The Unapologetic Mathematician

Mathematics for the interested outsider

Cauchy’s Condition for Uniform Convergence

As I said at the end of the last post, uniform convergence has some things in common with convergence of numbers. And, in particular, Cauchy’s condition comes over.

Specifically, a sequence f_n converges uniformly to a function f if and only if for every \epsilon>0 there exists an N so that m>N and n>N imply that |f_m(x)-f_n(x)|<\epsilon.

One direction is straightforward. Assume that f_n converges uniformly to f. Given \epsilon we can pick N so that n>N implies that |f_n(x)-f(x)|<\frac{\epsilon}{2} for all x. Then if m>N and n>N we have

|f_m(x)-f_n(x)|<|f_m(x)-f(x)|+|f(x)-f_n(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon

In the other direction, if the Cauchy condition holds for the sequence of functions, then the Cauchy condition holds for the sequence of numbers we get by evaluating at each point x. So at least we know that the sequence of functions must converge pointwise. We set f(x)=\lim\limits_{n\rightarrow\infty}f_n(x) to be this limit, and we’re left to show that the convergence is uniform.

Given an \epsilon the Cauchy condition tells us that we have an N so that n>N implies that |f_n(x)-f_{n+k}(x)|<\frac{\epsilon}{2} for every natural number k. Then taking the limit over k we find

|f_n(x)-f(x)|=\lim\limits_{k\rightarrow\infty}|f_n(x)-f_{n+k}(x)|\leq\frac{\epsilon}{2}<\epsilon

Thus the convergence is uniform.

September 8, 2008 Posted by | Analysis, Calculus, Functional Analysis | 6 Comments

   

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