Spectral decomposition of compact operators on hilbert spaces

Thesis Overview

<p> </p><p>LINEAR OPERATORS AND<br>BOUNDEDNESS<br>In this chapter, some well-known results which will be needed in sequel are provided.<br>1.1 Denitions<br>Denition 1.1. (Norm): A non-negative function jj:jj on a vector space X over R is called a norm<br>on X if and only if the following are satised.<br>(N1) jjxjj &gt; 0 8 x 2 X (positivity).<br>(N2) jjxjj = 0 if and only if x = 0 (Nondegeneracy).<br>(N3) jjxjj = jjjjxjj 8 x 2 X, for all 2 R (Homogeneity).<br>(N4) jjx + yjj jjxjj + jjyjj for all x; y 2 X (Sub-additivity).<br>A vector space X endowed with a norm jj:jj denoted by (X, jj:jj) is called a normed linear space<br>(or just a normed space).<br>Denition 1.2. A sequence (xn)n1 is said to be Cauchy if given ” &gt; 0 there exists N0 2 N such<br>that jjxn ô€€€ xmjj &lt; ” for all m; n N0.<br>Denition 1.3. A space (X; d), where d is a metric is said to be complete if every Cauchy sequence<br>in X converges to a point in it.<br>3<br>Remark<br>Completeness is a metric space concept. In a normed space, the metric is d(x; y) = jjxô€€€yjj where<br>it satises the following special properties:<br>(a) The underlying space is a vector space<br>(b) Homogenity: d(x; y) = jjd(x; y)<br>(c) Translation invariance d(x + z; y + z) = d(x; y)<br>Conversely, every metric satisfying those three conditions denes a norm: jjxjj = d(x; 0)<br>Denition 1.4. A complete normed vector space is called a Banach space.<br>Denition 1.5. Space C([a; b];R)<br>The space C([a; b];R) denotes the set of all real valued continous functions on [a; b] into R:<br>1.2 Examples of Banach spaces<br>1. The space C([a; b];R) endowed with the sup-norm is Banach.<br>Proof. Let (fn)n1 be a Cauchy sequence in C[a; b]. This implies for every x 2 [a; b] and for all<br>” &gt; 0 there exists an N 2 N such that<br>jjfn ô€€€ fmjjC[a; b] = sup jfn(x) ô€€€ fm(x)j ”<br>for all x 2 [a; b] and for all m; n N:<br>This implies jfn(x) ô€€€ fm(x)j ” for all x 2 [a; b] and m; n N,<br>thus (fn(x))n1 is a Cauchy sequence in R and since R is complete, it implies<br>fn(x) ô€€€! f(x) 2 R as n ô€€€! 1<br>this implies jfn(x) ô€€€ f(x)j ” for all x 2 [a; b] and for all n N we have sup jfn(x) ô€€€ f(x)j ”<br>for all x 2 [a; b] and for all n N<br>thus<br>jjfn ô€€€ fjjC[a;b] ” for all n N;<br>this implies f 2 C[a; b]<br>Hence C[a; b] endowed with the sup-norm is Banach.<br>2.The space Rn with jjxjjRn = (<br>P1<br>n=1 jxij2)1=2 is Banach.<br>4<br>1.3 Linear operators<br>Denition 1.6. Let T be an operator from a vector space X to a vector space Y, then the domain<br>D(T) is given by D(T) = f x 2 X : Tx exists in Y g<br>and the range R(T) is given by R(T) = fy 2 Y : 9 x 2 X such that Tx = yg.<br>Denition 1.7. (Null space)<br>Let T be an operator from a vector space X to a vector space Y, then the null space N(T) is given<br>by N(T) = f x 2 X : Tx = 0 g.<br>Denition 1.8. (Injectivity)<br>An operator T from X to a vector space Y is said to be injective if 8 x1; x2 2 D(T), such that<br>Tx1 = Tx2 implies x1 = x2.<br>Remark: If T is injective, then there exists an operator<br>Tô€€€1 : R(T) Y ô€€€! D(T) X such that Tô€€€1(y0) = x0 =) Tx0 = y0.<br>Denition 1.9. (Continuity)<br>An operator T from a vector space X to a vector space Y said to be continous at a point x0 2 X<br>if given any &gt; 0 9 &gt; 0 such that<br>jjx ô€€€ x0jj =) jjTx ô€€€ Tx0jj<br>Denition 1.10. (Linear Operators)<br>Let X and Y be vector spaces. Let T : X ô€€€! Y . Then T is said said to be linear if:<br>i. The domain D(T) is a vector space and the range R(T) lies in a vector space over the same<br>eld.<br>ii. 8 x; y 2 D(T) and scalars ,<br>T(x + y) = Tx + Ty (1.1)<br>T(x) = Tx (1.2)<br>1.3.1 Examples of linear operators<br>1. Dierential operator: Let X be the vector space of all polynomials on [a; b]. We dene a linear<br>operator T on X by setting Tx(t) = x0(t) 8 x 2 X, where the prime denotes dierentiation<br>with respect to t. This operator maps X into itself.<br>2. Integral operator: A linear operator T from C[a; b] into itself can be dened by Tx(t) =<br>R t<br>a x(s)ds; t 2 [a; b].<br>5<br>3. Multiplication by t: This is linear operator from C[a; b] into itself dened by: Tx(t) = tx(t).<br>Theorem 1.11. Let T : X ô€€€! Y be a linear operator space, then<br>(a) The range R(T) is a vector space.<br>(b) If dimX = n &lt; 1, then dimR(T) n.<br>(c) The null space N(T) is a vector space.<br>Proof. (a). Let y1; y2 2 R(T), we show that y1 + y2 2 R(T) for any scalars , . Since<br>y1; y2 2 R(T), we have y1 = Tx1; y2 = Tx2 for some x1; x2 2 D(T), and x1+x2 2 D(T) because<br>D(T) is a vector space. The linearity of T yields<br>T(x1 + x2) = Tx1 + Tx2 = y1 + y2:<br>Hence y1 + y2 2 R(T). Since y1; y2 2 R(T) were arbitrary and so were the scalars, this prove<br>that R(T) is a vector space.<br>(b). We choose n + 1 elements y1; y2; :::; yn+1 in R(T) arbitrary. Then we have<br>y1 = Tx1; :::; yn+1 = Txn+1 for some x1; x2; :::; xn+1 in X. Since dimX = n, the set fx1; :::; xn+1g<br>must be linearly dependent. Hence<br>1×1 + + n+1xn+1 = 0 (1.3)<br>for some scalars 1; ; n+1 not all zero. Since T is linear then T(0) = 0. Applying T to both<br>sides of (1.3) gives T(1×1 + + n+1xn+1) = 1y1 + ::: + n+1yn+1 = 0. This shows that<br>fy1; :::; yn+1g is linearly independent set because the {‘s are not all zero.<br>Remembering that this subset of R(T) was chosen arbitrary, we conclude that R(T) has no linearly<br>independent subsets of n+1 or more elements, this implies dimR(T) n.<br>(c). Let x1; x2 2 N(T), then Tx1 = Tx2 = 0. Since T is linear then for any ; we have<br>T(x1 + x2) = Tx1 + Tx2 = 0:<br>It implies x1 + x2 2 N(T). Hence N(T) is a vector space.<br>Theorem 1.12. ( Inverse of a linear operator)<br>Let X and Y be vector spaces over R. Let T : X ô€€€! Y be linear operator then:<br>(a) The inverse Tô€€€1 : R(T) ô€€€! X exists if and only if Tx = 0 =) x = 0 (T is injective).<br>(b) If Tô€€€1 exists, then it is a linear operator.<br>(c) If dimX = n &lt; 1 and Tô€€€1 exists, then dimR(T) = dimX<br>6<br>Proof. (a).Suppose Tx = 0 =) x = 0. Let Tx1 = Tx2. Since T is linear,<br>T(x1 ô€€€ x2) = Tx1 ô€€€ Tx2 = 0;<br>so that x1 ô€€€ x2 = 0 by hypothesis. Hence Tx1 = Tx2 =) x1 = x2 and Tô€€€1 exist by remark on<br>Denition 1.4. Conversely Tô€€€1 exists then remark on denition 1.4 holds.<br>From Denition 1.4 with x2 = 0, we obtain Tx1 = T0 = 0 =) x1 = 0.<br>(b). We assume Tô€€€1 exists and show that it is linear. The domain of Tô€€€1 is R(T) and it is a<br>vector space, then by Theorem 1.7a, we consider any x1; x2 2 D(T) and their images<br>y1 = Tx1 and y2 = Tx2, then x1 = Tô€€€1y1 and x2 = Tô€€€1y2. T is linear so that for any scalar<br>and , we have<br>y1 + y2 = Tx1 + Tx2 = T(x1 + x2):<br>It implies Tô€€€1(y1 + y2) = x1 + x2 = Tô€€€1y1 + Tô€€€1y2: It implies Tô€€€1 is linear.<br>(c). We have dimR(T) dimX by Theorem 1.7b and dimX dimR(T) by the same theorem<br>applied to Tô€€€1. Hence, dim X = dimR(T).<br>Lemma 1.13. (Inverse of product)<br>Let T : X ô€€€! Y and S : Y ô€€€! Z be bijective linear operator, where X; Y;Z are vector spaces.Then<br>the inverse (ST)ô€€€1 : Z ô€€€! X of the product (the composite) ST exists and (ST)ô€€€1 = Tô€€€1Sô€€€1.<br>Proof. The operator ST : X ô€€€! Z is bijective, so (ST)ô€€€1 exists. We have<br>(ST)(ST)ô€€€1 = IZ;<br>where IZ is the identity operator on Z. Applying Sô€€€1 and using Sô€€€1S = IY (the identity operator<br>on Y ), we obtain<br>Sô€€€1(ST)(ST)ô€€€1 = T(ST)ô€€€1 = Sô€€€1IZ = Sô€€€1:<br>Applying Tô€€€1 and using Tô€€€1T = IX, we obtain the desired result<br>Tô€€€1T(ST)ô€€€1 = (ST)ô€€€1 = Tô€€€1Sô€€€1:<br>Implies (ST)ô€€€1 = Tô€€€1Sô€€€1.<br>Theorem 1.14. Every linear operator on a nite dimensional vector space can be represented by<br>means of matrix.<br>Proof. Let X and Y be nite dimensional vector spaces over thesame eld. Let T : X ô€€€! Y<br>be a linear operator, let dimX = n and dimY = r, then there exists a basis fe1; e2; :::eng for X and<br>a basis fb1; b2; :::brg for Y.<br>Let x 2 X ) x =<br>Pn<br>i=1 iei where 0i<br>s are scalars . Since T is linear<br>y = T(x) =<br>Xn<br>i=1<br>iT(ei):<br>7<br>So T is uniquely determined if the images Tei 1 i n are prescribed. Since y and Tei are in<br>Y so y =<br>Pr<br>j=1 jbj and Tei =<br>Pr<br>j=1 jibj where j and ji are scalars, thus<br>y =<br>Xr<br>j=1<br>jbj =<br>Xn<br>i=1<br>iT(ei) =<br>Xn<br>i=1<br>i<br>Xr<br>j=1<br>jibj =<br>Xr<br>j=1<br>(<br>Xn<br>i=1<br>jii)bj :<br>Hence<br>j =<br>Xn<br>i=1<br>jii: 1 j r:<br>1.4 Bounded linear operators<br>Denition 1.15. (Bounded linear operator): Let X and Y be normed spaces and T:X ô€€€! Y be<br>linear operator. The operator T is said to be bounded if there exist a real number c &gt;0 such that<br>jjTxjj cjjxjj for all x 2 D(T).<br>Theorem 1.16. Let T : X ô€€€! Y be a bounded linear operator. Then<br>jjTjj := sup<br>x2X;jjxjj=1<br>jjTxjj = sup<br>x2X;x6=0<br>jjTxjj<br>jjxjj<br>:<br>Proof. Let jjxjj = a, set y = ( 1<br>a )x, where x 6= 0. Then jjyjj = jjxjj<br>a = 1. Since T is linear, then<br>sup<br>x2X;x6=0<br>jjTxjj<br>jjxjj<br>= sup<br>jjTxjj<br>a<br>= sup jjT(<br>1<br>a<br>)x)jj = sup<br>y2X;jjyjj=1<br>jjTyjj := jjTjj:<br>Remark: jj:jj denes a norm on X.<br>Theorem 1.17. (Finite dimension): If a normed space X is nite dimensional, then every linear<br>operator on X is bounded.<br>Proof. Let dim X = n and fe1; e2; : : : ; eng be a basis for X, then for all x 2 X;<br>x =<br>Xn<br>i=1<br>iei<br>i scalars. Since T is linear,<br>jjTxjj = jj<br>Xn<br>i=1<br>iTeijj<br>Xn<br>i=1<br>jijjjTeijj max<br>i<br>jjTeijj<br>Xn<br>i=1<br>jij = jjxjj1 where ( = max jjTeijj)<br>= cjjxjj (where c= k by equivalence of norms on nite dimensional vector space)<br>Theorem 1.18. (Continuity and boundedness): Let T : X ô€€€! Y be a linear operator, where X<br>and Y are normed spaces. Then:<br>(a) T is continuous if and only if T is bounded.<br>8<br>(b) If T is continuous at the origin, then T is continuous.<br>Proof. (a) For T = 0, the statement is trivial. Let T 6= 0,then jjTjj 6= 0. We assume T is<br>bounded and consider x0 2 X such that jjx ô€€€ x0jj &lt; where = =jjTjj, we obtain<br>jjTx ô€€€ Tx0jj = jjT(x ô€€€ x0)jj jjTjjjjx ô€€€ x0jj &lt; jjTjj = :<br>Since x0 2 X was arbitrary, this shows that T is continuous.<br>Conversely, assume that T is continuous at an arbitrary x0 2 X, then given any &gt; 0, there exist<br>&gt; 0 such that<br>jjTx ô€€€ Tx0jj<br>for all x 2 X satisfying<br>jjx ô€€€ x0jj :<br>We now take y 6= 0 2 X and set x = x0 +<br>jjyjjy. Then x ô€€€ x0 =<br>jjyjjy. Hence jjx ô€€€ x0jj = . Since<br>T is linear we have<br>jjTx ô€€€ Tx0jj = jjT(x ô€€€ x0)jj = jjT(</p><p>jjyjj<br>y)jj =</p><p>jjyjj<br>jjTyjj :<br>Thus jjTyjj<br>jjyjj; jjTyjj cjjyjj =) T is bounded, where c =</p><p>(b) Suppose T is continuous at a point x0 = 0, then it suces to show that T is bounded<br>(continuous). T is continuous at x0 = 0, take = 1, there exist &gt; 0. such that<br>jjxjj =) jjTxjj 1:<br>Let z 2 D(T) z 6= 0, then jj z<br>jjzjj</p><p>2 jj =<br>2 &lt; =) jjT( z<br>jjzjj</p><p>2 )jj &lt; 1<br>(By linearity of T) =) jjTzjj 2<br>jjzjj 8z 2 D(T) =) T is continuous.<br>Corollary 1.19. (Continuity and null space)<br>Let T : X ô€€€! Y be a bounded linear operator. Then<br>(a) xn ô€€€! x implies Txn ô€€€! Tx:<br>(b)The null space N(T) is closed.<br>Proof. (a) Suppose xn ô€€€! x in X i.e jjxn ô€€€ xjj ô€€€! 0. Since T is linear and bounded, then<br>jjTxn ô€€€ Txjj = jjT(xn ô€€€ x)jj kjTjjjjxn ô€€€ xjj ô€€€! 0:<br>It implies Txn ô€€€! Tx as n ô€€€! 1<br>(b) Let x 2 N(T) it implies there exist (xn)n1 N(T) such that xn ô€€€! x. Since T is bounded<br>by corollary 1.19a Txn ô€€€! Tx, but xn 2 N(T): It implies Txn = 0 8n 1, thus T(x) = 0:<br>It implies x 2 N(T). Hence N(T) is closed.<br>9<br>1.5 Examples of bounded operators on innite dimensional spaces<br>1. Let K : [0; 1] [0; 1] ô€€€! R be continuous. Let T : C([0; 1];R) ô€€€! C([0; 1];R) be dened by<br>T(f)(x) =<br>Z 1<br>0<br>K(x; y)f(y)dy:<br>Then T 2 L(C[0; 1]) for f 2 C[0; 1] and bounded.<br>Proof. Clearly T is linear. We next show boundedness.<br>jT(f)(x)j<br>Z 1<br>0<br>jK(x; y)jjf(y)jdy supjf(y)j<br>Z 1<br>0<br>jK(x; y)jdy jjfjj<br>Z 1<br>0<br>jK(x; y)jdy:<br>It implies<br>jjT(f)(x)jj1 cjjfjj1<br>where<br>R 1<br>0 jK(x; y)j c since K is continuous. It implies T is bounded.<br>2. Let p 1, we dene<br>lp = f(xn)n1 R :<br>1X<br>n=1<br>jxnjp &lt; 1g<br>Let T : lp ô€€€! lp be dened by<br>T((xn)n1) = (xn+1)n1 (The left shift operator) is bounded<br>where (xn)n1 = (x1; x2; x3; :::) and T((xn)n1) = (x2; x3; :::)<br>Proof.<br>jjT((xn)n1)jj = (<br>1X<br>n=2<br>jxnjp)1=p (<br>1X<br>n=1<br>jxnjp)1=p = jj(xn)n1jj:<br>Thus T is bounded with jjTjj 1<br>3. Let T : L2([0; 1];R) ô€€€! L2([0; 1];R) be dened by<br>(Tf)(t) = tf (t) for a:e t 2 [0; 1]:<br>Then T is bounded.<br>Proof. jjTfjj2<br>L2[0;1] =<br>R 1<br>0 j(Tf)(t)j2dt =<br>R 1<br>0 jtj2jf(t)j2dt<br>R 1<br>0 jf(t)j2dt = jjfjj2<br>L2[0;1]:<br>It implies jjTfjjL2[0;1] jjfjjL2[0;1]: Hence T is bounded with jjTjj 1<br>10<br>1.6 Hilbert spaces<br>Denition 1.20. Let E be a real vector space. An inner product on E is a function,<br>h:; :i : E E ô€€€! R such that<br>(a) jjxjj2 hx; xi 0 with equality jjxjj2 = 0 i x = 0<br>(b) hx; yi = hy; xi<br>(c)hax + by; zi = ahx; zi + bhy; zi i.e x ô€€€! hx; zi is linear.<br>A real vector space E endowed with the inner product i.e (E; h:; :i) is called an inner product space.<br>Lemma 1.21. (Cauchy-Schwartz Inequality) Let E be an inner product space. Then for arbitrary<br>x; y 2 E,<br>jhx; yij jjxjjjjyjj<br>Lemma 1.22. (The Parallelogram Law) Let E be a real inner product space. Then for arbitary<br>vector x; y 2 E,<br>jjx + yjj2 + jjx ô€€€ yjj2 = 2(jjxjj2 + jjyjj2):<br>Proof. Expanding the LHS jjx+yjj2+jjxô€€€yjj2 = hx; xi+2hx; yi+hy; yi+hx; xiô€€€2hx; yi+hy; yi<br>= 2(hx; xi + hy; yi) = 2(jjxjj2 + jjyjj2) = RHS.<br>Denition 1.23. A complete inner product space is called a Hilbert space.<br>Denition 1.24. Let x, y be vectors in a Hilbert space H, then we say that x and y are orthog-<br>onal,written x ? y, if hx; yi = 0. We say that subsets A and B are orthogonal, written A ? B, if<br>x ? y for every x 2 A and y 2 B. The orthogonal complement A? of a subset of A is the set of<br>vectors orthogonal to A,<br>A? = fx 2 H : x ? y for all y 2 Ag:<br>Denition 1.25. Let M and N be closed linear subspaces of a Hilbert space H, we dene the<br>orthogonal direct sum or simply the direct sum M<br>L<br>N of M and N by<br>M<br>M<br>N = fy + z : y 2M and z 2 Ng:<br>Denition 1.26. A subset U of nonzero vectors in a Hilbert space H is orthogonal if any two<br>distinct elements in U are othorgonal. A set of vectors U is orthonormal if it is orthogonal and<br>jjujj = 1 for all u 2 U.<br>1.7 Some properties of Hilbert spaces<br>Theorem 1.27. The orthogonal complement of a subset of a Hilbert space is a closed linear<br>subspace.<br>11<br>Proof. Let H be a Hilbert space and A a subset of H. if y; z 2 A? and ; 2 R:Then the<br>linearity of the inner product implies that<br>hx; y + zi = hx; yi + hx; zi = 0<br>for all x 2 A:<br>Therefore, y + z 2 A?, so A? is a linear subspace.<br>To show that A? is closed, we show that if (yn)n1 is a convergent sequence in A?, then the limit<br>y also belongs to A?. Let x 2 A then by continuity of inner product we have<br>hx; yi = hx; lim<br>nô€€€!1<br>yni = lim<br>nô€€€!1<br>hx; yni = 0:<br>Since hx; yni = 0 for every x 2 A and yn 2 A?. Hence y 2 A?<br>Theorem 1.28. Let M be a closed linear subspace of a Hilbert space H<br>(a) For every x 2 H there is a unique closest point y 2M such that<br>jjx ô€€€ yjj = minjjx ô€€€ zjj; z 2M<br>(b)The point y 2Mclosest to x 2 H is the unique element ofMwith the property that (xô€€€y) ?M<br>Proof. (a). Let d be the distance of x from M i.e<br>d = inffjjx ô€€€ zjj : z 2 Mg:<br>First, we prove that there is a closest point y 2 M at which this inmum is attained, meaning<br>that jjx ô€€€ yjj = d: From the denition of d, there is a sequence of elements yn 2M such that<br>lim<br>nô€€€!1<br>jjx ô€€€ ynjj = d:<br>Thus, for any ” &gt; 0; there is an N such that<br>jjx ô€€€ ynjj d + ” when n N:<br>We show that the sequence (yn)n1 is Cauchy. From the parallelogram law, we have<br>jjym ô€€€ ynjj2 + jj2x ô€€€ ym ô€€€ ynjj2 = 2jjx ô€€€ ymjj2 + 2jjx ô€€€ ynjj2:<br>Since (ym + yn)=2 2M; it implies that jjx ô€€€ (ym + yn)=2jj d. Thus for all m; n N<br>jjym ô€€€ ynjj2 = 2jjx ô€€€ ymjj2 + 2jjx ô€€€ ynjj2 ô€€€ jj2x ô€€€ ym ô€€€ ynjj2 4(d + “)2 ô€€€ 4d2 = 4″(2d + “):<br>Therefore,(yn)n1 is Cauchy.Since a Hilbert space is complete, there is a y such that yn ô€€€! y and<br>since M is closed, we have y 2M: By continuity of norm we have<br>jjx ô€€€ yjj = lim<br>nô€€€!1<br>jjx ô€€€ ynjj = d<br>12<br>We prove the uniqueness of the vector y 2Mthat minimizes jjxô€€€yjj. Suppose that y and y0 both<br>minimize the distance to x, meaning that jjx ô€€€ yjj = d, jjx ô€€€ y0jj = d:<br>Then the parallelogram law implies that<br>2jjx ô€€€ yjj2 + 2jjx ô€€€ y0jj2 = jj2x ô€€€ y ô€€€ y0jj2 + jjy ô€€€ y0jj2:<br>Since (y + y0)=2 2M,<br>jjy ô€€€ y0jj2 = 4d2 ô€€€ 4jjx ô€€€ (y + y0)=2jj2 0:<br>Therefore, jjy ô€€€ y0jj = 0 so that y = y0: (b) We show that the unique y 2M found above satises<br>the condition that the vector x ô€€€ y is orthogonal to M. Since y minimizes the distance to x, we<br>have for every 2 C and z 2M that<br>jjx ô€€€ yjj2 jjx ô€€€ y + zjj2:<br>Expanding the right-hand side of this equation, we obtain that<br>2Rehx ô€€€ y; zi jj2jjzjj2:<br>Suppose that hx ô€€€ y; zi = jhx ô€€€ y; zijei’: Choosing = “eô€€€i’; where ” &gt; 0 and dividing by “, we<br>get<br>2jhx ô€€€ y; zij “jjzjj2:<br>Taking the limit as ” ô€€€! 0+, we get hx ô€€€ y; zi = 0 so (x ô€€€ y) ?M.<br>Finally, we show that y is the only element in M such that (x ô€€€ y) ? M: Suppose that y0 is<br>another such element in M: Then y ô€€€ y0 2M; and for any z 2M; we have<br>hz; y ô€€€ y0i = hz; x ô€€€ y0i ô€€€ hz; x ô€€€ yi = 0:<br>In particular, we may take z = y ô€€€ y0 and therefore we have y = y0<br>Denition 1.29. Let<br>be an open set in Rn, then<br>Lp(<br>) = ff :<br>ô€€€! R; measurable :<br>R</p><p>jfjpdx &lt; 1g, 1 p &lt; 1.<br>Denition 1.30. Let<br>be an open set in Rn and n 2 N, the Sobolev space Hm(<br>) is dened by<br>Hm(<br>) = ff 2 L2(<br>);Df 2 L2(<br>); 2 Nn; jj mg where Df = @jj<br>@x<br>1<br>1 :::@xn<br>n<br>f, jj = 1 + ::: +<br>n and jjujjHm(<br>) = jjujjL2 +<br>P<br>:jjm jDujL2 ; u 2 Hm(<br>)<br>Denition 1.31. Let ‘ :<br>ô€€€! R be continous, then the support of ‘ is dened by<br>Supp(‘) = fx 2<br>: ‘(x) 6= 0g<br>Denition 1.32. D(<br>) the space of test functions is dened by<br>D(<br>) = ff 2 C1 : Supp(f) is compact in<br>g<br>13<br>Denition 1.33. A distribution is a continuous linear map T : D(<br>) ô€€€! R such that<br>lim<br>nô€€€!1<br>T(‘n) = T(‘)<br>for any sequence<br>‘n<br>D(<br>)<br>ô€€€! ‘:<br>The space of distribution on<br>is denoted by D0(<br>).<br>Denition 1.34. A sequence of distribution Tn 2 D0(<br>) is said to converge in the sense of<br>distribution to T 2 D0(<br>), if for every test function ‘ 2 D(<br>) one has<br>limnô€€€!1hTn; ‘i = hT; ‘i<br>1.7.1 Examples of Hilbert spaces<br>(a). L2(<br>) equipped with the norm jjfjjL2(<br>) = (<br>R</p><p>jfj2dx)1=2 is Hilbert.<br>(b). H1(<br>) equipped with the norm jjujj2<br>H1(<br>) =<br>R</p><p>u2dx +<br>R</p><p>jruj2dx is Hilbert, where<br>H1(<br>) = ff 2 L2(<br>) :<br>@f<br>@xi<br>2 L2(<br>)g<br>Proof. (a).Let (fn)n1 be a Cauchy sequence in L2(<br>), then we can nd a subsequence (fnk )k1<br>such that<br>jjfnk ô€€€ fnk+1jj &lt;<br>1<br>2k ; k = 1; 2; 3; :::<br>Choose a function g 2 L2(<br>). By the Schwartz inequality,<br>Z</p><p>jg(fnk ô€€€ fnk+1jd<br>jjgjj<br>2k :<br>Hence<br>1X<br>k=1<br>Z</p><p>jg(fnk ô€€€ fnk+1jd jjgjj:<br>Thus<br>jg(x)j<br>1X<br>k=1<br>jfnk (x) ô€€€ fnk+1(x)j &lt; +1<br>almost everywhere on X.<br>It implies<br>1X<br>k=1<br>jfnk (x) ô€€€ fnk+1(x)j &lt; +1<br>almost everywhere on X.<br>Since the kth partial sum of the series<br>P1<br>k=1(fnk (x)ô€€€fnk+1(x)) which converges almost everywhere<br>on X is fnk (x) ô€€€ fnk+1(x):<br>It implies<br>f(x) = lim<br>kô€€€!1<br>fnk (x):<br>14<br>Let ” &gt; 0 be given, there exists N0 2 N such that<br>jjf ô€€€ fnk jj lim inf<br>jô€€€!1<br>jjfnj ô€€€ fnk jj “:<br>Thus f ô€€€ fnk 2 L2(<br>), and since f = (f ô€€€ fnk ) + fnk , we see that f 2 L2(<br>).<br>Also, since ” is arbitrary,<br>lim<br>kô€€€!1<br>jjf ô€€€ fnk jj = 0:<br>Finally, the inequality<br>jjf ô€€€ fnjj jjf ô€€€ fnk jj + jjfnk ô€€€ fnjj<br>shows that (fn) converges to f in L2(<br>).<br>(b). Let (un)n1 be a Cauchy sequence in H1(<br>) then given ” &gt; 0 there exist n0 2 N such that<br>8m; n &gt; n0<br>jjun ô€€€ umjjH1(<br>) &lt; “:<br>which implies that<br>Z</p><p>jun ô€€€ umj2dx +<br>Z</p><p>jrun ô€€€ rumj2dx<br>1=2<br>&lt; ” ():<br>Thus (un)n1 be a Cauchy sequence in L2(<br>) and (run)n1 is also a Cauchy sequence in L2(<br>).<br>Since L2(<br>) is complete,<br>un ô€€€! u 2 L2(<br>) and run ô€€€! wi 2 L2(<br>):<br>We need to show that ru = wi.<br>But<br>un ô€€€! u 2 L2(<br>) ) un ô€€€! u 2 D0(<br>);<br>thus<br>run ô€€€! ru 2 D0(<br>):<br>By uniqueness of limit in D0(<br>), we have ru = wi<br>From (**), let n be xed and let m ô€€€! 1 we have<br>Z</p><p>jun ô€€€ uj2dx +<br>Z</p><p>jrun ô€€€ ruj2dx<br>1=2<br>&lt; “;<br>thus un ô€€€! u in H1(<br>). Hence H1(<br>) is Hilbert.<br>15<br>Theorem 1.35. (Riesz Theorem) Let H be a Hilbert space over R or C. If T is a bounded linear<br>functional on H i.e T is a bounded operator from H to the eld R or C, then there exists some<br>g 2 H such that for every f 2 H we have<br>T(f) = hf; gi . Moreover, jjTjj = jjgjj:<br>Proof. We can choose an orthonormal basis j ; j 1 for H. Let T be bounded linear<br>functional and set aj = T(j ). Choose f 2 H, let cj = hf; ji and dene<br>fn =<br>Xn<br>j=1<br>cjj :<br>Since j forms a basis we know that jjfn ô€€€ fjj ô€€€! 0 as n ô€€€! 1.<br>Since T is linear we have<br>T(fn) =<br>Xn<br>j=1<br>ajcj (1)<br>Since T is bounded, say with norm jjTjj &lt; 1 we have<br>jjT(fn) ô€€€ T(f)jj jjTjjjjfn ô€€€ fjj (2)<br>Because jjfn ô€€€ fjj ô€€€! 0 as n ô€€€! 1, we conclude from equations (1) and (2) that<br>T(f) = lim<br>nô€€€!1)<br>T(fn) =<br>1X<br>j=1<br>ajcj (3)<br>Infact, the sequence aj must itself be square-summable. To see this, rst note that since jT(f)j<br>jjTjjjjfjj we have<br>j<br>1X<br>j=1<br>ajcj j jjTjj(<br>1X<br>j=1<br>c2j<br>)1=2 (4)<br>Equation (4) must hold for any square-summable sequence cj (since any cj corresponds to some<br>elements in H).Fix a positive integer N and dene a sequence cj = aj for j N, cj = 0 for j N.<br>Clearly such a sequence is square-summable and equation (4) then yields<br>j<br>XN<br>j=1<br>a2j<br>j jjTjj(<br>XN<br>j=1<br>a2j<br>)1=2<br>or<br>(<br>XN<br>j=1<br>a2j<br>)1=2 jjTjj (5)<br>Thus aj is square-summable the function g =<br>P<br>j ajj is well dened as an element of H and<br>T(f) =<br>X<br>j<br>ajcj = hf; gi:<br>16<br>Finally, equation (5) makes it clear that jjgjj jjTjj. But from Cauchy-Schwartz we also have<br>jT(f)j = jhf; gij jjfjjjjgjj implying jjTjj jjgjj, so jjTjj = jjgjj.<br>17</p><p>&nbsp;</p> <br><p></p>