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The claim is the following:

Every family of bounded linear functions is equicontinuous if and only it is uniformly bounded.

Equicontinuity is defined here. Any suggestions about this?

I only have trouble figuring out the forward direction.

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If $\|f\| \le M$, then $|f(x) - f(y)| \le M \|x-y\|$. Can you see how to do it now? And for the forward direction, I don't think you can use the uniform boundedness principle, because the conclusion of that theorem is that such a family is equicontinuous. –  Christopher A. Wong Mar 2 '13 at 0:40
    
Yes! How did I miss that! What would be a way to go about it? –  user44069 Mar 2 '13 at 0:42
    
Try just using the definition of equicontinuity. –  Christopher A. Wong Mar 2 '13 at 0:47
    
It doesn't seem to be very helpful. –  user44069 Mar 2 '13 at 1:55
    
By linearity, a family of operators is equicontinuous if they are equicontinuous at the origin. Perhaps this can help you. –  Christopher A. Wong Mar 2 '13 at 2:12

1 Answer 1

  1. Suppose the family $F$ is bounded. Then there is $M > 0$ such that $\|f\| \le M$ for all $f \in F$, and hence by linearity $|f(x) - f(y)| = |f(x-y)| \le M \|x - y\|$. Thus we can choose $\delta = \epsilon /M$.

  2. Suppose $F$ is equicontinuous. Then $F$ must be equicontinuous at the origin. Then, for every ball $B_{\epsilon}$, there exists $\delta$ such that $f(B_{\delta}) \subset B_{\epsilon}$ for every $f \in F$. In particular, if $S$ is the unit sphere, then $f(S) \subset B_{\epsilon/\delta}$. Then, by linearity, for any $x$, we have $f(x) \subset \|x\| B_{\epsilon/\delta}$, which is equivalent to $|f(x)| \le \epsilon/\delta \|x\|$.

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