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Let $X$ be a Banach space and $A : X \rightarrow X$ linear operator such that for any $\phi\in X'$ operator $\phi \;\circ A$ is continuous. I want to prove that $A$ is also continuous.

My work so far

I wanted to use closed graph theorem (this situation is quite suitable for this theorem. We have linear operator between Banach spaces)

To show that $A$ is continuous using closed graph theorem I should prove that for any $(x_n) \subset X$, $x_n \rightarrow 0$, $A(x_n) \rightarrow y$ we have $y = 0$

I tried some tricks, firstly, becuase $\phi \circ A$ is bounded:

$$0 \le \|(\phi \circ A)(x_n)\| \le \|\phi\circ A\|\cdot \|x_n\| $$

but becuse $x_n \rightarrow 0$ then $(\phi\circ A )(x_n) \rightarrow 0$

Also because $A(x_n) \rightarrow y$ then $\phi(A(x_n)) \rightarrow \phi(y)$ (becuase $\phi$ is continuous).

Out of these two facts we have that $\phi(y) = 0$

And then I tried to somehow prove that $\phi(y) = y$ but I couldn't. I also tried to rewrite somehow

$\|y\| \le $ something that tends to $0$

but also I didn't figure out anything. Could you please give me a hand?

Lucian
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    Hint: If $y \neq 0$ try to use the Hahn--Banach theorem to show that there exists some $\phi\in X'$ such that $\phi(y) \neq 0$.... Note that $\phi(y)=y$ makes no sense as $y$ and $\phi(y)$ are in different sets. – N. S. Jun 22 '21 at 15:59
  • I understand the proof. But can you tell me how do you conclude by Hahn - Banach theorem that there has to be $\phi \in X'$ such that $\phi(y) \neq 0$? – Lucian Jun 22 '21 at 17:06
  • I understand from Hahn - Banach theorem that in our case it tells that: $\exists_{f : X \rightarrow X}: f|_{X'} = \phi$ – Lucian Jun 22 '21 at 17:07
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    Yes. Given a vector $y\neq0$, take the 1-dimensional subspace that is the linear span of $y$. define a functional by setting $\phi(\lambda y)=\lambda|y|$. Then extend this functional using Hahn-Banach. This satisfies $\phi(y)=|y|\neq0$. Is this clear? – Just dropped in Jun 22 '21 at 21:05

2 Answers2

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Two lemmas of the Hahn-Banach theorem are needed.

Lemma 1. Let $X$ be any norm linear space and $x_0 \neq 0$. Then $\exists~~f \in X'$ such that $f(x_0)=\|x_0\|$.

Proof: Let $Y:=\textit{span} \{x_0\}$. Define $g:Y \rightarrow \mathbb{K}$ $\hspace{0.3cm}$ ($\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$) $\hspace{0.2cm}$ by $g(\alpha x_0)=\alpha \|x_0\|$ for all $\alpha \in \mathbb{K}$. Then clearly $g$ is linear. Now \begin{align} |g(\alpha x_0)|=|\alpha| \|x_0\|=\|\alpha x_0\|~~~\forall~~\alpha \in \mathbb{K} \end{align} Thus $g$ is bounded and $\|g\|=1$. By Hahn-Banach theorem $\exists~~f \in X'$ such that $f|_Y=g$. Therefore, \begin{align} f(x_0)=g(x_0)=\|x_0\|. \end{align} Lemma 2. Let $X$ be any norm linear space and $x \in X$ is such that $f(x)=0$ for all $f \in X'$, then $x=0$.

Proof: Can be easily proved from Lemma 1.

To use the closed graph theorem let $\{(x_n, Ax_n)\}_n$ be any sequence in the graph of $A$ converging to some $(x, y) \in X \times X$. We need to show that $Ax=y$. Then, \begin{align} & x_n \rightarrow x \\ &\Rightarrow (\phi \circ A) (x_n) \rightarrow (\phi \circ A)(x) \end{align} and \begin{align} & Ax_n \rightarrow y \\ & \Rightarrow (\phi \circ A)(x_n) \rightarrow \phi (y) \end{align} (using the continuities). It must be that $ (\phi \circ A)(x)=\phi(y)$ which implies $\phi(Ax-y)=0$ and this is true for all $\phi \in X'$ so by Lemma 2. we get $Ax=y$. Thus graph of $A$ is closed and hence $A$ is continuous.

absolute0
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So you have shown that for all $\phi \in X'$ there holds $\phi(y)=0$. Assume that $y\neq 0$. Then your claim follows, if you can show that there is a $\phi \in X'$ with $\phi(y)\neq 0$. This is actually an easy corollary of Hahn-Banach: Consider the subspace $span(y) \subset X$. Assume w.l.o.g. that $\|y\|=1$. For every element in $v\in span(y)$ there is a unique $t\in K$ with $v = t y$. Thus we can define on $span(y)$ a linear functional $\phi_y$ as $\phi_y(v)=t$. Moreover it holds that $|\phi_y(v)\| =|t|=\|y\|$, hence $\phi_y$ is dominated by the nonnegative and sublinar function $\|.\|$. Thus with Hahn-Banach we can conclude that $\phi_y$ can be extended to a linear functional $\phi$ on $X$, which fulfills $\phi(v)=\phi_y(v)=t$ for all $v\in span(y)$. Thus $\phi(y)\neq 0$, a contradiction.

crankk
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