My CVHomework of Functional Analysis
Date: November 22nd
Name: Yuze Zhao
7.Set $X$ is a Banach space, $x,y\in X$ . If for all continuous linear functionals, $f(x)=f(y)$ , then $x,y$ .
proof:
Known that
\[f(x)-f(y)=f(x-y)=0.\]If $f$ is not zero functional, then $x=y$ .
8.Set $X$ is a Banach space, try to prove that for all $x\in X$ , $||x||=\sup{|f(x)|:f\in X’,||f||\leq 1}$ .
proof:
We shall prove it from both sides.
\[|f(x)|\leq \sup_{||f||\leq1}||f||\cdot||x||=||x||.\]According to lemma 2.1, $\exists f_0\in X’$ , subject to
\[\sup_{||f||\leq1}|f(x_0)|\geq |f_0(x_0)|=||x_0||\neq 0.\]| Then $|x|=\sup{ | f(x) | : f\in X’ , |f|\leq 1}$ . |
9.Set $p(x)$ is subbadditive and positively homogeneous in normal linear space $X$ . That is, for all $x,y\in X,\alpha \in \mathbb C$ ,
\[\begin{gather} p(x+y)\leq p(x)+p(y)\\ p(\alpha x)=|\alpha|\ p(x) \end{gather}\]If $p(x)$ is continuous at $x=0$ , then $p(x)$ continues at every point in $X$ .
proof:
Method one:
| We know it is continuous at 0., that is for any $\epsilon>0$ , $\exists \ \delta>0$ subject to all point $x\in X$ subject to $ | x | <\delta$ , |
| Take $x,y\in X$ which subject to $ | x-y | <\delta$ , Might as well set $z=x-y,z\in X$ . According to the conclusion above, |
Method tow:
| It is bounded at zero, that is $\exists\ \delta>0,\forall\ | x | \leq \delta$ , |
| We might set $y=\frac{\delta}{ | x | }x$ . Obviously, $ | y | =\delta$ . That is to say |
So $x$ is also bounded, that is to say $x$ continuous at any point in $X$ .
10.Set $p(x)$ is a subnorm of linear space, then ${x:p(x)<r}\ (r>0)$ is a convexy equilibrium and absorbing.
proof:
Take $x,y\ s.t.\ p(x)<r,\ p(y)<r$ ,
\[p(\alpha x+(1-\alpha)y)\leq |\alpha|\ p(x)+|1-\alpha|\ p(y)\leq r.\]So it is convex.
As it is positively homogeneous,
\[p(\lambda x)=|\lambda|p(x)<\lambda r<r.\]So it is equilibrium and absorbing.
11.Prove that the closed hull of a convex set is convex, the closed hull of a equilibrium set is equilibrium, the closed hull of an absorbing set is absorbing.
proof:
We set ${x_n}^\infty_{n=0},\ {y_n}^\infty_{n=0}$ subject to $x_n\to x,\ y_n\to y$ .
\[\alpha x+(1-\alpha )y=\lim_{n\to\infty}\alpha x_n+(1-\alpha )y_n\in\bar M.\]So it is convex.
\[|\lambda x|=\lim_{n\to\infty}|\lambda x_n|\in\bar M.\]So it is equilibrium and absorbing.
12.Compute the general form of bounded linear functional in $L^1[a,b]$ .
proof：
Comparing to the relationship between $l^1$ and $l^\infty$ , we suppose that the Dual space of $L^1[a,b]$ should be $L^\infty$ , and it should be reflexive. And the general form should be
\[f(x)=\int^b_a\ x(t)\ y(t)\ dt,\quad x(t)\in L^1[a,b],y(t)\in L^\infty[a,b].\] \[||y||=||f||=\inf_{m(E)=0}\sup_{t\in[a,b]\setminus E}y(t).\]As the similarity to the proof of theorem 4.4, we can also find it is well defined. So we just need to proof for every $f$ , $y\in L^\infty$ .
We denote a set $E_N$ , s.t.
\[E_N=\{t\in[a,b]:|y(t)|\leq N\}.\]And for all $\epsilon>0$ we denote A as
\[E=\{t\in[a,b]:|y(t)|>||f||+\epsilon\ \}\]We also denote $y_N(t)=\chi_{E_N\cap E}\cdot sgn\ y(t)$ , then
\[\left\| y_N(t) \right\|_1 = \int_a^b \chi_{E_N \cap E}(t)\, sgn\ y(t)\, dt = m(E_N \cap E)\] \[m(E_N\cap E)\ (\ ||f||+\epsilon\ )\leq\int_{E_N\cap A}|y(t)|dt\leq\int^b_ay_N(t)\cdot y(t)dt\leq||f||\ m(E_N\cap E).\]When $N\to\infty$
\[m(E)(||f||+\epsilon)\leq m(E)||f||.\]That is to say $m(E)=0$ . We can take $\inf_{m(E)=0} E$ , it is also subject to all properties.
\[||y||=\inf_{m(E)=0}\sup_{t\in[a,b]\setminus E}y(t)\leq ||f||+\epsilon.\]Also according to the arbitrary, take $\epsilon \to 0$
\[||y||=||f||=\inf_{m(E)=0}\sup_{t\in[a,b]\setminus E}y(t).\]12.A Banach space $X$ is reflexive, only if its duel space $X’$ is reflexive.
proof:
Necessity :
We denote the element of $X$ as $x$ , the element of $X’$ as $X’$ , the element of $X’’$ as $X’’$ and the element of $X’’’$ as $x’’’$ .
The typical map $\tau(x)$ , s.t.
\[\tau(x)=x''\]the in fact the element of $x’’’$ can be express as
\[x'''(x'')=x'''(\tau(x))=x_\Phi(x).\]$x_\Phi$ is the elemnet of $X’$ . And it is apparently linear.
\[|x_\Phi(x)|\leq ||x_\Phi||\cdot||x||=||x'''||\cdot||x||.\]So it is bounded. Now we try to find the typical map of $X’’’$ ,
\[x''(x_\Phi)=x_\Phi(\tau^{-1}(x''))=x'''(\tau(\tau^{-1}(x'')))=x'''(x'').\]That is to say $x’’‘(x’’)=x’‘(x_\Phi)$ , and $x_\Phi\in X’$ . So the typical map $\tau’(x_\Phi)=x’’’ $ .
Sufficiency :
| If $X$ is not reflexive, then $\tau(X)$ is the subset of $X’’$ . We denote $F=X’’-\tau(X)$ , then according to Hahn-Banach Theorem, exist $x’’‘\in X’’’,\ | x’’’ | \neq 0$, s.t. $x’’’(\tau(x))=x’’‘(x’’)=0$ . $X’$ is reflexive, so exist $\tau’(x_\Phi)=x’’’$ |
| That is to say $ | x_\Phi | =0$ . But actually $\tau’$ preserve the norm, so $x’’’$ must be also zero. That is comfit to the assumption. |
20.Set $X,Y$ are both Banach spaces, $T:X\to Y$ is a map preserving norm. Prove that Banach conjugate operator $T’$ is also norm preserving operator form $Y’$ to $X’$ . So $X\cong Y$ , then $X’\cong Y’$ . ‘ $\cong$ ‘ means norm-preserving isomorphism.
proof:
| Firstly, we shall prove that $T’$ is also preserve the norm. Now that $T$ is norm-preserving operator, exist $ | x_0 | = | y_0 | =1$ s.t. $Tx_0=y_0$ |
| That is to say $ | y’ | \geq | T’y | $ . |
According to the definition
\[\sup_{||y_0||=1}|y'(y_0)|=||y'||.\]| For all $\epsilon>0$ , exist $ | y_0 | =1$ s.t. |
So similarly
\[\sup_{||x_0||=1}|Y'y'(x_0)|=||T'y'||=|y'(y_0)|\geq||y'||-\epsilon.\]| So $ | T’y’ | = | y | $ . From theorem 5.1, we know $ | T | = | T’ | =1$ . So $T’$ preserves norm. Then we should prove it is both injective and surjective. |
Apparently, the map is injective. For it preserve the norm.
Now that $T$ is an bijection, we should prove for $\forall x’,\ \exists\ y’,\ s.t.\ T’y’=x’$ . Difine
\[y'(y)=x'(x).\] \[|y'(y)|\leq ||x||\cdot||T^{-1}||\cdot||y||.\]So $y’$ is bounded by the difinition.
\[x'(x)=y'(y)=T'y'(x).\]Then $Y’y’=x’$ . Q.E.D.
21.Set $X,Y$ are both Banach spaces, $T\in L(X,Y)$ . Prove that if $T$ is finite rank operator, then $T’$ is also finite rank operator.
proof:
According to the properties, we can take ${y_i}{i=1}^n$ s.t. $Tx_i=y_i$ . From Hahn-Banach theorem, we can take ${y’_i}{i=1}^n$ , s.t. $y_i’(y_j)=\delta_{ij}$ .
\[x'(x)=T'y'(x)=y'(Tx)=y'\left(\sum_{i=1}^n\alpha_iy_i\right).\]Because $R(T)$ has finite rank. We denote $\alpha_i=y_i’(Tx)$
\[x'(x)=\sum_{i=0}^n\alpha_iy'(y_i)=\sum_{i=0}^ny'(y_i)y_i'(Tx)=\sum_{i=0}^ny'(y_i)x_i'(x).\]If $x_i$ is linear relative
\[x'(x)=\sum_{i=0}^n\alpha_ix_i'(x)=\sum_{i=0}^n\alpha_iy_i'(Tx)=0.\]Take $Tx_j=y_j$ . Then $\alpha_j=0$ . So, $x’=span{x_i}$ . Q.E.D.
22.Prove:
$(1) \ \ ^0N(A’)=\overline{R(A)}; \quad (2)\overline{R(A’)}\subset N(A)^0$.
proof:
(1) $^0N(A’)=^0((\overline{R(A)})^0)=\overline{R(A)}$
(2)Take any one $x’\in \overline{R(A’)}$ . Then exist $x\in N(A),\ y’\in Y$ , s.t.
\[x'(x)=A'y'(x)=y'(Ax)=0\]So, $x’=N(A)^0$ . Q.E.D.