# If composition with each functional is bounded, then the operator is bounded

Let $$X,Y$$ be normed linear spaces and $$T:X\to Y$$ a linear operator such that for all $$\psi\in Y^*$$ we have $$\psi\circ T\in X^*$$. Prove that $$T$$ is continuous.

There is a "classical approach" of this problem that can be found here as well (See gerw's answer), I had another idea: If we had $$X,Y$$ being Banach spaces, then this would be quite easy: By the closed graph theorem, it suffices to show that for a sequence $$(x_n)\subset X$$ with $$x_n\to0$$ and $$Tx_n\to y$$, then $$y=0$$.

For $$\psi\in Y^*$$ we have $$\psi\circ T(x_n)\to0=\psi(y),$$ hence $$y\in\displaystyle{\bigcap_{\psi\in Y^*}\ker(\psi)=\{0\}}$$. This equality is justified by $$\|y\|=\displaystyle{\sup_{\|\psi\|_{Y^*}=1}\|\psi(y)\|}$$.

I was thinking that if we could embed $$X,Y$$ in their completions, maybe this could work. But we would have to show that $$\psi\circ T$$ is continuous for the functionals on the completion on $$Y$$ and we should also extend $$T$$ to the completion somehow. Do you think that this is possible? Moreover, do you believe that this is a simpler technique, in the sense that it comes more naturally?

• No, I don't see any hope for this. Completeness is essential for the proof and extending $T$ to the completion before knowing that it is continuous is not possible. – Kavi Rama Murthy Apr 28 at 0:02

Your idea works somehow if you take concrete completions of your spaces, namely the closure of (the canonical image of) the space in its bidual. The assumption you have precisely means that the transposed $$T^*:Y^*\to X^*$$ is well defined and (as in gerw's answer to the question you mention) $$T^*$$ is then continuous. If you restrict $$T^{**}: X^{**}\to Y^{**}$$ to the closure of $$X$$ in $$X^{**}$$ you get an extension of $$T$$ to the completions (however, it is then needless to refer to the closed graph theorem again because $$T^{**}$$ is clearly continuous).