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I'm having problems determining whether or not I can perform a limit inside an integrand, or must perform the integral outside. I'm seeing two types of problems:

$\lim_{n \rightarrow \infty} \int_0^1 f(x^n) dx$

and problems like:

$\lim_{n \rightarrow \infty} \int_0^1 f(\cos nx) dx$

Any help would be most appreciated!

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3  
what are the hypotheses on the functions you are integrating? –  John Martin Aug 25 '12 at 16:32
    
f is continuous on [0,1] –  Nik Kumar Aug 26 '12 at 18:48

1 Answer 1

up vote 2 down vote accepted

In general, for any sequence of functions $g_n$ and function $g$ we have \begin{align} \lim\limits_{n\to\infty}\int_0^1 g_n(x)dx=\int_0^1 g(x)dx &\iff \lim\limits_{n\to\infty}\int_0^1 (g(x)-g_n(x))dx=0 \end{align} so it suffices that $g_n\to g$ in $L^1$.

In your case, if we assume that $f$ is continuous we can let $g_n(x)=f(x^n)$ and note that $g_n\to f(0)$ in $L^1$. To see this, let $M$ be an upper bound for $|f|$ on $[0,1]$ and take any $\epsilon>0$. Then we have some $\delta>0$ such that $0\leq x<\delta\implies |f(x)-f(0)|<\epsilon/2$, and if we choose $N$ large enough that $(1-\epsilon/2M)^N<\delta$ then for $n\geq N$ we have $$\begin{align} \|g_n-f(0)\|_1 &= \int_0^1|f(x^n)-f(0)|dx\\ &\leq\int_0^{1-\epsilon/2M}|f(x^n)-f(0)|dx+\int_{1-\epsilon/2M}^{1}Mdx\\ &<\int_0^{1-\epsilon/2M}\epsilon/2 dx+\epsilon/2<\epsilon. \end{align}$$ Thus for any continuous $f$, you can move the limit inside the integral when calculating $\lim\limits_{n\to\infty}\int_0^1f(x^n)dx$.

However, you certainly can't do that in the second case for just any continuous $f$, because $\lim\limits_{n\to\infty}{f(\cos nx)}$ may not exist in the $L^1$ sense. For example, if $f$ is the identity function then the sequence of functions $f(\cos nx)=\cos nx$ is not even Cauchy, as $$\begin{align} \|\cos nx-\cos 2nx\|_1 &=\int_0^1|\cos nx-\cos 2nx|dx\\ &\geq \left\lfloor\frac{n}{\pi}\right\rfloor \int_0^{\pi/n}|\cos nx-\cos 2nx|dx\\ &\geq \left\lfloor\frac{n}{\pi}\right\rfloor \int_{\pi/2n}^{\pi/n}|\cos 2nx|dx\\ &\geq \left\lfloor\frac{n}{\pi}\right\rfloor \frac{1}{2n}\int_{\pi}^{2\pi}|\cos x|dx\\ &\geq \left\lfloor\frac{n}{\pi}\right\rfloor \frac{1}{n}\to\frac{1}{\pi} \end{align}$$ and so it cannot converge.

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Thanks Alex! In the second type of problem, are we not able to carry out the integral by performing a change of variables and then dividing the integral into k segments of width n/(2kpi) and then integrating to be left with 0 + sin(n)/n which tends to 0? I figure that for all n the graph will have a certain number of periods, and then a term at the end that has a width that tends to 0. –  Nik Kumar Aug 26 '12 at 13:57
    
The reason I asked these two questions is that I thought they were fundamentally different in the sense that in one the objective is to integrate then apply the limit, and in the other we want to apply the limit then integrate –  Nik Kumar Aug 26 '12 at 13:59
    
@NikKumar I don't quite follow your idea for evaluating the second type of problem. What I would do is change variables from $nx$ to $x$, then break up and evaluate the integrals, and then take the limit. Perhaps that is what you are saying as well. Keep in mind that you can't change variables (in the normal manner, anyway) from $x$ to $\cos nx$ because $\cos nx$ is not increasing on $(0,1)$. –  Alex Becker Aug 26 '12 at 19:12
    
we are saying the same thing about the change of variables. Are you saying we cannot change the variable nx to x and then integrate (1/n)cos(x) from 0 to n? And then proceed by breaking up and evaluating the integrals and applying the limit? –  Nik Kumar Aug 26 '12 at 21:41

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