# Equivalence of continuity and boundedness

1. Sep 16, 2007

I need a push with the following theorem, thanks in advance.

Let X and Y be normed spaces, and A : X --> Y a linear operator. A is continuous iff A is bounded.

So, let A be continuous. Then it is continuous at 0, and hence, for $\epsilon = 1$ there exists $\delta > 0$ such that for all x from X such that ||x - 0|| = ||x||<$\delta$, we have ||Ax - A0|| = ||Ax|| < 1. This is where I'm stuck.

For the other direction, let A be bounded. So, there exists some M > 0 such that ||Ax|| $\leq$ M, for all x in X. No further inspiration.

2. Sep 16, 2007

### morphism

You almost have them.

Let y be in X. Then ||Ay|| = ||A($\frac{\delta ||y||}{\delta ||y||} y$)||. Now use the fact that A is linear. And what do we know about the norm of y/||y||?

For the other direction, the inequality you want to use is ||Ax|| <= M||x||.

3. Sep 16, 2007

For the first one, we simply need to constuct a vector with a norm lesser than $\delta$. So, for some non-zero x, v = $\frac{\delta x}{2 ||x||}$, and, abviously, ||v|| = $\frac{\delta}{2} < \delta$. Hence ||Av||< $\epsilon$. Further on, we have ||Av||= $\frac{\delta ||Ax||}{2||x||}< \epsilon$, and hence ||Ax||<$\frac{2\epsilon ||x||}{\delta}$. This holds for any non-zero x, and ||Ax||$\leq \frac{2\epsilon ||x||}{\delta}$ holds for any x, so A is bounded.
Assume A is bounded, and let $\epsilon > 0$ be given. For any x', x'' in X, there exists some k > 0 such that ||A(x' - x'')|| = ||Ax' - Ax''|| < k||x'-x''||. Now, if we simply put $\delta = \frac{\epsilon}{k}$, we have the implication $||x' - x''|| < \delta \Rightarrow ||Tx' - Tx''||<\epsilon$. Thus, T is continuous as x'' in X. Even better, this shows that T is uniformly continuous on X !