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Equivalence of continuity and boundedness

  1. Sep 16, 2007 #1


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    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 [itex]\epsilon = 1[/itex] there exists [itex]\delta > 0[/itex] such that for all x from X such that ||x - 0|| = ||x||<[itex]\delta[/itex], 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|| [itex]\leq[/itex] M, for all x in X. No further inspiration.
  2. jcsd
  3. Sep 16, 2007 #2


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    You almost have them.

    Let y be in X. Then ||Ay|| = ||A([itex]\frac{\delta ||y||}{\delta ||y||} y [/itex])||. 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||.
  4. Sep 16, 2007 #3


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    morphism, thanks. I just figured it out.

    For the first one, we simply need to constuct a vector with a norm lesser than [itex]\delta[/itex]. So, for some non-zero x, v = [itex]\frac{\delta x}{2 ||x||}[/itex], and, abviously, ||v|| = [itex]\frac{\delta}{2} < \delta[/itex]. Hence ||Av||< [itex]\epsilon[/itex]. Further on, we have ||Av||= [itex]\frac{\delta ||Ax||}{2||x||}< \epsilon[/itex], and hence ||Ax||<[itex]\frac{2\epsilon ||x||}{\delta}[/itex]. This holds for any non-zero x, and ||Ax||[itex]\leq \frac{2\epsilon ||x||}{\delta}[/itex] holds for any x, so A is bounded.

    Assume A is bounded, and let [itex]\epsilon > 0[/itex] 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 [itex]\delta = \frac{\epsilon}{k}[/itex], we have the implication [itex]||x' - x''|| < \delta \Rightarrow ||Tx' - Tx''||<\epsilon[/itex]. Thus, T is continuous as x'' in X. Even better, this shows that T is uniformly continuous on X !

    (rubs sweat off)
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