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Let $B(X, Y)$ be the set of bounded linear maps from $X$ to $Y$ (i.e. such that $\sup_{||x|| \leq 1} L(x) < \infty$). Is $L \in B(X, Y)$ continuous? What about if $X$ is a Banach space? What about if $Y$ is a Banach space?

Thank you!

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I guess $X$ and $Y$ are normed space (a norm is missing after the $\sup$). Did you try to prove it, –  Davide Giraudo Nov 28 '11 at 17:57
    
Also consider the neat fact that in finite dimensional normed spaces all linear maps are bounded. –  Listing Nov 28 '11 at 18:02
    
Well I stared at the statement a bit but it seems kind of counterintuitive. You think I should try to prove it for realsies? –  badatmath Nov 28 '11 at 18:05

1 Answer 1

up vote 1 down vote accepted

Theorem 5.4 from Rudin's Real and Complex Analysis: For a linear transformation $\Lambda$ of a normed linear space $X$ into a normed linear space $Y$, the following are equivalent:

  • $\Lambda$ is bounded.
  • $\Lambda$ is continuous.
  • $\Lambda$ is continuous at one point of $X$.

The proof is very straightforward.

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Thanks, it's good to know that that's a theorem. I saw it in my notes but thought it might be wrong because continuity seems like a very strong condition. I thought continuity was $\exists c$ such that $||L(x)|| < c||x||$, and boundedness was $||L(x)|| < c$, so for small values of $||x||$ shouldn't continuity be stronger? –  badatmath Nov 28 '11 at 18:04
    
No, continuity is for every $\epsilon>0$ there exists $\delta>0$ such that $\|x_2-x_1\|_X<\delta$ implies that $\|Lx_2-Lx_1\|_Y<\epsilon$. Linearity is the glue which makes all of these things equivalent. –  dls Nov 28 '11 at 18:06
    
Well I guess when you put it that way, $||Lx_2 - Lx_1|| < ||L|| ||x_2 - x_1|| \to 0$ for bounded $L$. Thanks! I think I was just staring at the wrong condition. –  badatmath Nov 28 '11 at 18:09

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