The the image of the unit ball in X is weak-* dense in the unit ball of X**

Question

I am trying to prove the following theorem and am stuck.

Let $X$ be a Banach space. The the image of the unit ball in $X$ is weak-* dense in the unit ball of $X^{**}$.

My proof idea

Assume $X^{**}$ is equipped with its weak-* topology. Let $Q$ be the isometric map from $X$ to $X^{**}$ such that $Q(x)(\pi) = \pi(x)$. Denote by $Q(B_x)$ the image of the unit ball $B_x$ in $X$. Now for every $ y \in Q(B_x)$ if $\|y\| < 1$ then $Q(B_x)\subset B_{x^{**}}$ where $B_{x^{**}}$ is the unit ball in $X^{**}$. But this is equivalent to showing that every vector in the complement of the closure of the $Q(B_x)$ has norm greater than $1$. (*)

Let $Q'(B_x)$ denote the closure of $Q(B_x)$ in $X^{**}$. Take $ \pi \in X^{**}\setminus Q'(B_x)$. As $Q'(B_x)$ is closed and $\{\pi \}$ is compact, by the Hahn-Banach seperation theorem, we have that there exists a continuous linear functional $\omega': X^{**} \to \mathbb{C}$ and $\alpha \in \mathbb{R}$ such that $\text{Re}(\omega'(\pi)) \leq \alpha \leq \text{Re}(\omega'(y))$ for all $y \in Q'(B_x)$.

I am stuck here.

I know that for $x \in X$ we have $Q(x)(\pi) = \pi(x)$ and $ \pi(x) = \|x\|$. So if $ x \notin B_x$ then $ \pi(x) > 1$ but I cannot find a way of putting these together. I would appreciate if someone could tell me how to proceed. I thank you for your time.

(*) Cannot use a direct approach as there are possibly infinite number of vectors in $X$.

Edit: I would also appreciate if you would give me suggestions towards writing a formal proof.

Answer 1

You are sort of two inches from the finish line.

We know that $Q'(B_x)$ is a convex, balanced, and (weak*) closed subset of $X^{\ast\ast}$ (that is contained in $B_{x^{\ast\ast}}$).

By (one of) the Hahn-Banach theorem(s), for every $\pi \notin Q'(B_x)$, there is a continuous linear form $\lambda$ on $X^{\ast\ast}$ (endowed with the weak* topology) with $\lvert\lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$, and $\lvert\lambda(\pi)\rvert > 1$.

The difference to what you have is (apart from specifying $\alpha = 1$) the strict inequality for $\lvert\lambda(\pi)\rvert$, where you only had a weak inequality. That difference is crucial, however.

Now, by definition of the weak* topology on $X^{\ast\ast}$, the space of continuous linear forms is $X^\ast$.

Then $\lvert \lambda(\eta)\rvert \leqslant 1$ for all $\eta \in Q'(B_x)$ implies $\lvert\lambda(x)\rvert \leqslant 1$ for all $x\in B_x$, which means $\lVert \lambda\rVert \leqslant 1$.

But then

$$1 < \lvert \pi(\lambda)\rvert \leqslant \lVert \pi\rVert\cdot\lVert\lambda\rVert \leqslant \lVert\pi\rVert.$$

So $\pi \notin B_{x^{\ast\ast}}$. That holds for all $\pi \notin Q'(B_x)$, hence

$$Q'(B_x) = B_{x^{\ast\ast}}.$$

Answer 2

this is exercise (52), part (d) of [real analysis, Folland]. actually this exercise is:

$\color{red}{\textbf{problem:}}\,$

Let $X$ be a Banach space and let $f_1, ...,f_n$ be linearly independent elements of $X^*$.

$\color{red}{\textbf{(a)}}\;$Define $f:X\rightarrow \mathbb{C}^n$ by $Tx=\Big(f(x_1),...,f(x_n)\Big)$. if $N=\{x|Tx=0\}$ and $M$ is linear span of $f_1,...,f_n$, then $M=N^0$, where $N^0=\Big\{f\in X^*\Big|\; f|_N=0\Big\}$ and $M$ is isomorphic with $\big(\frac{X}{N}\big)^*$.

$\color{red}{\textbf{(b)}}\;$ if $F\in X^{**}$, for eny $\epsilon>0$, there exists $x\in X$ such that for $j=1,...,n$, we have $F(f_j)=f_j(x)$ and $\|x\|\leq (1+\epsilon)\|F\|$.

$\vdots$

$\color{red}{\textbf{(d)}}\;$ in the weak$^*$ topology on $X^{**}$, the closed unit ball in $X$ is dense in the closed unit ball in $X^{**}$.

$\color{black}{\textbf{Before the proof, we prove this lemma:}}\;$

$\color{blue}{\textbf{ lemma:}}\;$

$\color{blue}{\textbf{(a)}}\;$ Suppose that $X$ and $Y$ are normed vector spaces and $T\in\mathscr{B}(X,Y)$. define $T^{\dagger}:\underset{f\longrightarrow f\circ T}{Y^*\rightarrow X^*}$. then $T^{\dagger}\in\mathscr{B}(Y^{\dagger},X^{\dagger})$ and $\|T^{\dagger}\|=\|T\|$.

$\color{blue}{\textbf{(b)}}\;$ Suppose $X$ is a Banach space and $N$ is a proper closed subspace of $X$ and $\pi:X\rightarrow\frac{X}{N}$ be quotient map. then $\pi^{\dagger}:\big(\frac{X}{N}\big)^*\rightarrow X^*$ is an isometric isomorphism from $ \big(\frac{X}{N}\big)^* $ onto $N^0$, where $\frac{X}{N}$ has the quotient map norm.

$\color{blue}{\textbf{proof of (a):}}\;$

$\|T^{\dagger}f\|=\|f\circ T\|\leq \|T\|\|f\|$

so $\|T^{\dagger}\|\leq\|T\|$.

on the other hand,

by Hahn Banach theorem, for every $x\in X$ such that $Tx\neq 0$, there exists $f_x\in X^*$ such that $f_x(Tx)=\|Tx\|$ and $\|f_x\|=1$. therefore,

\begin{align} \|T\|&=\underset{\|x\|=1}{\sup}\|Tx\|\\&=\underset{\underset{\|Tx\|\neq 0}{\|x\|=1}}{\sup}\|Tx\|\\&= \underset{\underset{\|Tx\|\neq 0}{\|x\|=1}}{\sup}|f_x(Tx)| \\&=\underset{\underset{\|Tx\|\neq 0}{\|x\|=1}}{\sup}|f_x\circ T(x)| \end{align}

and this implies

\begin{align} \|T^{\dagger}\|&= \underset{\|f\|=1}{\sup}\|T^{\dagger}f\|\\&= \underset{\|f\|=1}{\sup}\|f\circ T\|\\&= \underset{\|f\|=1}{\sup} \underset{\|x\|=1}{\sup}|f\circ T(x)|\\&= \underset{\|f\|=1}{\sup} \underset{\underset{\|Tx\|\neq 0}{\|x\|=1}}{\sup}|f\circ T(x)| \\&\geq \underset{\underset{\|Tx\|\neq 0}{\|x\|=1}}{\sup}|f_x\circ T(x)|\\&= \|T\| \end{align}

thus, $ \|T^{\dagger}\|\geq \|T\| $ and hence $\|T^{\dagger}\|=\|T\|$.

$\color{blue}{\textbf{proof of (b):}}\;$

by

Norm of the quotient map for a normed space

we know that $\|\pi\|=1$. thus $\|\pi^{\dagger}\|=1$. therefore $\|\pi^{\dagger}f\|\leq\|f\|$. on the other hand

$\|f\|=\underset{\|x+N\|<1}{\sup}f(x+N)$

but by definition of norm of quoutient space, there exists $n\in N$ such that $\|x+n\|<1$. also for every $\epsilon>0$, there exists some $x\in X$ such that

$ \|f\|-\epsilon<\|x+N\| $

therefore \begin{align} \|f\|-\epsilon&< f(x+N)\\&= f\big((x+n)+N\big)\\&\leq \underset{\|x\|<1}{\sup}f(x+N)\\&= \underset{\|x\|<1}{\sup}f(\pi(x))\\&= \underset{\|x\|<1}{\sup}f\circ \pi(x)\\&= \|f\circ\pi\|\\&= \|\pi^{\dagger}(f)\| \end{align}

so $\|f\|\leq\|\pi^{\dagger}(f)\|$ and hence $\|\pi^{\dagger}(f)\|=\|f\|$.

$\color{red}{\textbf{proof of (a):}}\;$

$M\subseteq N^{0}$ is obvious. we prove $N^{0}\subseteq M$.

for $n=1$, this proved by this link: $\ker T\subset \ker S\Rightarrow S=rT$ when $S$ and $T$ are linear functionals

Now, assume $n>1$. for $j=1,...,n$, let $N_j=\{x|f_j(x)=0\}$. then $N=\underset{j=1}{\overset{n}{\bigcap}}N_j$. since $\{f_1,...,f_n\}$ are linearly independent, from above link, for any $j=1,...,n$, we have $N_j\subsetneqq\underset{\underset{i\neq j}{i=1}}{\overset{n}{\bigcap}}N_i$. so for every $j=1,...,n$, there exists $x_j\in \underset{\underset{i\neq j}{i=1}}{\overset{n}{\bigcap}}N_i$ such that $x_j\notin N_j$ and since the sets $\{N_j\}$ are vector spaces, we can choose $x_j$ such that $f_i(x_j)=\delta_{ij}$, where $\delta_{ij}$ is the kronocker delta.

The set $\{x_1,...,x_n\}$ is linearly independent, for if $c_1x_1+...+c_nx_n=0$, then $c_i=f_i(c_1x_1+...c_nx_n)=0$.

Now, if $\{v_{\alpha}\}_{\alpha\in\mathcal{A}}$ be a basis for $N$, then $\Big(\{v_{\alpha}\}_{\alpha\in\mathcal{A}}\Big)\bigcup\Big(\{x_1,...,x_n\}\Big)$ is a basis for $X$, for if $x\in X$, then

$x-\Big(f_1(x)x_1+...f_n(x)x_n\Big)\in N$$\qquad\qquad(*)$

Now, let $f\in N^0$ and $x\in X$. from relation $(*)$ in above, we have $x= \Big(f_1(x)x_1+...f_n(x)x_n\Big)+n$, where $n\in N$. so $f(x)=f(x_1)f_1(x)+...+f(x_n)f_n(x)\in M$

then $M=N^0$.

$\color{red}{\textbf{proof of (b):}}\;$

by above lemma, the quotient map $\pi:X\rightarrow\frac{X}{N}$, induce the isometric isomorphism map

$\pi^{\dagger}:\underset{\Lambda\xrightarrow{\hspace{.7cm}} \Lambda\circ\pi}{\big(\frac{X}{N}\big)^*\rightarrow N^{\circ}}=M$

also, the map $ \pi^{\dagger} $, induce the map

$(\pi^{\dagger})^{\dagger}:M^*\rightarrow \big(\frac{X}{N}\big)^{**}\color{red}{=}\widehat{(\frac{X}{N})}$

equality in red holds, because $\big(\frac{X}{N}\big)^*\cong M$ and $M$ is finite dimentional. thus $\big(\frac{X}{N}\big)^*$ is finite dimentional and hence is reflexive and by

A Banach space is reflexive if and only if its dual is reflexive

$\frac{X}{N}$ is reflexive.

Now, since $F\in X^{**}$, we have $F|_M\in M^*$. thus

$F|_M\circ\pi^{\dagger}=(\pi^{\dagger})^{\dagger}(F|_M)\in\big(\frac{X}{N}\big)^{**}= \widehat{(\frac{X}{N})} $

thus, there exist $x'\in X$, such that

$F|_M\circ\pi^{\dagger}= \widehat{x'+N} $

Now, let $f_j\in\{f_1,...,f_n\}\subseteq M$. then there exists $\Lambda\in{\big(\frac{X}{N}\big)^*}$ such that $\pi^{\dagger}(\Lambda)=f_j$.

Now, we have

\begin{align} F(f_j)&=F|_M(f_j)\\&= F|_M(\pi^{\dagger}(\Lambda))\\&= F|_M\circ\pi^{\dagger}(\Lambda)\\&= \widehat{x'+N}(\Lambda)\\&= \Lambda(x'+N)\\&= \Lambda(\pi(x'))\\&= \Lambda\circ\pi(x')\\&= (\pi^{\dagger}(\Lambda))(x')\\&= f_j(x') \end{align}

Now, if $\|x'+N\|+\epsilon\|F\|$, then there exists $n\in N$ such that $\|x'+n\|<\|x'+N\|+\epsilon\|F\|$. so

\begin{align} \|x'+n\|& <\|x'+N\|+\epsilon\|F\|\\& =\|\widehat{x'+N}\|+\epsilon\|F\|\\& =\|F|_M\circ \pi^{\dagger}\|+\epsilon\|F\|\\& \leq\|F|_M\|.\|\pi^{\dagger}\|+\epsilon\|F\|\\& \leq\|F\|+\epsilon\|F\|\\& =(1+\epsilon)\|F\| \end{align}

since $\|\pi^{\dagger}\|=1$.

Now, if choose $x=x'+n$, since

$f_j(x)=f_j(x')=F(f_j)$

then $\color{red}{\text{(b)}}$ is proved.

$\color{red}{\textbf{proof of (d):}}\;$

if we denote the closed unit ball in $X$ with

$B=\big\{x\in X\big|\,\|x\|\leq 1\big\}$

and the closed unit ball in $X^{**}$ with

$B^{**}=\big\{F\in X^{**}\big|\,\|F\|\leq 1\big\}$

we have to prove that $\overline{\widehat{B}}=B^{**}$. it is clear that $\widehat{B}\subseteq B^{**}$ and since by Banach–Alaoglu theorem, $ B^{**} $ is weak$^*$-compact, then $\overline{\widehat{B}}\subseteq \overline{B^{**}}=B^{**}$. we have to prove $ B^{**}\subseteq \overline{\widehat{B}}$. for proof, it is suffice to show $(B^{\circ})^{**} \subseteq\overline{\widehat{B}} $, where

$(B^{\circ})^{**}=\big\{F\in X^{**}\big|\,\|F\|<1\big\}$

Now, if $F\in (B^{\circ})^{**}$, we construct a net $\widehat{x}_{U}\in \overline{\widehat{B}} $ such that

$\widehat{x}_{U}\overset{weak^*}{\xrightarrow{\hspace{.9cm}}}F$

if we consider all finite subsets of $X^*$ with $\mathscr{F}$. let for every $U,V\in \mathscr{F}$, define the relation $U\leq V$ if $U\subseteq V$. then $(\mathscr{F},\leq)$ is a directed set.

but by $\color{red}{\textbf{(b)}}$, for every finite set $U\in\mathscr{F}$, there exists $x_U\in X$ such that

$\|x_U\|\leq (1+\epsilon)\|F\|$

and since $\|F\|< 1$, if we choose $\epsilon>0$ such that $(1+\epsilon)\|F\|<1$, then $\|x_U\|< 1$.

Now, we know

$\widehat{x}_{U}\overset{weak^*}{\xrightarrow{\hspace{.9cm}}}F\quad$ iff $\quad\widehat{x_U}(f)\longrightarrow F(f)\quad (\forall f\in X^*)$

so if $f\in X^*$, for every $U\geq\{f\}$, by $\color{red}{\textbf{(b)}}$, we have $f(x_U)=F(f)$ and hence $\widehat{x_U}(f)\longrightarrow F(f) $. so

$\widehat{x}_{U}\overset{weak^*}{\xrightarrow{\hspace{.9cm}}}F$

and $\color{red}{\textbf{(d)}}$ is proved.

$\textbf{REMARK:}$ note that it seems, this problem is true in every normed vector space, since we did not use Banach space property in the proof.