Uniform convergence of composition of functions and integration

Question

I've been wondering about this problem for a bit, it came up in class but its not really homework.

Let $f:[0,1]->R$ be continuous and non-negative.

We know $f(x^n)\to f(0)$ for $x \in [0,1)$ and $f(x^n)\to f(1)$ for $x=1$. Does $f(x^n)\to f(0)$ uniformly? What I would like to do is to use that in showing that the following holds $$ \lim_{n\to\infty} \int^{1}_{0} f(x^n) =f(0) $$ by splitting up $[0,1]$ into $[0,1-\epsilon]$ and $[1-\epsilon,1]$...or something like that. But I've been really stuck on that first part. I hope the latex works on my phone..

Answer 1

The pointwise convergence $f(x^n) \rightarrow f^*(x) = \bigg\{ \begin{eqnarray} f(0), & x < 1 \\ f(1), & x = 1 \end{eqnarray}$ as $n \rightarrow \infty$ is not necessarily uniform. However, there is a majorant $g(x) = \max_{x \in [0,1]} |f(x)|$ that satisfies for $f_n(x) = f(x^n)$ the inequality $g(x) \geq |f_n(x)|$ and is integrable, that is, its integral converges. Hence you can apply the Lebesgue dominated convergence theorem (the second equation below). We have \begin{eqnarray} \lim_{n \rightarrow \infty} \int_0^1 f(x^n) dx & = & \lim_{n \rightarrow \infty} \int_0^1 f_n(x) dx = \int_0^1 f^*(x) dx = f(0) \ . \end{eqnarray} Note that $f^*(x)$ can be calculated as follows. Assume first $x<1$. Then by continuity of $f$ we have \begin{equation} f(0) = f(\lim_{n \rightarrow \infty} x^n) = \lim_{n \rightarrow \infty} f(x^n) \end{equation} Assume then $x=1$. We have \begin{equation} f(1) = \lim_{n \rightarrow \infty} f(1) = \lim_{n \rightarrow \infty} f(1^n) = \lim_{n \rightarrow \infty} f(x^n). \end{equation}

Answer 2

You can also use the theorem of dominated convergence (as $f$ is continuous, it is also bounded on $[0,1]$) to conclude using the pointwise convergence.

Answer 3

Your idea works great. The graph over the second interval has length $\epsilon$ and bounded height, so goes to 0, while the first part needs a little more work. Pick $\delta$ so that if $0<u<\delta$, $|f(u)-f(0)|<\eta$ for any $\eta$. Then pick $n$ so $(1-\epsilon)^n<\delta$.