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This seems like a simple enough problem, but I'm having some trouble. Here it is:

Suppose $f:[0,\infty)\rightarrow \mathbb{R} $ is continuous, and $\lim\limits_{x\rightarrow \infty}f(x)=L.$ Then $\lim\limits_{n \rightarrow \infty}\int_0^1 f(nx)dx=L.$

My thoughts:

  1. First, I think since $f(x)$ is continuous on $[0,\infty)$, $f(nx)$ is continuous on $[0,1]$ (I haven't proved this), and hence integrable on $[0,1]$.

  2. Then, if we can show $f(nx)$ converges to $L$ uniformly on $[0,1]$, then we may interchange the limit and integral as $$\lim\limits_{n \rightarrow \infty}\int_0^1 f(nx)dx=\int_0^1 \lim\limits_{n \rightarrow \infty}f(nx)dx=\int_0^1Ldx=L. $$

Your thoughts? Thanks in advance for your suggestions.

(This is not homework)

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The sequence $(f(nx))$ converges to the function $h$ defined by $$h(x)=\left\{\begin{array}{cl}\\L&\quad\text{if}\ 0<x\leq1\\ f(0)&\quad\text{if}\ x=0 \end{array}\right.$$ and since the continuous function has a finite limite at $+\infty$ then it's bounded on the interval $[0,+\infty)$ so there's $M>0$ s.t. $$|f(nx)|\leq M \quad \forall\ 0\leq x\leq 1$$ hence we apply the dominated convergence theorem and we have $$\lim_{n\to\infty}\int_0^1f(nx)dx=\int_0^1h(x)dx=L.$$

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  • $\begingroup$ Where'd you get your extensive gravatar wardrobe! You look "smashing"! $\endgroup$
    – amWhy
    Jun 2 '14 at 12:20
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We can change varibles to $y=nx$ to rewrite the integral as $$S_n=\frac{1}{n}\int_0^n f(y)dy.$$ If you define $$a_k=\int_{k-1}^k f(y)dy,$$ we get $S_n=\frac{1}{n}\sum_{k=1}^n a_k$. This is known as the Cesaro Mean (or Cesaro sum) of the sequence $a_k$.

Now, it's easy to show that $a_k\rightarrow L$ and there's a result that says that, when this is the case, $S_n\rightarrow L$, as we wanted to show.

This result is not so hard to prove: given $\epsilon>0$ take $K$ such that $k\geq K\Rightarrow |a_k-L|<\epsilon/2$. For $n\geq K$, we can write $$|S_n-L|\leq\frac{1}{n}\sum_{k=1}^{K-1} |a_k-L|+\frac{1}{n}\sum_{k=K}^n |a_k-L|.$$ Now, the term on the right is less than $\epsilon/2$ by our choice of $K$. We can then choose $n$ sufficiently large to make the term on the left as small as we wish, say smaller than $\epsilon/2$. Thus we get that, for $n$ sufficiently large, we have $|S_n-L|<\epsilon$, which proves the result.

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$(f(nx))$ need not be a uniformly convergent sequence. $f(x)=\arctan x$ furnishes a counterexample.

But, one way to solve your problem is to make the substitution $u=nx$ in the integral. This leaves you with having to evaluate $\lim\limits_{n\rightarrow\infty} {1\over n}\int_0^n f(u)\,du$. As $f$ is continuous, L'Hôpital's rule can be applied here (the rule applies when the denominator has infinite limit), and proves fruitful.


Another, and perhaps better, way to approach the problem is outlined as follows:

Let $\epsilon>0$.

$\ \ \ 1)$ The hypotheses imply that $f$ is a bounded function. Thus, there is a $\delta>0$ so that $$\biggl|\,\int_0^\delta f(nx)\, dx\,\biggr| <\epsilon/2$$ $\ \ \ \phantom{1)}$ for all $n$.

$\ \ \ 2)$ Now use the hypothesis that $\lim\limits_{x\rightarrow\infty} f(x)=L$ to choose $N$ so large that for $n>N$ you have $$L-\epsilon/2<\int_\delta^1 f(nx)\,dx<L+\epsilon/2.$$

Now put $1)$, $2)$, and the arbitrary nature of $\epsilon$ together.

(Sketching the graphs of $f(nx)$ for $f(x)=\arctan x$ might make the motivation behind the ideas in the above clear.)

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