# Continuity of a linear transformation

Let $$X,Y$$ be normed spaces and $$T:X\to Y$$ be a linear map.

I need to show that $$T$$ is continuous $$\iff$$ for all $$x_n\to 0$$ in $$X$$ , $$sup_n||Tx_n|| \lt \infty$$.

If $$T$$ is continuous then if $$x_n\to 0$$ then $$T(x_n) \to T(0)$$ and in particular $$sup ||T(x_n)||$$ is finite.

Im not sure how to do the other direction.

Any ideas?

Thanks for helping.

If $$T$$ is not continuous, then the set$$\{T(x)\,|\,\lVert x\rVert=1\}$$is unbounded. So, there is a sequence $$(x_n)_{n\in\mathbb N}$$ of unitary vectors such that $$(\forall n\in\mathbb{N}):\bigl\lVert T(x_n)\bigr\rVert\geqslant n$$. So, consider the sequence$$\left(\frac{x_n}{\sqrt n}\right)_{n\in\mathbb N}.$$
• Nice! and we have that sequence because we can take $x_n$ s.t $||x_n|| =1$ and $||T(x_n)|| \to \infty$ (by assumption ) and then take a sub sequence that satisfy $||y_n|| \ge n$ , right? – user335501 Feb 9 at 20:13
• Since the set $\{T(x)\,|\,\lVert x\rVert=1\}$ is unbounded, then, for each natural $n$, there is a vector $x_n$ with norm $1$ such that $\bigl\lVert T(x_n)\bigr\rVert\geqslant n$. – José Carlos Santos Feb 9 at 20:50