# Bounded functional which composed with an unbounded operator becomes unbounded

While working on a problem, I came up with a certain lemma, however I'm not sure whether it's true and I'd be grateful for some insight.

Let $X$ and $Y$ be normed spaces over reals, where $X$ is not Banach, and $L:X \to Y$ a linear operator, which is unbounded. Then there is a bounded linear functional $f:Y \to \mathbb{R}$ such that $f \circ L$ is not bounded.

Proposition: if for any functional $\varphi \in Y^*$, $\varphi \circ L$ is bounded, than L is bounded.
let $(x_n) \subset X, \|x_n \| \to 0$, than for any $\varphi \in Y^*$, $\varphi \circ L(x_n) \to 0$. Hence $L(x_n) \to_{weak} 0$ in Y. Since every weakly convergent sequence is bounded, we have $\|L(x_n) \| <m$ for some $m>0$. So $\|x_n \| \to 0$ implies boundedness of $\|L(x_n) \|$.
Now assume that L is unbounded, tnan there is a bounded sequence $(x_n), \|L(x_n) \| \to \infty$, hence for $x'_n= x_n/ \|L(x_n)\|^{1/2}$, $x'_n \to 0, \|L(x'_n)\| \to \infty$ , so $L(x'_n)$ is not bounded, contradiction.
• I think this works, but you need to add that without loss of generality you can assume that $Y$ is complete (by Hahn-Banach). – tomasz Jan 21 '15 at 8:15