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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.

Now I think this problem is trivial once you have the notion of weak topology. But without that notion, I'm not sure how to approach this. I tried using the inequalities $|(f\circ T)(x)|\leq(||f\circ T||)(||x||)$ and $|(f\circ T)(x)|\leq(||f|||)(||Tx||)$, but I don't see how that gets us any closer to find what $||Tx||$ is less than or equal to. Maybe a judicious choice of $f$ would help?

Keshav Srinivasan
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1 Answers1

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I would approach this using the closed graph theorem. So take some sequence $(x_n,Tx_n)$ for $(x_n)\subseteq X$ such that $(x_n)$ converges to some $x\in X$ and $(Tx_n)$ converges to some $y\in Y$. Now you need to show $Tx=y$. So, assume that this is false and try to get to a contradiction to the continuity assumption for some 'good' $f\in Y'$.

Jonas Lenz
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