Prove that operator between Banach spaces is continuous 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?
 A: 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.
A: 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.
