This is what I am trying to prove:
Thrm: Let $T:X \to Y$ be a continuous linear map between normed spaces, then T is bounded.
Pf: By definition we have that $$ \left\lVert T \right\rVert = \sup_{ \left\lVert x\right\rVert \leq 1 } \left\lVert Tx\right\rVert _Y $$ , then put $f : X \to \mathbb{R} $ where $f(x) = \left\lVert Tx\right\rVert _Y$ then as norms are continuous and $T$ is continuous then $f$ is the composition of continuous functions so we have that $f$ is continuous. Therefore, because the set of $x \in X$ such that $ \left\lVert x\right\rVert \leq 1 $ is compact we have that we have that $\sup_{ \left\lVert x\right\rVert \leq 1} f(x) < \infty $ by the extreme value theorem, hence we conclude that $\left\lVert T \right\rVert < \infty $ $\square$
Something feels off, it feels a little too simple, I guess i just want to double check that I am not missing anything.
Thank you!
Best Answer
My favorite way is to argue by contradiction, although $\varepsilon$-$\delta$ argument with $\varepsilon=1$ is simple as well. Assume $T$ is not bounded. Then $$\sup_{\|x\|\le 1}\|Tx\|=\infty$$ Therefore for every $ n$ there exists $x_n$ such that $\|x_n\|\le 1 $ and $\|Tx_n\|\ge n^2.$ Hence $y_n:={x_n\over n}\to 0$ and $\|Ty_n\|\ge n.$ Thus $T$ is not continuous at $0. $