-spaces<br>Theorem 1.1.2.2 (Minkowski’s Inequality) . If 1 p +1 and f; g 2 Lp(<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 and p &gt; 1<br>(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>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 k 1, there<br>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 is Cauchy in<br>R. Let fn(x) = limn fn(x) forall x 2<br>ô€€€ A. Letting m goes to +1 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>7<br>+ Lp-spaces<br>Case2. Assume that 1 p &lt; +1. Let (fn)n1 be a Cauchy sequence in Lp(<br>), then<br>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 g(x) almost<br>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>It follows that (fn(x)) is Cauchy in R and converges to some f(x) a.e. Letting m goes to<br>+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 (xn) has a<br>convergence subsequence, then it converges to the same limit.<br>The preceding proof contains a result which is interesting enough to be stated separetely:<br>Theorem 1.1.3.2 (Convergence criteria for Lp functions) Let 1 p &lt; +1. Let (fn)<br>and f in Lp(<br>) such that (fn) converges to f in Lp(<br>). Then there exists a subsequence<br>(fnk) of (fn) and h 2 Lp(<br>) such that fnk(x) ! f(x) for a.e, x 2<br>and fnk(x) h(x), a.e<br>x 2<br>.<br>Remark 1.1.3.1 It is in general not true that the entire sequence itself converge pointwise to<br>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. Then kfnkp !<br>0, but fn(x) does not converge for any x 2 [0; 1].<br>8<br>+ Lp-spaces<br>Example 1.1.3.2 Let<br>= R, and for n 2 N, set fn = X[n; n + 1]. Then fn(x) ! 0 as<br>n ! 1; but kfnkp = 1 for p 2 [0;1). Thus fn converge pointwise but not in norm.<br>Theorem 1.1.3.3 Let 1 p &lt; 1. Let ffng be a sequence in Lp such that fn(x) ! f(x)<br>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>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; p &lt; 1. So it<br>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[ where<br>ak; bk 2 Q and<br>. Let E be the Q-vector space spanned by the 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 (Oi)i2I<br>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>9<br>+ Lp-spaces<br>1.1.4 Dual Space<br>Theorem 1.1.4.1 (Riesz representation theorem.) Let 1 &lt; p &lt; +1 and let 2 (Lp)0.<br>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 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 E is closed<br>and dense in (Lp). E is closed by using the fact that kTgk = kgkp0 and Lp0 is Banach. For<br>density we will show that if L 2 (Lp)00 and L = 0 on E then L = 0 on (Lp)0. Since Lp is<br>reflexive, we identify (Lp)00 to Lp through the canonical embeding. Thus there exists f 2 Lp<br>such that hL; i = h; fi, for all 2 (Lp)0. So L = 0 on E leads hTg; fi = 0 for all g 2 Lp0<br>and this implys that f = 0 so L is.<br>Theorem 1.1.4.2 (Dual space of L1). Let 2 (L1)0, then there exists a unique g 2 L1<br>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>Indeed assume that L1 is reflexive and let<br>open such that assume that 0 2<br>. Let<br>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 enough, we<br>have B(0; 1=n)<br>. By reflexivity, ffng has a weakly convergence subsequence fnk to some<br>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>10<br>+ Lp-spaces<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 follows that<br>Z</p><p>f’ dx = 0; 8′ 2 Cc(<br>ô€€€ f0g):<br>Thus f = 0 a.e on<br>. On the other hand, taking ‘ 1 in (1.8) leads to<br>Z</p><p>f dx = 1.<br>Contradiction. So L1(<br>) is not reflexive.<br>Since a Banach space is reflexive if and only if its dual E0 is reflexive, then L1(<br>) is not<br>reflexive.<br>Remark 1.1.4.2 Since (L1)0 = L1, then from Banach-Alaogulu theorem any bounded sequence<br>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 there is no<br>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 extension<br>theorem, 0 can be extended to a continuous linear form on L1(<br>), say . We summarize<br>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>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 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>11<br>+ Lp-spaces<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 product<br>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 (X;A; dx) and<br>(Y; B; dy) be -finite measure spaces. Let F be a measurable function on the product space<br>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>in the sense that the integral on the left hand side exists if the one on the right hand side is<br>finite, and in this case the inequality holds. Note that the inequality 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>12<br>+ Lp-spaces<br>by invariance of Lebesgue’s measure under translation. Thus f g(x) exists a.e 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 obtain<br>Z<br>RN<br>jF(x; y)j dy &lt; 1 a.e x 2 RN<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 know that<br>for a.e x 2 RN fixed, y 7! jf(xô€€€y)jjg(y)jp is integrable or equivalently y 7! jf(xô€€€y)j1=pjg(y)j<br>is in Lp(RN). Since y 7! jf(x ô€€€ y)jq is in Lq(RN), we have 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>). The family<br>of functions ; &gt; 0, is called a mollifier with kernel . Note that<br>Z<br>RN<br>dx = 1.<br>13<br>+ Lp-spaces<br>Definition If f is a function on RN and a 2 RN, we define the translation of f by a, af<br>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 for each<br>&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>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 on RN.<br>Then f is continous on RN for each &gt; 0 and for each x 2 RN we 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) a.e. Since<br>2 L1(RN) and the function x ! f(xô€€€y)(y) is continuous a.e y 2 RN then by Lebesgue<br>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>14<br>+ Lp-spaces<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>It follows that for all such &gt; 0, we can write the integral as a sum over jyj &lt; and jyj<br>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 of open<br>subsets of<br>with union U. Let E U. If E Uj is a set of Lebesgue measure 0 for each<br>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 radius<br>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 subset U<br>of<br>on which f is 0 almost everywhere, just take the union of open sets on which f vanishes.<br>Definition The complement of U is called the support of f in<br>and is denoted by supp(f).<br>Proposition 1.1.6.2 If f :<br>! R is continuous then the support of f in<br>is 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 functions<br>on RN with compact support in<br>. We denote by D(<br>) the set of infinitely continuously<br>differentiable functions with compact support in</p><p>15<br>+ Lp-spaces<br>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 mollifier<br>and moreover supp () is the -Ball B0(0; ).<br>Lemma 1.1.6.1 (Uryshon.) Let<br>be an open subset of RN and K<br>be a compact set.<br>Then there exists 2 Cc(<br>) such that 0 1 and = 1 on some neighborhood of K.<br>Proof. Let be a continuous mollifier as above and let L be the closed -neighborhood of<br>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 of K and<br>so has compact support in<br>, 0 1 and = 1 on the ( ô€€€ )-neighborhood of K.<br>Theorem 1.1.6.1 (Density of Cc(<br>) in Lp(<br>) ) . Let<br>be an open subset of RN and let<br>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 the simple<br>functions are dense in Lp(<br>) for finite p, it suffices to show that we can approximate the<br>characteristic function XA of a measurable set A of finite measure by function in Cc(<br>). Let<br>&gt; 0. By the regularity of Lebesgue measure there exits a compact set K A and an open<br>set U, A U such that m(U ô€€€ K) &lt; p. From Uryshon’s Lemma, there is 2 Cc(U) such<br>that 0 1 and = 1 on 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>), and Theorem<br>1.1.3.1 shows that Lp(<br>) is complete. Thus Lp(<br>) is the completion of the metric space<br>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 abstractly<br>as equivalent classes of Cauchy sequence in S. The important point in the present situation is<br>that the various Lp-completion of Cc(<br>) again turn out to be spaces of functions on<br>.<br>The case p = +1 differs from the cases p &lt; 1. The L1-completion of Cc(<br>) is not L1(<br>),<br>but is C0(<br>), the spaces of all continuous functions on<br>which vanish at infinity.<br>16<br>+ Lp-spaces<br>Definition A function f :<br>! R is said to vanish at infinity if for every &gt; 0, there exists<br>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 at infinity.<br>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 by the<br>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 metric space<br>if the distance between f and g is taken to be kf ô€€€ gk1. We have to show that (i) Cc(<br>) is<br>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 that jf(x)j &lt;<br>outside K. Uryshon’s lemma gives us that there exists a function ‘ 2 C0(<br>) such that<br>0 ‘ 1 and ‘(x) = 1 on K. Put h = ‘f. Then 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 of Cauchy<br>sequence and supremum norm, we can assume that ffng converges uniformly. Then its<br>pointwise limit function f is continuous. Given &gt; 0, there exists an N so that kfN ô€€€fk1 &lt;<br>=2 and there exists a compact set K so that jfN(x)j &lt; =2 outside K. Hence jf(x)j &lt;<br>outside K, and we have proved that 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; +1 and f 2<br>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 y; z 2 RN and<br>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, then the<br>support of vg stays in a fixed compact set K for kvk 1. Since g is bounded we have<br>jvgj M XK:<br>It follows that<br>jvg ô€€€ gjp (2M)pXK 2 L1(RN); for kvk 1<br>17<br>+ Lp-spaces<br>since g is continuous we have vg ! g as v ! 0 pointwise. By the dominate convergence<br>theorem<br>Z<br>RN<br>jvg ô€€€ gjp dx ! 0 as v ! 0. Thus there exists &gt; 0 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(RN), 1 p &lt; +1 and let f 2 Lp(RN). If<br>Z<br>RN<br>dx = 1 then<br>f ! f in Lp(RN) as ! 0. If<br>Z<br>RN<br>dx = 0 then 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 continuity of the<br>translation. Thus the Lebesgue dominated convergence theorem 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 f 2 Lp(RN) and<br>g 2 Lq(RN) then f g is uniformly continuous.<br>Proof. We have<br>f g(x)ô€€€f g(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, that is,<br>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>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€€€ kxk2); x 2 RN.<br>Then 2 C1(RN) and (x) = 0 if kxk 1. Moreover, for a suitable choice of the constant<br>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>18<br>+ Lp-spaces<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 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 fact we<br>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 on K as<br>! 0<br>Proof. The convolution exists for each x because the mollifier has compact support. Note<br>that f (x) =<br>Z<br>RN<br>(x ô€€€ y)f(y) dy implies f 2 C1(RN) by standard results on<br>differentiating under the integral sign (since has compact support). The second is obvious<br>and the third follows from<br>kf ô€€€ fkp<br>Z<br>RN<br>kfyf ô€€€ fkp(y) dy supkykkyf ô€€€ fkp<br>Assume that K the set of continuity points of f is compact. Then f is uniformly continuous<br>on K and this shows a little bit more: let &gt; 0, then there exists &gt; 0 such that if x 2 K,<br>z 2 RN and kx ô€€€ zk &lt; then it follows that jf(x) ô€€€ f(z)j &lt; . Note that we do not require<br>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>19<br>+ Lp-spaces<br>Corollary 1.1.7.1 Let<br>be an open subset of RN and K is a compact subset of RN then<br>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 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 C1(RN) has<br>its support in the closed 2-neighborhood of K and so has compact support in<br>. We have<br>that 0 1 and = 1 on the ( ô€€€ )-neighborhood of K.<br>Theorem 1.1.7.2 (Density of D(<br>) in Lp(<br>)) . Let<br>be an open subset of RN and let<br>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>) such that<br>kf ô€€€ gkp &lt;</p><p>2<br>. Let &gt; 0 and define g = g . Then g 2 C1(<br>) and 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 &lt; . Now<br>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 let (Uj)j2J<br>be a locally finite open cover of<br>such that each Uj has compact closure in<br>. Then there<br>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>Proof. There exists an open covering (wj) of<br>such that wj Uj U<br>j for all j 2 Uj .<br>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 j =<br>j</p><p>The j are called a smooth partion of unity subordinate to the locally finite open cover<br>(Uj).<br>Theorem 1.1.7.4 (Finite partition of unity) . Let K be a compact subset of RN and let<br>(Uj)j=1; ;N be a finite open cover of K. Then there exists functions j 2 D(Uj) such that<br>j 0 and<br>XN<br>j=1<br>j = 1<br>in a neighborhood of K.<br>20<br>+ Lp-spaces<br>Proof. For x 2 K, let Vx be an open neighborhood of x such that Vx is a compact subset<br>of Uj with x 2 Uj . Since K is compact there exsit a finite set x1; ; xm 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 is compact,<br>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 of Kj and<br>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 To obtain<br>the equality on a neighborhood of K, we would enlarge K a bit.<br>Theorem 1.1.7.5 (Dubois-Reymond) . Let<br>be an open subset of RN. If f 2 L1(<br>; loc)<br>and<br>Z</p><p>f(x)(x) = 0 for each 2 D(<br>) then f = 0 a.e in</p> <br><p></p>