Problems involving reflexivity and weak / weak-* convergence

Question

Consider the following two questions where throughout $X$ is a Banach space, $X'$ denotes its dual and $(f_n)_{n=1}^\infty\subset X'$. We also denote the canonical map $J_X:X\to X''$ where for all $f\in X'$, $(J_Xx)(f)=f(x)$.

$1.$ Using the canonical map, show that if $f_n\rightharpoonup f$ in $X'$ then also $f_n\overset{\star}{\rightharpoonup} f$ in $X'$.

$2.$ Show that if $X$ is reflexice then $f_n\overset{\star}{\rightharpoonup} f$ in $X'$ implies that $f_n\rightharpoonup f$ in $X'$.

Here are my attempts to each question:

Answer to $1:$

To say that $f_n\rightharpoonup f$ in $X'$ means that $\forall\, \psi\in X'',\,\psi(f_n)\to\psi(f)$ in $\mathbb R$ as $n\to\infty$.

But by definition of the second dual, $\forall\, \psi\in X''\,\exists \,x\in X: \psi=J_Xx$ so that $\psi(f)=(J_Xx)(f)=f(x)$.

Consider then that, $\forall\,x\in X$,

$$0\le\|f_n(x)-f(x)\|=\|\psi(f_n)-\psi(f)\|$$

Now $\lim_{n\to\infty}\|\psi(f_n)-\psi(f)\|=0$ so that by the Sandwich theorem we can infer that $\|f_n(x)-f(x)\|\to0$ as $n\to\infty$. But notice,

$$\lim_{n\to\infty}\|f_n(x)-f(x)\|=0\iff f_n(x)\to f(x)\,\,\text{as}\,\, n\to\infty\,\forall\, x\in X$$

But this precisely means that $f_n\overset{\star}{\rightharpoonup} f$ which is what we wanted to show.

Answer to $2:$

To say that $X$ is reflexive means that $\forall\,\psi\in X''\,\exists x\in X:\psi=J_Xx$.

We also note that saying $f_n\overset{\star}{\rightharpoonup} f$ in $X'$ means that $\forall\,x\in X,\,f_n(x)\to f(x)$ in $\mathbb F$ as $n\to \infty$.

Now since $f_n\overset{\star}{\rightharpoonup} f$ we have that,

$$0\le\|f_n(x)-f(x)\|=\|\psi(f_n)-\psi(f)\|$$

By definition of $\psi\in X''$. Now $\forall\, x\in X$ consider the following,

$$0\le\|\psi(f_n)-\psi(f)\|=\|f_n(x)-f(x)\|$$

Taking the limit of the right most term, as we did for the first problem, we see that it converges to zero as $n\to\infty$. So applying the Sandwich theorem again we are able to deduce that,

$$\lim_{n\to\infty}\|\psi(f_n)-\psi(f)\|=0\implies\psi(f_n)\to\psi(f)\,\,\text{as}\,\,n\to\infty$$And this precisely means that $f_n\overset{\star}{\rightharpoonup} f$ which is what we wanted to show.

Is what I have done correct? The part where I am most unsure is where I have invoked the Sandwich theorem.

Answer 1

What you stated is the case when $X$ is reflexive.

When $X$ is reflexive, one corollary is that we know $X'$ is also reflexive. So when we apply the reflexive definition twice, we get $X\cong X'$ and $X'\cong X''$, we have $X$ and $X''$ are isometrically isomorphic (as you said with the map $J_X$).

However in general, $X''$ is the "bigger" space, that is we can have the case when the injective map $J_X$ is not onto, so $J_X(X)\subsetneq X''$.

For example, $L^1(\mathbb{R})$ is not reflexive, so we have $(L^1(\mathbb{R}))'' = (L^\infty(\mathbb{R}))'$ is not (isomorphic) $L^1(\mathbb{R})$. To see this, the easiest way is to use another theorem that $X'$ is separable implies $X$ is separable. If $(L^\infty(\mathbb{R}))' = L^1(\mathbb{R})$ which is separable, then we would have $L^\infty(\mathbb{R})$ is separable (which is false).

Explicit example of can be constructed using Hahn-Banach theorem, see Dual of $l^\infty$ is not $l^1$.