Equivalence of continuity and boundedness

[radou]

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 &gt; 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.

 

[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||.

 

[radou]

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||.

morphism, thanks. I just figured it out.

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} &lt; \delta. Hence ||Av||< \epsilon. Further on, we have ||Av||= \frac{\delta ||Ax||}{2||x||}&lt; \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 &gt; 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&#039; - x&#039;&#039;|| &lt; \delta \Rightarrow ||Tx&#039; - Tx&#039;&#039;||&lt;\epsilon. Thus, T is continuous as x'' in X. Even better, this shows that T is uniformly continuous on X !

(rubs sweat off)