My Math Forum uniform convergence of continuous functions.

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 January 1st, 2012, 06:15 PM #1 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 uniform convergence of continuous functions. Hi, here's my question: Let $f:[0,1]\to\mathbb{R}$ be continuous, and $f(0)\not=f(1)$. Let $f_n(x)=f(x^n)$. I claim then that $f_n$ does not converge uniformly on $[0,1]$ and that $\lim_{n\to\infty} \int_0^1 f_n(x)dx=f(0)$. Attempt at proof: So the first bit isn't bad; either it can be shown by contradiction (suppose $f_n$ converges uniformly, and get a contradiction), or by noting that if $f_n$ did converge uniformly, then $\lim_{n\to\infty} f_n(x)=\lim_{n\to\infty} f(x^n)=\begin{cases}f(0), &x=<1\\f(1),=&x=1\end{cases}$, which is a discontinuous function (since $f(0)\not=f(1)$). This contradicts the fact that if a sequence of continuous functions converges uniformly, then their limit is continuous as well. As for the second bit, I'm having some trouble with the details. I want to do something like, for some $1>\delta>0$: $\lim_{n\to\infty}\int_0^1 f_n(x)dx=\lim_{n\to\infty}\int_0^{1-\delta}f(x^n)dx+\int_{1-\delta}^1f(x^n)dx=\int_0^{1-\delta}f(0)dx+\lim_{n\to\infty}\int_{1-\delta}^1f(x^n)dx...$ but not quite sure how to proceed. Thanks for any help!
 January 2nd, 2012, 12:35 PM #2 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 670 Re: uniform convergence of continuous functions. The first integral (in your end expression) = f(0)(1-?). The second integral is bounded by ?M, where M = max(|f(x)|) over the interval. Continuity implies M is finite. Let ? -> 0 and get your result.

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