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I'm trying to understand this problem on MSE.

Let $X$ and $Y$ be Banach spaces, and let $T:X\rightarrow Y$ be a linear map such that $f\circ T$ is continuous for all $f\in Y'$. Show that $T$ is continuous.

Jonas provides a good simple start of the problem but I was not able to physically show the contradiction. I have found a similar problem here which they took a completely different route and user38355 didn't clearly give proof of the direction I'm trying to understand.

user640718
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Well I am going to follow Jonas' approach with some more details.

Both the spaces are Banach, so we can apply the closed graph theorem. This means, we take a sequence $(x_n)\subset X$ such that $x_n\to0$ and we assume that $T(x_n)\to y$ where $y\in Y$ is some element. If we manage to show that $y=0$, then by the closed graph theorem we can conclude that $T$ is bounded (this is the standard application).

Assume that $y\in Y$ is non-zero. By the Hahn-Banach theorem we can find a functional $f\in Y^*$ so that $f(y)\neq0$. But now $f\circ T$ is (linear, of course) and continuous by assumption, so $|f\circ T(x_n)|\leq \|f\circ T\|\cdot\|x_n\|$. Letting $n\to\infty$ we conclude that $f\circ T(x_n)\to0$. But since $f$ itself is continuous and $T(x_n)\to y$, we have that $f\circ T(x_n)=f(T(x_n))\to f(y)$, so $f(y)=0$, a contradiction.

Just dropped in
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  • How does the boundedness of $T$ guarantee continuity? – user640718 Apr 28 '21 at 15:15
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    @user640718 A linear operator $T:X\to Y$ between normed spaces is continuous if and only if there exists a constant $C>0$ so that $|T(x)|\leq C|x|$ (this is what it means for $T$ to be bounded). This can be found on any introductory book on functional analysis. For a quick proof of what you are asking: if $T$ is bounded and $x_n\to x$, then $|T(x_n)-T(x)|=|T(x_n-x)|\leq C|x_n-x|\to0$, so $Tx_n\to Tx$ so $T$ is continuous. – Just dropped in Apr 28 '21 at 15:18