Let $X$ and $Y$ be normed spaces, $T:X \rightarrow Y$ a linear operator. Show that $T$ is continuous if $y' \circ T$ is continuous for every $y' \in Y'$. My idea is the following:
Suppose $T$ is not continuous, then there exists a sequence $x_n$ with $\|x_n\|=1$ such that $\|Tx_n\|>n$. If I could assure there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ are linearly independent I think I could construct an element $z'$ of $Y'$ such that $|(z' \circ T)x_{n_k}|>n_k$.
The first thing is, can I assure the existence of that subsequence? And if I can, how do I conclude applying H-B?
I'd like a solution with this same approach, but if there is a simpler proof i'd like to see it too.
Best Answer
This follows from the uniform boundedness principle. Consider the closed unit ball $B$ in $X$. Let $E=\{Tx:x\in X, \|x\|\leq1\}$. Then $y'(E)=(y' \circ T)(B)$ is bounded $\forall y' \in Y'$ by the continuity of $y' \circ T$. By the uniform boundedness principle, this implies that $E$ must be bounded, proving the continuity of $T$.