1. Let (X,d) be a metric space. Let x\in X, A\subset X, and r>0.

(1) Prove that B(x,r)=\{z\in E:d(z,x)<r\} is open.

  • \forall y\in B(x,r), let r'=r-d(x,y). Notice that \forall z\in B(y,r'), we have d(x,z)\leq d(x,y)+d(y,z)<d(x,y)+r'=r, which means that z\in B(x,r). Therefore \forall y\in B(x,r),\exist r'>0,s.t.\enspace B(y,r')\subset B(x,r). So B(x,r) is open.

(2) Prove that \operatorname{int} A is open and A is open iff A=\operatorname{int}A.

  • Suppose that \operatorname{int}A\neq\emptyset, otherwise it's trivial that \operatorname{int} A is open. According to the defination of \operatorname{int}A, we know \forall x\in\operatorname{int}A,\exists r>0,s.t.\enspace B(x,r)\subset A. Now consider y\in B(x,r) and let r'=r-d(x,y). As B(y,r')\subset B(x,r)\subset A, we have y\in\operatorname{int}A. Therefore B(x,r)\subset \operatorname{int}A, which means \operatorname{int}A is open.
  • \Rightarrow: Suppose that A is open. Therefore \forall x\in A, we know x\in\operatorname{int}A, which means A\subseteq\operatorname{int} A. Consider by the defination, we also have \operatorname{int} A\subseteq A. So, A=\operatorname{int}A.
    \Leftarrow: Suppose that A=\operatorname{int}A, which means that \forall x\in A,\exists r>0,s.t.\enspace B(x,r)\subset A. It's clear that A is open.
    Q.E.D.

(3) Prove that A is closed iff A has the following property: If (x_k)_{k\ge1} is a sequence in A satisfying \lim_{k\rightarrow\infty}x_k=\overline x\in X, then \overline x\in A.

  • \Rightarrow: As A is closed, A^c is open. Suppose that \overline x\in A^c. Then \exists r>0,s.t.\enspace B(\overline x,r)\subset A^c. As \lim_{k\rightarrow\infty}d(x_k,\overline x)=0, we get \exists x_{k_0},d(x_{k_0},\overline x)<r, which means x_{k_0}\in A^c. It's a contradiction that we also have x_k\in A. Therefore the supposition is false, and \overline x\in A.
    \Leftarrow: Suppose that x\in A^c\setminus\operatorname{int}A^c. So x\in A^c and x\notin \operatorname{int}A^c, which means that \forall r>0,\exists x_r\in B(x,r),s.t.\enspace x_r\notin A^c. Let r=\frac{1}{k}, we get x_k\in B(x,\frac{1}{k})\Rightarrow \lim_{k\rightarrow\infty}x_k=x, x_k\in A. So we should have x\in A and x\in A^c. It's a contradiction, which means x doesn't exist. Therefore A^c=\operatorname{int}A^c\Rightarrow A^c is open \Rightarrow A is closed.
    Q.E.D.

(4) Let (x_k)_{k\ge1} be a sequence converging to some point x\in X. Prove that the following statements are true.

(i) (x_k)_{k\ge1} is a Cauchy sequence.

  • As \lim_{k\rightarrow\infty}d(x_k,x)=0, we have \forall\epsilon >0,\exists N\in \N^*,s.t.\enspace\forall n>N,d(x_n,x)<\frac{\epsilon}{2}. So \forall i,j>N,d(x_i,x_j)\leq d(x_i,x)+d(x_j,x)<\epsilon, which means (x_k)_{k\ge 1} is a Cauchy sequence.

(ii) (x_k)_{k\ge1} can not converge to points other than x.

  • If \exists x'\in X,s.t.\enspace \lim_{k\rightarrow\infty}d(x_k,x')=0, consider 0\leq d(x,x')\leq d(x,x_k)+d(x_k,x')\leq\lim_{k\rightarrow\infty}(d(x,x_k)+d(x_k,x'))=0. So x'=x.

2. Let (X,d) be a metric space, and A\subset X. We say x is an accumulation point (or a limit point) of A, if x can be written as x = \lim_{k\rightarrow\infty}x_k, where x_k\in A are distinct points. The set of accumulation points of A is denoted by A', and we call A' the derived set of A.

(1) Prove that A is closed iff A'\subset A.

  • \Rightarrow: According to 1.(3), we have \forall x_k\in A(,s.t.\enspace x=\lim_{k\rightarrow\infty}x_k\in X), x\in A. As \forall x\in A', x can be written as x = \lim_{k\rightarrow\infty}x_k, where x_k\in A are distinct points. Therefore x\in A, which means A'\subset A.
    \Leftarrow: Let (x_k)_{k\ge1} be a sequence in A satisfying \lim_{k\rightarrow\infty}x_k=\overline x\in X. Consider if there exists (x_{k_l})_{l\ge 1},x_{k_l}=x,\forall l, then x must in A. Otherwise if there exists finate same points x_t\neq x, we remove other x_t from (x_k) in order to only left one x_0. So by this new sequence x\in A'\subset A. According to 1.(3), A must be closed.

(2) Prove that A' is closed, and A\cup A' is closed.

  • (Let i\neq j.) Consider x\in (A')', \exists x_k\in A',x_i\neq x_j,s.t.\enspace x=\lim_{k\rightarrow\infty}x_k\in X. As x_k\in A'\Rightarrow \exists x_{k_l}\in A,x_k=\lim_{l\rightarrow\infty}x_{k_l},x_{k_i}\neq x_{k_j}, so now consider (x_{k_k}). As x_i\neq x_j, we can get that when k increases, x_{i_i}\neq x_{j_j},x=\lim_{k\rightarrow\infty}x_{k_k}. So, x\in A', which means (A')'\subset A'. According to (1), A' is closed.
  • (Let i\neq j.) Consider x\in (A\cup A')', \exists x_k\in A\cup A',x_i\neq x_j,s.t.\enspace x=\lim_{k\rightarrow\infty}x_k\in X. As x_k\in A\cup A'\Rightarrow \exists x_{k_l}\in A,x_k=\lim_{l\rightarrow\infty}x_{k_l}, so now consider (x_{k_k}). As x_i\neq x_j, we can get that when k increases, x_{i_i}\neq x_{j_j},x=\lim_{k\rightarrow\infty}x_{k_k}. So, x\in A', which means (A\cup A')'\subset A'\subset A\cup A'. According to (1), A' is closed.

(3) Show that A\cup A' is the smallest closed set containing A.

  • Consider A\subset B\subsetneq A\cup A'. So there exists x_0\in (A\cup A')\setminus B. As if x_0\in A, x_0 must in B, we get x_0\in A'\setminus B. Notice that x_0\in A' means that exists distinct points x_k\in A\subset B,s.t.\enspace x=\lim_{k\rightarrow\infty}x_k. According to 1.(3), by this sequence x\notin B proves that B is not closed. Therefore, A\cup A' is the smallest closed set containing A.

(4) Show that A\cup A' =\overline A, where \overline A is the closure of A defined by \overline A=(\operatorname{int}A^c)^c.

  • According to (3), A\subset\overline A, and \overline A is a closed set, we get A\cup A'\subset\overline A. Now suppose that there exists x_0\in \overline A\setminus(A\cup A'). As x_0\in (A\cup A')^c, which is an open set, we know that \exists r>0,s.t.\enspace B(x_0,r)\subset (A\cup A')^c\subset A^c. Notice that it means x_0\in\operatorname {int} A^c, and x_0\notin\overline A, which is a contradiction. So \overline A\setminus(A\cup A')=\emptyset. Therefore, A\cup A' =\overline A.

3. Let (E,\| \cdot\| ) be a normed vector space. Suppose that B is the unit ball in E and B^* is the unit ball in E^*, namely B :=\{x \in E :\|x\|_E \leq1\}, B^* :=\{f\in E^* :\|f\|_{E^*}\leq1\}. We define the polar set of a set A \subset E as follows: A^\circ := \{f \in E^*:|\langle f,x\rangle|\leq 1, \forall x \in A\}. Prove that :

(1) B is convex and B = -B.

  • Let x,y\in B,\lambda\in[0,1]. Consider that \|\lambda x+(1-\lambda) y\|\leq\lambda\|x\|+(1-\lambda)\|y\|\leq 1\Rightarrow \lambda x+(1-\lambda) y\in B. So B is convex.
  • Let x\in B. Notice that \|-x\|=\|x\|\leq 1\Rightarrow-x\in B. Therefore B=-B.

(2) B^* = B^\circ.

  • Notice that \|f\|_{E^*}=\sup_{x\in E,\|x\|\leq 1}f(x)=\sup_{x\in B}f(x)=\sup_{x\in B=-B}-f(x)=\sup_{x\in B}|f(x)|. Therefore, \|f\|_{E^*}\leq1\Leftrightarrow\sup_{x\in B}|f(x)|\leq 1\Leftrightarrow|f(x)|\leq 1,\forall x\in B, which means B^* = B^\circ.

4. Suppose that K\subset (\R^n,\|\cdot\|) is a convex, bounded and closed set with o \in \operatorname{int}K and K = −K. We define \|x\|_K = \inf\{\lambda > 0 : x \in λK\}. Prove that \|x\|_K is a norm on \R^n.

  • Let x,y\in \R^n,\alpha\in\R.
    Triangle Inequality: \|x+y\|_K=\inf\{\lambda>0:x+y\in\lambda K\}\leq\inf\{\lambda>0:x\in\lambda K\}+\inf\{\lambda>0:y\in\lambda K\}=\|x\|_K+\|y\|_K. (x+y\in\lambda K,x\in\lambda_1K,y\in\lambda_2K\Rightarrow x+y\in(\lambda_1+\lambda_2)K\Rightarrow\inf\lambda\leq\inf(\lambda_1+\lambda_2)\leq\inf\lambda_1+\inf\lambda_2).
    Positive Homogeneity: \|\alpha x\|_K=\inf\{\lambda>0:\alpha x\in\lambda K\}=\inf\{|\alpha|\lambda>0:\alpha x\in|\alpha|\lambda K\}=|\alpha|\inf\{\lambda>0:\alpha x\in|\alpha|\lambda K\}\xlongequal{x\in\lambda K\Leftrightarrow\alpha x\in|\alpha|\lambda K}|\alpha|\inf\{\lambda>0:x\in\lambda K\}=|\alpha|\|x\|_K
    Positive Definiteness: \|x\|_K=\inf\{\lambda > 0 : x \in λK\}>0 if x\neq 0.
    Q.E.D.

5. Let \Omega\subset\R^n be an open bounded subset. C(\overline\Omega) := \{uniformly continuous functions f : \Omega \rightarrow\R\}. Show that C(\overline\Omega,\|\cdot\|_1) and C(\overline \Omega,\| \cdot \|_\infty) are all normed vector space.

  • It's obvious that C(\overline\Omega) is a vector space. Let f,g\in C(\overline\Omega),\alpha\in\R.
    Triangle Inequality: \|f+g\|_1=\int_{\Omega}|(f+g)(x)|\mathrm dx\leq\int_{\Omega}|f(x)|\mathrm dx+\int_{\Omega}|g(x)|\mathrm dx=\|f\|_1+\|g\|_1; \|f+g\|_\infty=\sup_{x\in\Omega}|(f+g)(x)|\leq\sup_{x\in\Omega}(|f(x)|+|g(x)|)=\sup_{x\in\Omega}|f(x)|+\sup_{x\in\Omega}|g(x)|=\|f\|_\infty+\|g\|_\infty.
    Positive Homogeneity: \|\alpha f\|_1=\int_{\Omega}|\alpha f(x)|\mathrm dx=|\alpha|\int_{\Omega}|f(x)|\mathrm dx=|\alpha|\|f\|_1; \|\alpha f\|_\infty=\sup_{x\in\Omega}|\alpha f(x)|=|\alpha|\sup_{x\in\Omega}|f(x)|=|\alpha|\|f\|_\infty.
    Positive Definiteness: \|f\|_1=\int_{\Omega}|f(x)|\mathrm dx>0 if f\neq 0; \|f\|_\infty=\sup_{x\in\Omega}|f(x)|>0 if f\neq0.
    Q.E.D.

6. Let (X,d) be a metric space. Prove that: if X is compact, X must be complete. Furthermore, the reverse is not necessarily true, please give an example.

  • Consider a cauchy sequence (x_k) in X. As X is compact, we know that (x_k) has a convergent subsequence (x_{k_l}),s.t.\enspace \lim_{l\rightarrow\infty}x_{k_l}=x_0\in X. Notice that \forall\epsilon>0,\exists N_0\in\N^*,\forall i,j>N_0,d(x_i,x_j)<\epsilon/2;\exists N_1\in\N^*,\forall l>N_1,d(x_{k_l},x_0)<\epsilon/2, therefore \forall k>N_0,\exists l>N_1,d(x_k,x_0)\leq d(x_k,x_{k_{l}})+d(x_{k_{l}},x_0)=\epsilon, which means x_0=\lim_{k\rightarrow\infty}x_k. So, X is complete.
  • Consider \R is a complete metric space and (n) as a sequence in \R. Notice that (n) doesn't have any convergent subsequence, which means \R is not precompact. Therefore, \R is not compact.

7. Let \Omega\subset \R^n be an open bounded subset. For each f\in C^1(\overline\Omega), define \|f\|_{1,1}=\int_\Omega|f(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_i f(x)|\mathrm dx. Prove that (C^1(\overline\Omega),\|\cdot\|_{1,1}) is a normed vector space.

  • It's obvious that C^1(\overline\Omega) is a vector space. Let f,g\in C^1(\overline\Omega),\alpha\in\R.
    Triangle Inequality: \|f+g\|_{1,1}=\int_\Omega|(f+g)(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_i (f+g)(x)|\mathrm dx=\int_\Omega|f(x)+g(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_if(x)+\partial_ig(x)|\mathrm dx\leq\int_\Omega|f(x)|\mathrm dx+\int_\Omega|g(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega(|\partial_i f(x)|+|\partial_ig(x)|)\mathrm dx=\|f\|_{1,1}+\|g\|_{1,1}.
    Positive Homogeneity: \|\alpha f\|_{1,1}=\int_\Omega|\alpha f(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_i (\alpha f(x))|\mathrm dx=|\alpha|\int_\Omega|f(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\alpha\partial_i f(x)|\mathrm dx=|\alpha|(\int_\Omega|f(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_i f(x)|\mathrm dx)=|\alpha|\|f\|_{1,1}.
    Positive Definiteness: \|f\|_{1,1}=\int_\Omega|f(x)|\mathrm dx+\sum_{i=1}^n\int_\Omega|\partial_i f(x)|\mathrm dx>0 if f\neq 0.
    Q.E.D.

8. Let E be a n.v.s.

(i) Show that (E^*,\|\cdot\|_{E^*}) is a n.v.s.

  • It's obvious that E^* is a vector space. Let f,g\in E^*,\alpha\in\mathcal F.
    Triangle Inequality: \|f+g\|_{E^*}=\sup_{x\in E}\frac{|(f+g)(x)|}{\|x\|}\leq\sup_{x\in E}\frac{|f(x)|+|g(x)|}{\|x\|}=\sup_{x\in E}\frac{|f(x)|}{\|x\|}+\sup_{x\in E}\frac{|g(x)|}{\|x\|}=\|f\|_{E^*}+\|g\|_{E^*}.
    Positive Homogeneity: \|\alpha f\|_{E^*}=\sup_{x\in E}\frac{|\alpha f(x)|}{\|x\|}=\sup_{x\in E}|\alpha|\frac{|f(x)|}{\|x\|}=|\alpha|\sup_{x\in E}\frac{|f(x)|}{\|x\|}=|\alpha|\|f\|_{E^*}.
    Positive Definiteness: \|f\|_{E^*}=\sup_{x\in E}\frac{|f(x)|}{\|x\|}>0 if f\neq 0.
    Q.E.D.

(ii) Show that E^* is complete, and hence E^* is a Banach space.

  • Let \epsilon >0,x_0\in E. Consider a cauchy sequence (f_k) in E^* and f:=\lim_{k\rightarrow\infty}f_k. So, \exists N_f\in\N^*,s.t.\enspace\forall k>N_f,\|f-f_k\|_{E^*}<\epsilon. Notice that f(x+y)=\lim_{k\rightarrow\infty}f_k(x+y)=\lim_{k\rightarrow\infty}(f_k(x)+f_k(y))=\lim_{k\rightarrow\infty}f_k(x)+\lim_{k\rightarrow\infty}f_k(y)=f(x)+f(y). So, f is linear. Let k>N_f,\delta=\frac{\epsilon}{\epsilon+\|f_k\|_{E^*}}, we have \forall \|x-x_0\|_E<\delta,|f(x)-f(x_0)|=|(f-f_k)(x-x_0)+f_k(x-x_0)|\leq(\|f-f_k\|_{E^*}+\|f_k\|_{E^*})\|x-x_0\|<(\epsilon+\|f_k\|_{E^*})\cdot\frac{\epsilon}{\epsilon+\|f_k\|_{E^*}}=\epsilon, which means f is continuous. Therefore, f\in E^*, which means E^* is complete, and hence E^* is a Banach space.

9. Suppose that (X,d_1),(Y,d_2) are two metric spaces and f : X \rightarrow Y is continuous. If K \subset X is compact, show that f(K) is also compact. Consequently, if Y = \R and d_2 is the Euclidean metric, then f attains \sup and \inf on K, i.e., \exists x_0,y_0\in K, such that \sup_{x\in K} f(x) = f(x_0), \inf_{x\in K}f(x) = f(y_0).

  • Consider an open cover \{O_i\}_{i\in I} of f(K). With f is a continuous map, we know \{f^{-1}(O_i)\}_{i\in I} is also open, and it's an open cover of K. As K is compact, \{f^{-1}(O_i)\}_{i\in I} has a finite subcover \{f^{-1}(O_j)\}_{j\in J}. So \{O_j\}_{j\in J}=\{f(f^{-1}(O_j))\}_{j\in J} is an finite subcover of \{O_i\}_{i\in I}. Therefore f(K) is compact.
  • If Y=\R, d_2 is the Euclidean metric, we know that compact set in (\R,d_2) is bounded and closed, which means f(K) has \sup and \inf in f(K). Therefore, f attains \sup and \inf on K.

10. Let E be a n.v.s., and let f : E\rightarrow \R be a linear function. We say f is bounded if \sup_{\|x\|\leq 1}|f(x)|<\infty. Prove that f is continuous if and only if f is bounded

  • \Rightarrow: As f is continuous. Let x_0\in E,\epsilon_0>0,\exists\delta_0>0\forall\|x-x_0\|<\delta_0,|f(x)-f(x_0)|=|f(x-x_0)|<\epsilon_0. Consider x=\delta_0 z+x_0, we get \forall \|z\|<1,|f(z)|=\frac{|f(\delta_0 z)|}{\delta_0}<\frac{\epsilon_0}{\delta_0}\Rightarrow\sup_{\|z\|\leq 1}|f(z)|\leq\frac{\epsilon_0}{\delta_0}<\infty, which means f is bounded.
    \Leftarrow: Consider t=\sup_{\|x\|\leq1}|f(x)|, \epsilon>0,x_0\in E. If t=0, f=0, which is continuous. Suppose that t>0. Let \delta=\frac{\epsilon}{t}, we have \forall\|x-x_0\|<\delta, |f(x)-f(x_0)|=|f(x-x_0)|=|\|x-x_0\|f(\frac{x-x_0}{\|x-x_0\|})|\leq\|x-x_0\|\sup_{\|x\|\leq1}|f(x)|<\delta t=\epsilon, which means f is continuous.
    Q.E.D.

11. Let (X,d_1), (Y,d_2) be metric spaces. Show that f : X \rightarrow Y is continuous iff f^{−1}(A) is open in X for any open sets A \subset Y.

  • \Rightarrow: Let A\subset Y be an open set. As f is contiunous, we know \forall\epsilon>0\exists\delta>0,f(B(x_0,\delta))\subset B(f(x_0),\epsilon). Suppose that f^{-1}(A) is not open, which means \exists x_0\in f^{-1}(A),s.t.\enspace\forall r>0\exists k\in B(x_0,r)\setminus f^{-1}(A). Let r<\delta, we have f(k)\in f(B(x_0,r)\setminus f^{-1}(A))\subset f(B(x_0,\delta))\setminus A\subset B(f(x_0),\epsilon)\setminus A, which actually means \forall \epsilon>0,\exists y\in B(f(x_0),\epsilon)\setminus A. It's a contradiction with A is open, so f^{-1}(A) must be open.
    \Leftarrow: Let x_0\in X,y_0=f(x_0),\epsilon>0. As f^{-1}(B(y_0,\epsilon)) is an open set in X and x_0\in f^{-1}(B(y_0,\epsilon)), we know \exists \delta>0,s.t.\enspace B(x_0,\delta)\subset f^{-1}(B(y_0,\epsilon))\Rightarrow f(B(x_0,\delta))\subset B(y_0,\epsilon), which means f is continuous.
    Q.E.D.

12. Let E be a n.v.s. and f : E \rightarrow \R be a linear function that does not vanish identically. Let \alpha\in\R. Show that f is continuous iff H = [f = \alpha] is closed.

  • \Rightarrow: As f is linear and continuous, f\in E^*,\|f\|_{E^*}<\infty. Consider (x_k) in H, x=\lim_{k\rightarrow\infty}x_k\in E. Therefore, |f(x)-\alpha|=|f(x)-f(x_k)|=|f(x-x_k)|\leq\|f\|_{E^*}\|x-x_k\|\rightarrow 0,k\rightarrow0\Rightarrow f(x)=\alpha, which means x\in H and hence H is closed.
    \Leftarrow: Suppose that \sup_{x\in E,\|x\|=1}|f(x)|=\infty, which means that there exists a sequence (x_k) in E,s.t.\enspace \|x_k\|\leq 1,|f(x_k)|\rightarrow\infty,k\rightarrow\infty. Consider y_k=\frac{x_k}{f(x_k)}, we know f(y_k)=\frac{f(x_k)}{f(x_k)}=1,\|y_k\|=\frac{\|x_k\|}{|f(x_k)|}\rightarrow0,k\rightarrow\infty. As H=[f=1] is closed, f(0)=f(\lim_{k\rightarrow\infty}y_k)=1, which is a contradiction with f is linear. Therefore, \sup_{x\in E,\|x\|=1}|f(x)|<\infty, which means f is continuous.
    Q.E.D.

13. Let (X,d_1),(Y,d_2) be metric spaces. We say f : X \rightarrow Y is closed if G(f) := \{(x,f(x)) \in X \times Y : x \in X\} is closed in the product space X \times Y, which is equipped with the product metric \tilde d((x_1,y_1),(x_2,y_2)) := d_1(x_1,x_2) + d_2(y_1,y_2). Show that:

(1) If f is a continuous function, then f is closed.

  • Consider a sequence ((x_k,f(x_k)))_{k\ge1} in G(f) converging to (x_0,y_0)\in X\times Y. As \lim_{k\rightarrow\infty}\tilde d((x_k,f(x_k)),(x_0,y_0))=\lim_{k\rightarrow\infty}(d_1(x_k,x_0)+d_2(f(x_k),y_0))=0\xRightarrow{d_1,d_2\ge0}\lim_{k\rightarrow\infty}d_1(x_k,x_0)=0,\lim_{k\rightarrow\infty}d_2(f(x_k),y_0)=0, we get that the sequence (x_k) converges to x_0\in X, and (f(x_k)) converges to y_0\in Y. Because f is continuous, which means \forall\epsilon>0\exists\delta>0,s.t.\enspace\forall d_1(x,x_0)<\delta,d_2(f(x),f(x_0))<\epsilon, and \forall\delta>0\exists N_x\in\N^*,s.t.\enspace\forall k>N_x,d_1(x_0,x_k)<\delta, we have \forall\epsilon>0\exists N_x\in\N^*,s.t.\enspace\forall k>N_x,d_2(f(x_k),f(x_0))<\epsilon, which means that \lim_{k\rightarrow\infty}f(x_k)=f(x_0)=y_0. Therefore, (x_0,y_0)=(x_0,f(x_0))\in G(f). So G(f) is closed and hence f is closed.

(2) Let C^{1}[a,b] := \{x|x \in C[a,b], x has continuous derivatives\}. We define a map T: C^{1}[a,b] \rightarrow C[a,b] as follows: T(x(t)) := \frac{\mathrm d}{\mathrm dt}x(t), \forall x \in C^1[a,b]. Verify that T is closed but not continuous. (As usual, C[a,b] is equipped with the norm \| \cdot \|_\infty and here we view C^1[a, b] \subset C[a, b] as a subspace.)

  • Consider a sequence ((x_k,T(x_k))) in G(T) converging to (x_0,y_0)\in C^1[a,b]\times C[a,b]. Same with (1), we know when k\rightarrow\infty, \|x_0-x_k\|_{\infty}\rightarrow0, \|y_0-T(x_k)\|_\infty\rightarrow 0. Notice that \lim_{k\rightarrow\infty}\sup_{t\in[a,b]}|x_0-x_k|=0\Rightarrow x_k\rightrightarrows x_0\xRightarrow{x'_k\rightrightarrows y_0} x'_k\rightrightarrows x'_0, which means T(x_0)=y_0. Therefore, (x_0,y_0)=(x_0,T(x_0))\in G(T), which means G(T) is closed, and T is closed.
    Consider a sequence (x_k)=(\frac{1}{k}\sin[\frac{2\pi k}{b-a}(t-a)]) in C^1[a,b]. By calculating, we know that when k\rightarrow\infty, we have \|x_k\|_{\infty}=\sup_{t\in[a,b]}\frac{1}{k}|\sin[\frac{2\pi k}{b-a}(t-a)]|=\frac{1}{k}\rightarrow 0,\|T(x)\|_{\infty}=\sup_{t\in[a,b]}\frac{2\pi}{b-a}|\cos[\frac{2\pi k}{b-a}(t-a)]|\rightarrow\frac{2\pi}{b-a}, which means T is not continuous at 0.

14. Let \Omega\subset\R^n be open bounded. For u,v \in C^1(\overline\Omega), define (u, v)_{H^1} = \int_\Omega u(x)v(x)\mathrm dx +\int_\Omega\nabla u(x)\cdot \nabla v(x)\mathrm dx. Show that (C^1(\overline\Omega),(\cdot,\cdot)_{H^1}) is an inner product space.

  • It's obvious that C^1(\overline\Omega) is a vector space. Let u,v,w\in C^1(\overline\Omega),\lambda_1,\lambda_2\in\R.
    Symmetric: (u,v)_{H^1}=\int_\Omega u(x)v(x)\mathrm dx +\int_\Omega\nabla u(x)\cdot \nabla v(x)\mathrm dx=\int_\Omega v(x)u(x)\mathrm dx +\int_\Omega\nabla v(x)\cdot \nabla u(x)\mathrm dx=(v,u)_{H^1}.
    Positive Definite: (u,u)_{H^1}=\int_\Omega u^2(x)\mathrm dx+\int_\Omega|\nabla u(x)|^2\mathrm dx\ge0, and (u,u)_{H^1}=0\Leftrightarrow u=0.
    Linear: (\lambda_1u+\lambda_2v,w)_{H^1}= \int_\Omega (\lambda_1u(x)+\lambda_2v(x))w(x)\mathrm dx +\int_\Omega\nabla (\lambda_1u(x)+\lambda_2v(x))\cdot \nabla w(x)\mathrm dx= \lambda_1[\int_\Omega u(x)w(x)\mathrm dx +\int_\Omega\nabla u(x)\cdot \nabla w(x)\mathrm dx]+ \lambda_2[\int_\Omega v(x)w(x)\mathrm dx +\int_\Omega\nabla v(x)\cdot \nabla w(x)\mathrm dx]=\lambda_1(u,w)_{H^1}+\lambda_2(v,w)_{H^1}.
    Q.E.D.