Functional Analysis Assignment NO.2

1. Let $(X,d)$ be a complete metric space, and let $K\subset X$ be precompact. Prove that $K$ is separable.

According to the Step 1 and Step 2 in the necessary part proof of Theorem 1.1, we know that $K$ is precompact can get $K$ is totally bounded. So $\forall \epsilon>0\exists m\in \N^*,x_i\in K,s.t.\enspace K\subset\bigcup_{i=1}^mB(x_i,\epsilon)$ . Let $\epsilon=\frac{1}{n},X_n=\{x_i\},S=\bigcup_{n=1}^\infty X_i$ . As $|X_n|$ is finite, $S\subset K$ is countable. And for any $x\in K$ , we know $\forall n>0 \exists x_n\in X_n\subset S,s.t.\enspace x\in B(x_n,\frac{1}{n})$ . So $\lim_{n\rightarrow\infty}d(x_n,x)=0$ , which means $S$ is dense subset. Therefore $S$ is a countable dense subset of $K$ , hence $K$ is separable.

2. Denote $\R^n =\{(x^1, \dots ,x^n)^T : x^k \in \R\}$ .

(i) Let $M$ be a symmetric positive definite $n\times n$ matrix. Prove that $(x,y) := x^TMy$ , $x,y \in \R^n$ , defines an inner product.

Let $x_i,x,y\in \R^n,\lambda_i\in \R$ .
Symmetric: $(x,y)=x^TMy=(x^TMy)^T=y^TM^Tx=y^TMx=(y,x)$ .
Positive definite: $(x,x)=x^TMx\ge0$ , and $(x,x)=x^TMx=0$ iff $x=0$ . (As $M$ is positive definite.)
Linear: $(\lambda_1x_1+\lambda_2x_2,y)=(\lambda_1x_1+\lambda_2x_2)^TMy=\lambda_1x_1^TMy+\lambda_2x_2^TMy=\lambda_1(x_1,y)+\lambda_2(x_2,y)$ .
Q.E.D.

(ii) Suppose $(\cdot,\cdot)$ is an inner product on $\R^n$ . Prove that there is a unique symmetric positive definite $n\times n$ matrix $M$ , such that $(x, y) = x^TMy$ , $x,y \in \R^n$ .

Consider a matrix $M$ , and $M_{ij}=(e_i,e_j)$ . As $x=\sum_{i=1}^nx_ie_i,y=\sum_{i=1}^ny_ie_i$ , we know $x^TMy=(\sum_{i=1}^nx_ie_i)^TM(\sum_{i=1}^ny_ie_i)=\sum_{i=1}^n\sum_{j=1}^nx_iy_je_i^TMe_j=\sum_{i=1}^n\sum_{j=1}^nx_iy_jM_{ij}=\sum_{i=1}^n\sum_{j=1}^nx_iy_j(e_i,e_j)=(\sum_{i=1}^nx_ie_i,\sum_{i=1}^ny_ie_i)=(x,y)$ . And as $(\cdot,\cdot)$ is symmetric and positive definite, $M$ is symmetric positive definite.
Now suppose that $M'$ such that $x^TMy=x^TM'y$ . Therefore $x^T(M-M')y=0,\forall x,y\in\R^n$ , which means $M-M'$ must be $0$ . So $M$ is unique.

3. Let $H$ be a Hilbert space. Let $D \subset H$ be a subset such that the linear space spanned by $D$ is dense in $H$ . Let $(E_n)_{n\ge1}$ be a sequence of closed subspaces in $H$ that are mutually orthogonal. Assume that $\sum_{n=1}^\infty|P_{E_n}u|^2=|u|^2,\forall u\in D$ . Prove that $H$ is the Hilbert sum of the $E_n$ 's.

Suppose that $h_k=u-\sum_{n=1}^k P_{E_n}u$ . As $E_n$ is mutually orthogonal, $P_{E_n}u$ must be orthogonal with each other. And $\langle h_k,P_{E_i}u\rangle=\langle u-\sum_{n=1}^k P_{E_n}u,P_{E_i}u\rangle=\langle u,P_{E_i}u\rangle-\langle P_{E_i}u,P_{E_i}u\rangle=0$ for all $1\leq i\leq k$ , which means $h_k$ is orthogonal with $P_{E_i}u$ for all $1\leq i\leq k$ . So $|u|^2=|\sum_{n=1}^k P_{E_n}u+h_k|^2=\sum_{n=1}^k|P_{E_n}u|^2+|h_k|^2$ . Notice that $\sum_{n=1}^\infty|P_{E_n}u|^2=|u|^2$ , therefore $\lim_{k\rightarrow \infty}|h_k|^2=0$ , which means $\lim_{k\rightarrow \infty}h_k=0$ . Therefore $u=\sum_{n=1}^\infty P_{E_n}u$ , which means $D\subseteq\operatorname{Span}\bigcup_{n=1}^\infty E_n$ . As the linear space spanned by $D$ is dense in $H$ means that $H=\overline{\operatorname{Span}D}$ , we have $H=\overline{\operatorname{Span}D}\subseteq\overline{\operatorname{Span}\bigcup_{n=1}^\infty E_n}\Rightarrow\overline{\operatorname{Span}\bigcup_{n=1}^\infty E_n}=H$ . So, $H$ is the Hilbert sum of the $E_n$ 's.

4. Let $E$ be an n.v.s., and $F$ be a Banach space. Show that $\mathcal L(E,F)$ is a Banach space.

It's obvious that $\mathcal L(E,F)$ is an n.v.s., now consider a Cauchy sequence $(f_i)$ in $\mathcal L(E,F)$ , which means $\forall\epsilon>0\exists N_0\in \N^*\forall m,n>N_0,s.t.\enspace d(f_m,f_n)=\|f_m-f_n\|_{\mathcal L(E,F)}=\sup_{\|u\|_E\leq1}\|(f_m-f_n)u\|_F<\epsilon$ . As $\|f_m(u)-f_n(u)\|_F\leq\|f_m-f_n\|\|u\|$ , $(f_i(u))$ must be a Cauchy sequence in $F$ . And notice that $F$ is a Banach space, which means it's complete, we have $\lim_{k\rightarrow\infty}f_k(u)\in F$ . Therefore we can let $f:u\mapsto\lim_{k\rightarrow\infty}f_k(u)$ . Because $f_i$ is linear, $f$ must be linear too. Now consider $\|f-f_m\|=\sup_{\|u\|\leq1}\|\lim_{k\rightarrow\infty}f_k(u)-f_m(u)\|\leq\limsup_{k\rightarrow\infty,\|u\|\leq1}\|f_k(u)-f_m(u)\|\leq\lim\sup_{n\rightarrow\infty}\|f_n-f_m\|\leq\epsilon$ , which means $f-f_m\in\mathcal L(E,F)$ . Notice that $\mathcal L(E,F)$ is a vector space, we get $f\in\mathcal L(E,F)$ . At last, we check that $\lim_{n\rightarrow\infty}d(f,f_n)=\lim_{n\rightarrow\infty}\|f-f_n\|\leq\lim_{n\rightarrow\infty}\limsup_{m\rightarrow\infty}\|f_m-f_n\|=0$ , which means $f=\lim_{k\rightarrow\infty}f_k\in\mathcal L(E,F)$ . Therefore, $\mathcal L(E,F)$ is a Banach space.

5. Let $E,F,G$ be Banach spaces.

(1) Let $A \in \mathcal L(E,F)$ . $A \in \mathcal K(E,F)$ is equivalent to that $\overline{A(D)}$ is compact for any bounded set $D\subset E$ . In other words, $A \in\mathcal K(E,F)$ is equivalent to that the sequence $(Ax_n)$ in $F$ has a convergent subsequence for every bounded sequence $(x_n)$ in $E$ .

$\Rightarrow:$ As $A\in\mathcal K(E,F)$ means that for a sequence $(y_i)$ in $A(B_E)$ , it has a convergent subsequence. Now consider a bounded sequence $(x_i)$ in $E$ , let $\lambda=\sup_{i\ge1}\|x_i\|\in \R$ . Therefore $(\frac{x_i}{\lambda})$ is a sequence in $B_E$ , and $(A\frac{x_i}{\lambda})$ is a sequence in $A(B_E)$ that has a convergent subsequence. Notice that $Ax_i=\lambda A\frac{x_i}{\lambda}$ and $\lambda<\infty$ . So, $(Ax_i)$ must has a convergent subsequence too.
$\Leftarrow:$ Consider a sequence $(y_i)$ in $A(B_E)$ , there must exist a sequence $(x_i)$ in $B_E$ such that $Ax_i=y_i$ . As $x_i\in B_E$ , we get $(x_i)$ is a bounded sequence. By the assumption, $(y_i)=(Ax_i)$ must has a convergent subsequence. Therefore, $A(B_E)$ is precompact, hence $A\in\mathcal K(E,F)$ .
Q.E.D.

(2) If $A \in \mathcal L(E,F)$ and $B\in\mathcal L(F,G)$ , and at least one of $A$ or $B$ is compact, then $B \circ A$ is compact.

First consider that $A$ is compact. By (1), we know for every bounded sequence $(x_i)$ in $E$ , $(Ax_i)$ in $F$ has a convergent subsequence $(Ax_{i_j})\rightarrow Ax_{i_0}$ . As $\|B\circ A(x_{i_j})-B(Ax_{i_0})\|=\|B(A(x_{i_j})-A(x_{i_0}))\|\leq\|B\|\|Ax_{i_j}-Ax_{i_0}\|\rightarrow0$ , $(B\circ A x_i)$ has a convergent subsequence $(B\circ A x_{i_j})$ . Therefore, by (1), $B\circ A$ is compact.
Then consider that $B$ is compact. Consider a bounded sequence $(x_i)$ in $E$ . As $\|Ax_i\|\leq\|A\|\|x_i\|$ , $(Ax_i)$ is also a bounded sequence. By (1), $(B(Ax_i))=(B\circ A(x_i))$ has a convergent subsequence, which means $B\circ A$ is compact.
Q.E.D.

(3) Let $(T_n)$ be a sequence in $\mathcal K(E,F)$ that converges to $T \in\mathcal L(E,F)$ . Prove that $T$ is compact.

As $\lim_{n\rightarrow\infty}\|T-T_n\|=\lim_{n\rightarrow\infty}\sup_{\|u\|\leq1}\|T(u)-T_n(u)\|=0$ , we have $\lim_{n\rightarrow\infty}\|T(u)-T_n(u)\|=\lim_{n\rightarrow\infty}\|u\|\|(T-T_n)(\frac{u}{\|u\|})\|\leq\lim_{n\rightarrow\infty}\|u\|\sup_{\|u\|\leq1}\|T(u)-T_n(u)\|=0$ , hence $T(u)=\lim_{n\rightarrow\infty}T_n(u)$ . Consider a bounded sequence $(x_i)$ in $E$ . For $T_1$ , there exists a subsequence $(x_i^{(1)})$ in $E$ such that $(T_1x_i^{(1)})$ converges. And for $T_{n+1}$ , there exists a subsequence $(x_{i}^{(n+1)})$ of $(x_i^{(n)})$ in $E$ such that $(T_{n+1}x_i^{(n+1)})_{i\ge1}$ converges. Notice that by this way, $(x_i^{(n)})$ is a subsequence of $(x_i^{(k)})$ , and $(T_kx_i^{(n)})_{i\ge1}$ is convergent for all $k\leq n$ . Now consider the sequence $(x_i^{(i)})$ is a subsequence of $(x_i)$ , and $\|Tx_i^{(i)}-T x_j^{(j)}\|=\|(T-T_n)(x_i^{(i)}-x_j^{(j)})+T_n(x_i^{(i)}-x_j^{(j)})\|\leq\|T-T_n\|\|x_i^{(i)}-x_j^{(j)}\|+\|T_nx_i^{(i)}-T_nx_j^{(j)}\|$ . Let $n\rightarrow\infty$ , we have $\|Tx_i^{(i)}-T x_j^{(j)}\|\leq\|T_nx_i^{(i)}-T_nx_j^{(j)}\|$ . As $(T_nx_i^{(i)})$ is convergent, $(Tx_i^{(i)})$ must be convergent, hence $T$ is compact.

6.

(1) The inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is defined by $i(x) = x$ for all $x \in\ell^2$ . Show that the inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is bounded but not compact.

As $\sup_{x\in\ell^2,x\neq0}\frac{\|i(x)\|_{\ell^\infty}}{\|x\|_{\ell^2}}=\sup_{x\in\ell^2,x\neq0}\frac{\sup_{n\ge1}|x_n|}{\sqrt{\sum_{n\ge1}{|x_n|^2}}}=\sup_{x\in\ell^2,x\neq0}\frac{1}{\sqrt{\sum_{n\ge1}{(\frac{|x_n|}{\sup_{n\ge1}|x_n|})^2}}}\leq1$ , $i$ is bounded. Consider the sequence $(e_i)$ in $\ell^2$ , such that $e_i=(0,\cdots,1,\cdots)$ , $(e_i)$ is also a sequence in $\ell^\infty$ , and $\|e_i-e_j\|_{\ell^\infty}=\delta_{ij}$ shows that $(i(e_i))$ has no convergent subsequence in $\ell^\infty$ , hence $i$ is not compact.

(2) Let $E$ be a Banach space. Show that the identity operator $I : E \rightarrow E$ is compact if and only if $E$ is finite-dimensional.

$P\Leftarrow Q$ : Suppose that $E$ is finite-dimensional, which means there exists $(e_i)_{1\leq i\leq n}$ to be a basis of $E$ and $\|e_i\|=1$ . Then consider a bounded sequence $(u_i)=(\sum_{j=1}^n x_{i_j}e_j)$ in $E$ . Notice that $(u_i)$ is bounded shows that $(x_{i,j})_{i\ge1}$ is bounded. Therefore, there exists a subsequence $(u_i^{(1)})$ of $(u_i)$ such that $(x_{i,1}^{(1)})$ is convergent by the precompactness of a bounded set in $\R$ . So following the choosing strategy in 5.(3), we can get a subsequence $(u_i^{(n)})_{i\ge1}$ such that $(x_{i,j}^{(n)})_{i\ge1}$ is convergent for all $j$ . Therefore, $(u_i^{(n)})_{i\ge1}$ is convergent, hence $I$ is compact.
$\overline P\Leftarrow \overline Q$ : Suppose that $E$ is infinite-dimensional. Consider $u_0\in E$ , such that $\|u_0\|=1$ , and $U_0=\operatorname{Span}\{u_0\}$ . As $U_0\neq E$ and $U_0$ is closed, there exists $x_1\in E\setminus U_0$ and $d_1=d(x_1,U_0)>0$ . Choose a point $v_1\in U_0$ such that $d_1<\|x_1-v_1\|<2d_1$ , and let $u_1=\frac{x_1-v_1}{\|x_1-v_1\|}$ , we have $d(u_1,U_0)=\inf_{v\in U_0}\|u_1-v\|=\inf_{v\in U_0}\frac{1}{\|x_1-v_1\|}\|x_1-v_1-(\|x_1-v_1\|)v\|\ge\frac{1}{\|x_1-v_1\|}d(u_1,U_0)\ge\frac{d_1}{2d_1}=\frac{1}{2}$ . By this strategy, let $U_i=\operatorname{Span}\{u_1,\cdots,u_i\}$ , we can get a sequence $(u_i)$ in $E$ such that $\|u_i\|=1$ and $d(u_i,u_j)\ge\frac{1}{2}$ for all $i\neq j$ . Therefore, $I$ is not compact by this bounded sequence $(u_i)$ .
Q.E.D.

7. Denote $B_E = \{x \in E : \|x\| \leq 1\}$ to be the unit ball of a normed vector space $(E,\|\cdot\|)$ . Prove that $B_E$ is compact if and only if $E$ is of finite dimension. (Hints: Prove and use the following Riesz’s Lemma.)

Lemma(Riesz). Suppose $E_0 \subset E$ is a closed subspace with $E_0\neq E$ . Then for every $0 < \epsilon < 1$ , there exists $x_0 \in E,s.t.\enspace\|x_0\| = 1$ and $\|x_0 −x\| > \epsilon$ for all $x \in E_0$ .

As $E_0\neq E$ , there exists $y\in E\setminus E_0$ . As $E_0$ is closed, we get $d:=d(y,E_0)>0$ . Choose $z\in E_0$ , such that $d\leq \|y-z\|\leq\frac{d}{\epsilon}$ . Now let $x_0:=\frac{y-z}{\|y-z\|}$ . Notice that for all $x\in E_0$ , $\|x_0-x\|=\frac{1}{\|y-z\|}\|y-(z+\|y-z\|x)\|\ge\frac{1}{\|y-z\|}d(y,E_0)\ge\frac{\epsilon}{d}\cdot d=\epsilon$ and $\|x_0\|=1$ . Hence we have proved the Lemma.
$P\Leftarrow Q$ : Suppose that $E$ is finite-dimensional, which means there exists $(e_i)_{1\leq i\leq n}$ to be a basis of $E$ and $\|e_i\|=1$ . Then consider a sequence $(u_i)=(\sum_{j=1}^n x_{i_j}e_j)$ in $B_E$ . Notice that $(u_i)$ in $B_E$ shows that $(x_{i,j})_{i\ge1}$ is bounded. Therefore, there exists a subsequence $(u_i^{(1)})$ of $(u_i)$ such that $(x_{i,1}^{(1)})$ is convergent by the precompactness of a bounded set in $\R$ . So following the choosing strategy in 5.(3), we can get a subsequence $(u_i^{(n)})_{i\ge1}$ such that $(x_{i,j}^{(n)})_{i\ge1}$ is convergent for all $j$ . Therefore, $(u_i^{(n)})_{i\ge1}$ is convergent, hence $B_E$ is precompact. As $B_E$ is also closed, $B_E$ is compact.
$\overline P\Leftarrow \overline Q$ : Suppose that $E$ is infinite-dimensional. Consider $u_0\in E$ , such that $\|u_0\|=1$ , and $U_0=\operatorname{Span}\{u_0\}$ . As $U_0\neq E$ and $U_0$ is closed, there exists $u_1\in E,s.t.\enspace\|u_1\|=1$ and $d(u_1,U_0)>\frac{1}{2}$ . By this strategy, let $U_i=\operatorname{Span}\{u_1,\cdots,u_i\}$ , we can get a sequence $(u_i)$ in $E$ such that $\|u_i\|=1$ and $d(u_i,u_j)\ge\frac{1}{2}$ for all $i\neq j$ . Therefore, $B_E$ is not precompact by this sequence $(u_i)$ in $B_E$ , hence $B_E$ is not compact.
Q.E.D.

8. Let $H$ be a Hilbert space, and $K : H \rightarrow H$ be a compact operator. Suppose $x_n \rightharpoonup x_0$ and $y_n \rightharpoonup y_0$ as $n \rightarrow\infty$ , prove that $(x_n,Ky_n) \rightarrow (x_0,Ky_0)$ , as $n \rightarrow\infty$ .

As $\lim_{n\rightarrow\infty}(x,y_n)=(x,y_0)$ , $(y_n)$ must be bounded. Notice that $\lim_{n\rightarrow\infty}(x,Ky_{n})=\lim_{n\rightarrow\infty}(K^*x,y_{n})=(K^*x,y_0)=(x,Ky_0)$ , which means $Ky_n\rightharpoonup Ky_0$ , and $Ky_{n_i}\rightarrow Ky_0$ . Now assume that $K y_{n}$ is not strongly convergent to $Ky_0$ , which means there exists a subsequence $(Ky_{n}^{(1)}),s.t.\|Ky_{n}^{(1)}-Ky_0\|>\epsilon,\forall i$ . But as $(y_{n_i})$ is bounded and $K$ is compact, $(Ky_{n}^{(1)})$ must have a strongly convergent subsequence that converging to $z$ . Notice that $Ky_n^{(1)}\rightharpoonup Ky_0$ and $Ky_n^{(1)}\rightarrow z$ , we have $Ky_n^{(1)}\rightarrow Ky_0$ , which is a contradiction. Therefore, $Ky_n\rightarrow Ky_0$ . Now consider $|(x_n,Ky_n)-(x_0,Ky_0)|=|(x_n-x_0,Ky_n)+(x_0,Ky_n-Ky_0)|=|(x_n-x_0,Ky_n-Ky_0)+(x_n-x_0,Ky_0)+(x_0,Ky_n-Ky_0)|\leq\|x_n-x_0\|\|Ky_n-Ky_0\|+|(x_n-x_0,Ky_0)|+|(x_0,Ky_n-Ky_0)|\rightarrow0$ when $n\rightarrow\infty$ , we get $(x_n,Ky_n) \rightarrow (x_0,Ky_0)$ .

9. Let $H$ be a Hilbert space. For each $u \in H\setminus\{0\}$ , and $\alpha \in \R$ , define the hyperplane $M_{u,α}$ by $M_{u,α} = \{v \in H :(u,v) = \alpha\}$ . Suppose $K$ is a closed convex subset of $H$ , and $x_0 \notin K$ . Prove that (without the use of Hahn-Banach Theorem) there is a hyperplane $M_{u,\alpha}$ strictly separates $K$ and $\{x_0\}$ : $\exists\epsilon > 0$ , such that $(u, x_0) > \alpha+\epsilon$ , and $(u,v) < \alpha−\epsilon$ , $\forall v \in K$ . (Hint: Use the projection mapping to find a “normal vector”.)

As $x_0\notin K$ and $K$ is closed, let $x_1=\operatorname{Proj}_K(x_0),s.t.\enspace\|x_0-x_1\|=\inf_{v\in K}\|x_0-v\|$ , $u:=x_0-x_1\neq0$ and $\alpha_0=(u,x_1)$ . Therefore, $(u,x_0)-\alpha_0=\|u\|^2>0$ . Consider for all $v\in K$ , and a real function $f(t)=\|x_0-x_1-t(v-x_1)\|^2=\|u\|^2-2t(u,v-x_1)+t^2\|v-x_1\|^2$ . As we know that $\|x_0-x_1\|=\inf_{v\in K}\|x_0-v\|$ , we have $f'(0)=0$ , which means $(u,v-x_1)\leq0$ . Therefore, $(u,v)-\alpha_0=(u,v-x_1)\leq0$ . So, we have $(u,x_0)>\alpha_0$ and $(u,v)\leq\alpha_0$ . Now let $\alpha=\frac{1}{2}((u,x_0)+\alpha_0)>\alpha_0$ , $0<\epsilon<\min\{\alpha-\alpha_0,(u,x_0)-\alpha\}$ , we can get $(u,x_0)>\alpha+\epsilon$ and $(u,v)\leq\alpha_0<\alpha-\epsilon$ , which means $M_{u,\alpha}$ strictly separates $K$ and $\{x_0\}$ .

10. Suppose $A[a,b]$ is the space of absolutely continuous functions on $[a,b]$ . Define $\|f\| = \|f\|_1 +\|f'\|_1$ , where $f \in A[a,b]$ and $f'$ denotes its derivative. Prove that $(A[a,b],\| \cdot \|)$ is a separable Banach space.

It's obvious that $A[a,b]$ is a vector space. Let $f,g\in A[a,b]$ , $\lambda\in\R$ .
Triangle Inequality: $\|f+g\|=\|f+g\|_1+\|f'+g'\|_1=\|f\|_1+\|f'\|_1+\|g\|_1+\|g'\|_1=\|f\|+\|g\|$ .
Positive Homogeneity: $\|\lambda f\|=\|\lambda f\|_1+\|\lambda f'\|_1=|\lambda|(\|f\|_1+\|f'\|_1)=|\lambda|\|f\|$ .
Positive Definiteness: $\|f\|=\|f\|_1+\|f'\|_1=\int_a^b|f(x)|\mathrm d x+\int_a^b|f'(x)|\mathrm dx>0$ if $f\neq 0$ .
Therefore, $(A[a,b],\|\cdot\|)$ is a n.v.s.
Now consider a Cauchy sequence $(f_i)$ in $A[a,b]$ . As $A[a,b]\subset L^1([a,b])$ and $\|f\|\rightarrow0\Rightarrow\|f\|_1\rightarrow0,\|f'\|_1\rightarrow0$ , $(f_i)$ and $(f'_i)$ are also Cauchy sequences in $L^1[a,b]$ . Notice that $L^1[a,b]$ is complete, let $f=\lim_{k\rightarrow\infty}f_k,g=\lim_{k\rightarrow\infty}f'_k$ . Now we consider $f_0(x):=f(a)+\int_a^xg(t)\mathrm dt$ is an absolutely continuous function in $A[a,b]$ , by the theorm in Real Analysis. As $\|f_k\|_1\rightarrow \|f_0\|_1$ and $\|f_k'\|_1\rightarrow\|f_0'\|_1$ , we have $f_k\rightarrow f_0$ in $(A[a,b],\|\cdot\|)$ . Therefore, $(A[a,b],\| \cdot \|)$ is complete, hence a Banach space.
Following the instruction in *Remark 10*, consider simple functions of the form $g_i=\sum_{k=1}^m q_k\chi_{R_k}$ , where $R_k =\prod_{i=1}^n(a_{i_k},b_{i_k})$ and $q_k,a_{i_k},b_{i_k}\in \mathbb Q$ . By the theorms of measure theory in Real Analysis, we know that $\{g_i\}$ is dense and countable in $L^1[a,b]$ . Now consider that $A[a,b]$ can be defined as $\{f\in C[a,b]|\exists f'\in L^1[a,b],f(x)=f(a)+\int_a^xf'(t)\mathrm dt\}$ . Therefore, let $\mathcal D=\{f_n(x)=c_n+\int_a^xg_n(t)\mathrm dt|c_n\in \mathbb Q\}$ . As $\forall f\in A[a,b],\exists c_n\in\mathbb Q,g_n,s.t.\enspace c_n\rightarrow f(a),g_n\rightarrow f'$ , we have $f_n\rightarrow f$ , hence $\mathcal D$ is countable and dense in $A[a,b]$ and $(A[a,b],\| \cdot \|)$ is separable.
Finally, we get $(A[a,b],\| \cdot \|)$ is a separable Banach space.

11. Prove the Arzela-Ascoli Theorem: Let $X$ be a compact metric space. If a sequence $(f_n)$ in $C(X)$ is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.

First, we know that $(f_n)$ is uniformly bounded means that $\exists M>0,s.t.\enspace\forall n\in\N,x\in X,|f_n(x)|\leq M$ , and equicontinuous means that $\forall\epsilon>0\exists\delta>0,s.t.\enspace\forall f_n,d(x,y)<\delta\Rightarrow|f_n(x)-f_n(y)|<\epsilon$ . As $X$ is compact, we know it's totally bounded, hence there exists a countable dense subset $D=\{x_k\}$ . For a point $x_k$ , let's consider $(f_n(x_m))_{n\ge1}$ , which is bounded and hence precompact. Therefore, there exists $(f_n^{(k)})_{n\ge1}$ such that $(f_n^{(k)}(x_k))_{n\ge1}$ is convergent. Let $f_n^{(k+1)}$ choose from $f_n^{(k)}$ , and $g_m=f_m^{(m)}$ , we have that $(g_m)$ is uniformly bounded, equicontinuous and convergent for all $x_k$ . Then, for all $\epsilon>0$ , there exists $\delta>0,s.t.\enspace d(x,y)<\delta\Rightarrow|g_k(x)-g_k(y)|<\epsilon/3$ (by equicontinuous), a finite group $U=\{x_k\}$ such that $\forall x\in X,\exists x_k\in U,s.t.\enspace d(x,x_k)<\delta$ (by $X$ is totally bounded), and $\exists N_0\in\mathbb N^*,s.t.\enspace\forall m,n>N_0,x_k\in U,|g_m(x_k)-g_n(x_k)|<\epsilon/3$ (by $(g_m(x_k))_{m\ge 1}$ is convergent and $k$ is finite). Now we have $|g_m(x)-g_n(x)|\leq|g_m(x)-g_m(x_k)|+|g_m(x_k)-g_n(x_k)|+|g_n(x_k)-g_n(x)|\leq\epsilon$ , which means $g_k$ is uniformly Cauchy sequence, hence a uniformly convergent subsequence.

12. Let $K : [a,b]^2 \rightarrow \R$ be continuous, where $a < b$ . Define an operator $T : L^1 \rightarrow L^1$ as follows: $Tf(x)=\int_a^bK(x,y)f(y)\mathrm dy$ . Show that $T$ is a compact operator. (Hint: use the Arzela-Ascoli Theorem.)

Consider a bounded sequence $(f_n)$ in $L^1$ , we have $\exists M>0,s.t.\enspace\forall f_n,\|f_n\|<M$ . Notice that $K$ is continuous on a close and compact set, hence $K$ must be bounded and uniformly continuous.Therefore, $\|Tf_k\|=\int_a^b\int_a^bK(x,y)f_k(y)\mathrm dy\mathrm dx\leq\int_{a}^b\int_a^bK(x,y) M\mathrm dx\mathrm dy\leq(b-a)^2M\sup K$ , which means $(Tf_n)$ is uniformly bounded. As $K$ is uniformly countinuous , we have $\forall \epsilon>0,\exists\delta>0,s.t.\enspace \|x_1-x_2\|<\delta,\|y_1-y_2\|<\delta\Rightarrow|K(x_1,y_1)-K(x_2,y_2)|<\epsilon$ . So, $|Tf_n(x_1)-Tf_n(x_2)|=|\int_a^b (K(x_1,y)-K(x_2,y))f(y)\mathrm dy|\leq|\int_a^b\epsilon M\mathrm dy|=(b-a)M\epsilon$ , which means $(Tf_n)$ is equicontinuous. Therefore, according to Arzela-Ascoli Theorem, we have $(Tf_n)$ has a uniformly convergent subsequence, hence $T$ is compact.

13. Let $E$ be a Banach space, and let $F \subset E$ be a closed subspace. Prove that the quotient space $E/F$ , equipped with the norm $\|u +F\|:= \inf_{v\in F}\|u +v\|_E$ , is also a Banach space.

It's obvious that $E/F$ is a vector space, let $u+F,u_1+F,u_2+F\in E/F,\lambda\in\R$ .
Triangle Inequality: $\|u_1+F+u_2+F\|=\|u_1+u_2+F\|=\inf_{v_1,v_2\in F}\|u_1+u_2+v_1+v_2\|_E\leq\inf_{v_1,v_2\in F}(\|u_1+v_1\|_E+\|u_2+v_2\|_E)=\|u_1+F\|+\|u_2+F\|$ .
Positive Homogeneity: $\|\lambda (u+F)\|=\|\lambda u+F\|=\inf_{v\in F}\|\lambda u+v\|_E=\inf_{v\in F}\|\lambda u+\lambda v\|_E=|\lambda|\inf_{v\in F}\|u+v\|_E=|\lambda|\|u+F\|$ .
Positive Definiteness: $\|u+F\|=\inf_{v\in F}\|u+v\|_E>0$ if $u+F\neq 0+F$ .
Therefore, $(E/F,\|\cdot\|)$ is a n.v.s.
Now consider a Cauchy sequence $(u_n+F)$ in $E/F$ , which means $\forall\epsilon>0\exists N_0\in\mathbb N^*,s.t\enspace m,n>N_0\Rightarrow\|(u_n+F)-(u_m+F)\|<\epsilon$ . As $\|(u_n+F)-(u_m+F)\|=\inf_{v\in F}\|u_n-u_m+v\|_E<\epsilon$ , let $x_1=u_1$ and $x_n=u_n+v_n$ defined by $\exists v_{n}\in F,s.t.\enspace\|x_n-x_{n-1}\|_E=\|u_n-u_{n-1}+v_n-v_{n-1}\|_E<\epsilon+\frac{1}{n}$ . Therefore, $(x_n)$ is a Cauchy sequence in $E$ and it converges to $x\in E$ . Notice that $\|(u_k+F)-(x+F)\|=\|(x_k+F)-(x+F)\|=\inf_{v\in F}\|x_k-x+v\|_E\rightarrow0$ as $0\in F$ and $x_k\rightarrow x$ . So, $u_k+F\rightarrow x+F\in E/F$ , hence $(E/F,\|\cdot\|)$ is a Banach space.

14. Let $E$ be an n.v.s. Let $V \subset E$ be a closed subspace, and let $W \subset E$ be a finite-dimensional subspace. Prove that $V + W$ is closed.

Consider a map $\varphi:E\rightarrow E/V,x\mapsto x+V$ . According to 13, we know that $E/V$ is a n.v.s. Notice that $\|\varphi\|=\sup_{\|x\|=1}\|\varphi(x)\|=\sup_{\|x\|=1}\inf_{v\in V}\|x+v\|\leq\sup_{\|x\|=1}\|x\|=1$ , which means $\varphi$ is continuous. As $V+W=\varphi^{-1}\circ\varphi(W)$ , and $W$ is finite-demensional gives that $W$ is closed, we have $V+W$ is closed.

15. Let $E$ be a n.v.s. Let $L \subset E$ be a closed subspace, and let $M \subset E$ be a finite-dimensional subspace, such that $E = L\oplus M$ . Suppose $T : E \rightarrow E$ is a bounded linear map, and denote $N$ to be the kernel of $T$ . Prove that $L+N$ is closed.

Consider a map $\varphi:E\rightarrow E/L,x\mapsto x+L$ . According to 14, we have $E/L$ is a n.v.s, and $\varphi$ is continuous. As $T$ is a bounded linear map, $N=\ker(T)$ is a subspace of $E$ . Notice that $E/L\cong M$ , by $E=L\oplus M$ and a bijective linear map $\sigma:M\rightarrow E/L,m\mapsto m+L;\sigma(m_1)=\sigma(m_2)\Rightarrow m_1-m_2\in L\Rightarrow m_1=m_2;P_E(x)\in E,\sigma(P_E(x))=P_E(x)+L=x+L$ . Therefore $\varphi(N)\subseteq E/L\cong M$ must be a finite-dimensional subspace, hence $\varphi (N)$ is closed. So, $L+N=\varphi^{-1}(\varphi(N))$ is closed.

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Functional Analysis Assignment NO.2

1. Let $(X,d)$ be a complete metric space, and let $K\subset X$ be precompact. Prove that $K$ is separable.

2. Denote $\R^n =\{(x^1, \dots ,x^n)^T : x^k \in \R\}$ .

(i) Let $M$ be a symmetric positive definite $n\times n$ matrix. Prove that $(x,y) := x^TMy$ , $x,y \in \R^n$ , defines an inner product.

(ii) Suppose $(\cdot,\cdot)$ is an inner product on $\R^n$ . Prove that there is a unique symmetric positive definite $n\times n$ matrix $M$ , such that $(x, y) = x^TMy$ , $x,y \in \R^n$ .

4. Let $E$ be an n.v.s., and $F$ be a Banach space. Show that $\mathcal L(E,F)$ is a Banach space.

5. Let $E,F,G$ be Banach spaces.

(2) If $A \in \mathcal L(E,F)$ and $B\in\mathcal L(F,G)$ , and at least one of $A$ or $B$ is compact, then $B \circ A$ is compact.

(3) Let $(T_n)$ be a sequence in $\mathcal K(E,F)$ that converges to $T \in\mathcal L(E,F)$ . Prove that $T$ is compact.

6.

(1) The inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is defined by $i(x) = x$ for all $x \in\ell^2$ . Show that the inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is bounded but not compact.

(2) Let $E$ be a Banach space. Show that the identity operator $I : E \rightarrow E$ is compact if and only if $E$ is finite-dimensional.

7. Denote $B_E = \{x \in E : \|x\| \leq 1\}$ to be the unit ball of a normed vector space $(E,\|\cdot\|)$ . Prove that $B_E$ is compact if and only if $E$ is of finite dimension. (Hints: Prove and use the following Riesz’s Lemma.)

Lemma(Riesz). Suppose $E_0 \subset E$ is a closed subspace with $E_0\neq E$ . Then for every $0 < \epsilon < 1$ , there exists $x_0 \in E,s.t.\enspace\|x_0\| = 1$ and $\|x_0 −x\| > \epsilon$ for all $x \in E_0$ .

8. Let $H$ be a Hilbert space, and $K : H \rightarrow H$ be a compact operator. Suppose $x_n \rightharpoonup x_0$ and $y_n \rightharpoonup y_0$ as $n \rightarrow\infty$ , prove that $(x_n,Ky_n) \rightarrow (x_0,Ky_0)$ , as $n \rightarrow\infty$ .

10. Suppose $A[a,b]$ is the space of absolutely continuous functions on $[a,b]$ . Define $\|f\| = \|f\|_1 +\|f'\|_1$ , where $f \in A[a,b]$ and $f'$ denotes its derivative. Prove that $(A[a,b],\| \cdot \|)$ is a separable Banach space.

11. Prove the Arzela-Ascoli Theorem: Let $X$ be a compact metric space. If a sequence $(f_n)$ in $C(X)$ is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.

12. Let $K : [a,b]^2 \rightarrow \R$ be continuous, where $a < b$ . Define an operator $T : L^1 \rightarrow L^1$ as follows: $Tf(x)=\int_a^bK(x,y)f(y)\mathrm dy$ . Show that $T$ is a compact operator. (Hint: use the Arzela-Ascoli Theorem.)

13. Let $E$ be a Banach space, and let $F \subset E$ be a closed subspace. Prove that the quotient space $E/F$ , equipped with the norm $\|u +F\|:= \inf_{v\in F}\|u +v\|_E$ , is also a Banach space.

14. Let $E$ be an n.v.s. Let $V \subset E$ be a closed subspace, and let $W \subset E$ be a finite-dimensional subspace. Prove that $V + W$ is closed.

15. Let $E$ be a n.v.s. Let $L \subset E$ be a closed subspace, and let $M \subset E$ be a finite-dimensional subspace, such that $E = L\oplus M$ . Suppose $T : E \rightarrow E$ is a bounded linear map, and denote $N$ to be the kernel of $T$ . Prove that $L+N$ is closed.

Functional Analysis Assignment NO.1

Functional Analysis Assignment NO.3

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1. Let (X,d) be a complete metric space, and let K\subset X be precompact. Prove that K is separable.

2. Denote \R^n =\{(x^1, \dots ,x^n)^T : x^k \in \R\}.

(i) Let M be a symmetric positive definite n\times n matrix. Prove that (x,y) := x^TMy, x,y \in \R^n, defines an inner product.

(ii) Suppose (\cdot,\cdot) is an inner product on \R^n. Prove that there is a unique symmetric positive definite n\times n matrix M, such that (x, y) = x^TMy, x,y \in \R^n.

4. Let E be an n.v.s., and F be a Banach space. Show that \mathcal L(E,F) is a Banach space.

5. Let E,F,G be Banach spaces.

(1) Let A \in \mathcal L(E,F). A \in \mathcal K(E,F) is equivalent to that \overline{A(D)} is compact for any bounded set D\subset E. In other words, A \in\mathcal K(E,F) is equivalent to that the sequence (Ax_n) in F has a convergent subsequence for every bounded sequence (x_n) in E.

(2) If A \in \mathcal L(E,F) and B\in\mathcal L(F,G), and at least one of A or B is compact, then B \circ A is compact.

(3) Let (T_n) be a sequence in \mathcal K(E,F) that converges to T \in\mathcal L(E,F). Prove that T is compact.

6.

(1) The inclusion map i : \ell^2 \rightarrow \ell^\infty is defined by i(x) = x for all x \in\ell^2. Show that the inclusion map i : \ell^2 \rightarrow \ell^\infty is bounded but not compact.

(2) Let E be a Banach space. Show that the identity operator I : E \rightarrow E is compact if and only if E is finite-dimensional.

7. Denote B_E = \{x \in E : \|x\| \leq 1\} to be the unit ball of a normed vector space (E,\|\cdot\|). Prove that B_E is compact if and only if E is of finite dimension. (Hints: Prove and use the following Riesz’s Lemma.)

Lemma(Riesz). Suppose E_0 \subset E is a closed subspace with E_0\neq E. Then for every 0 < \epsilon < 1, there exists x_0 \in E,s.t.\enspace\|x_0\| = 1 and \|x_0 −x\| > \epsilon for all x \in E_0.

8. Let H be a Hilbert space, and K : H \rightarrow H be a compact operator. Suppose x_n \rightharpoonup x_0 and y_n \rightharpoonup y_0 as n \rightarrow\infty, prove that (x_n,Ky_n) \rightarrow (x_0,Ky_0), as n \rightarrow\infty.

10. Suppose A[a,b] is the space of absolutely continuous functions on [a,b]. Define \|f\| = \|f\|_1 +\|f'\|_1, where f \in A[a,b] and f' denotes its derivative. Prove that (A[a,b],\| \cdot \|) is a separable Banach space.

11. Prove the Arzela-Ascoli Theorem: Let X be a compact metric space. If a sequence (f_n) in C(X) is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.

12. Let K : [a,b]^2 \rightarrow \R be continuous, where a < b. Define an operator T : L^1 \rightarrow L^1 as follows: Tf(x)=\int_a^bK(x,y)f(y)\mathrm dy. Show that T is a compact operator. (Hint: use the Arzela-Ascoli Theorem.)

13. Let E be a Banach space, and let F \subset E be a closed subspace. Prove that the quotient space E/F, equipped with the norm \|u +F\|:= \inf_{v\in F}\|u +v\|_E, is also a Banach space.

14. Let E be an n.v.s. Let V \subset E be a closed subspace, and let W \subset E be a finite-dimensional subspace. Prove that V + W is closed.

15. Let E be a n.v.s. Let L \subset E be a closed subspace, and let M \subset E be a finite-dimensional subspace, such that E = L\oplus M. Suppose T : E \rightarrow E is a bounded linear map, and denote N to be the kernel of T. Prove that L+N is closed.

Functional Analysis Assignment NO.1

Functional Analysis Assignment NO.3

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1. Let $(X,d)$ be a complete metric space, and let $K\subset X$ be precompact. Prove that $K$ is separable.

2. Denote $\R^n =\{(x^1, \dots ,x^n)^T : x^k \in \R\}$ .

(i) Let $M$ be a symmetric positive definite $n\times n$ matrix. Prove that $(x,y) := x^TMy$ , $x,y \in \R^n$ , defines an inner product.

(ii) Suppose $(\cdot,\cdot)$ is an inner product on $\R^n$ . Prove that there is a unique symmetric positive definite $n\times n$ matrix $M$ , such that $(x, y) = x^TMy$ , $x,y \in \R^n$ .

4. Let $E$ be an n.v.s., and $F$ be a Banach space. Show that $\mathcal L(E,F)$ is a Banach space.

5. Let $E,F,G$ be Banach spaces.

(2) If $A \in \mathcal L(E,F)$ and $B\in\mathcal L(F,G)$ , and at least one of $A$ or $B$ is compact, then $B \circ A$ is compact.

(3) Let $(T_n)$ be a sequence in $\mathcal K(E,F)$ that converges to $T \in\mathcal L(E,F)$ . Prove that $T$ is compact.

(1) The inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is defined by $i(x) = x$ for all $x \in\ell^2$ . Show that the inclusion map $i : \ell^2 \rightarrow \ell^\infty$ is bounded but not compact.

(2) Let $E$ be a Banach space. Show that the identity operator $I : E \rightarrow E$ is compact if and only if $E$ is finite-dimensional.

7. Denote $B_E = \{x \in E : \|x\| \leq 1\}$ to be the unit ball of a normed vector space $(E,\|\cdot\|)$ . Prove that $B_E$ is compact if and only if $E$ is of finite dimension. (Hints: Prove and use the following Riesz’s Lemma.)

Lemma(Riesz). Suppose $E_0 \subset E$ is a closed subspace with $E_0\neq E$ . Then for every $0 < \epsilon < 1$ , there exists $x_0 \in E,s.t.\enspace\|x_0\| = 1$ and $\|x_0 −x\| > \epsilon$ for all $x \in E_0$ .

8. Let $H$ be a Hilbert space, and $K : H \rightarrow H$ be a compact operator. Suppose $x_n \rightharpoonup x_0$ and $y_n \rightharpoonup y_0$ as $n \rightarrow\infty$ , prove that $(x_n,Ky_n) \rightarrow (x_0,Ky_0)$ , as $n \rightarrow\infty$ .

10. Suppose $A[a,b]$ is the space of absolutely continuous functions on $[a,b]$ . Define $\|f\| = \|f\|_1 +\|f'\|_1$ , where $f \in A[a,b]$ and $f'$ denotes its derivative. Prove that $(A[a,b],\| \cdot \|)$ is a separable Banach space.

11. Prove the Arzela-Ascoli Theorem: Let $X$ be a compact metric space. If a sequence $(f_n)$ in $C(X)$ is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.

12. Let $K : [a,b]^2 \rightarrow \R$ be continuous, where $a < b$ . Define an operator $T : L^1 \rightarrow L^1$ as follows: $Tf(x)=\int_a^bK(x,y)f(y)\mathrm dy$ . Show that $T$ is a compact operator. (Hint: use the Arzela-Ascoli Theorem.)

13. Let $E$ be a Banach space, and let $F \subset E$ be a closed subspace. Prove that the quotient space $E/F$ , equipped with the norm $\|u +F\|:= \inf_{v\in F}\|u +v\|_E$ , is also a Banach space.

14. Let $E$ be an n.v.s. Let $V \subset E$ be a closed subspace, and let $W \subset E$ be a finite-dimensional subspace. Prove that $V + W$ is closed.

15. Let $E$ be a n.v.s. Let $L \subset E$ be a closed subspace, and let $M \subset E$ be a finite-dimensional subspace, such that $E = L\oplus M$ . Suppose $T : E \rightarrow E$ is a bounded linear map, and denote $N$ to be the kernel of $T$ . Prove that $L+N$ is closed.