# $f\in C[0,\infty]$ and $\lim\limits_{x\to \infty}f(x)=L<\infty$. Compute $\lim\limits_{n\to \infty} \int_{0}^{2} f(nx)dx$

I'd really love your help with this

Let $f$ be a continuous function in $[0,\infty )$ and assume that $\lim\limits_{x\to \infty} f(x)=L<\infty$. I need to compute:

$$\lim_{n\to \infty} \int_{0}^{2} f(nx)dx.$$

Because of the fact that $f$ is continuous I want to insert the $\lim$ into the integral (Am I allowed to? What conditions?), and basically to write : $\lim\limits_{n\to \infty} \int_{0}^{2} f(nx)dx=\int_{0}^{2}\lim\limits_{n\to \infty} f(nx)dx=\int_{0}^{2}\ L dx=2L$. I'm quite sure that it's wrong. What are my mistakes, and why am I not allowed to do so? what is the correct solution?

Thanks a lot!

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You can solve this problem just using definition of the limit.

Let us denote $I_n = \int\limits_0^2f(nx)\,dx = \frac1n\int\limits_0^{2n} f(x)dx$. Since $\lim\limits_{x\to\infty}f(x) = L<\infty$, for any $\varepsilon>0$ there is $x(\varepsilon)$ s.t. for all $x>x(\varepsilon)$: $$L-\varepsilon\leq f(x)\leq L+\varepsilon.$$

Let us fix $\varepsilon>0$ and pick up any $n>x(\varepsilon)$, then $$I_n = \frac1n\int\limits_{0}^{x(\varepsilon)}f(x)dx+\frac1n\int\limits_{x(\varepsilon)}^{2n}f(x)dx\quad (1)$$ and so $$\frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L-\varepsilon)\leq I_n\leq \frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L+\varepsilon)\quad (2)$$ where $J(\varepsilon)= \int\limits_{0}^{x(\varepsilon)}f(x)dx$. In other words, $$\frac1n(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))-2\varepsilon\leq I_n-2L\leq \frac1n(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))+2\varepsilon.$$

For any $\delta$ we pick up $\varepsilon<\delta$ and $N>\frac{3}{\delta}(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))$, so for any $n\geq N$ we have $$|I_n-2L|\leq \delta,$$ so $\lim\limits_{n\to\infty}I_n = 2L$.

Let me show how $(1)$ implies $(2)$: $$I_n = \frac1n J(\varepsilon)+\frac1n\int\limits_{x(\varepsilon)}^{2n}f(x)dx\leq \frac1n J(\varepsilon) +\frac{2n-x(\varepsilon)}{n}(L+\varepsilon)=\frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L+\varepsilon).$$ where the inequality holds because $f(x)\leq L+\varepsilon$ for all $x\geq x(\varepsilon)$. The other inequality in $(2)$ is obtained in a similar way since $f(x)\geq L-\varepsilon$ for all $x\geq x(\varepsilon)$.

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Thank you for the answer. Can you please explain what did you do after the "and so" in the 7Th row? – Jozef Nov 29 '11 at 12:42
@S.D.: I was wondering whether I didn't already cover the part "version 2" in my answer? – Rudy the Reindeer Nov 29 '11 at 12:53
@Matt: oh, I am sorry - seems like you put it 7 minutes before I finished editing my answer, so I've missed it. I will eliminate Version 2 from my answer not to clash with yours. – S.D. Nov 29 '11 at 13:11
@Jozef: please, see the bottom of the edited version. – S.D. Nov 29 '11 at 13:20
@S.D.: That's nice of you, thanks! – Rudy the Reindeer Nov 29 '11 at 13:26

You can use Lebesgue's dominated convergence theorem. For this you need that $f$ is dominated by some integrable function $g$, i.e. $\int_0^2 |g| dx < \infty$ and $| f_n (x) | := |f(nx) | \leq g(x)$ for all $n$ and for all $x$ in the set over which you integrate, in your case all $x \in [0,\infty)$.

$f$ is continuous and bounded. So you can pick $g(x) := \sup\limits_{x \in [0,\infty)} |f(x)|$; then,

$$f_{n}(x) \leq |f_{n}(x)| \leq g(x)$$

and by the dominated convergence theorem,

$$\lim_{n \rightarrow \infty} \int_0^2 f(nx) dx = \int_0^2 \lim_{n \rightarrow \infty} f(nx) dx = \int_0^2 L dx = 2L$$

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Since $f$ is continuous and since $\lim\limits_{x\rightarrow\infty} f(x)=L$, it follows that $|f|$ is bounded by some constant $M$ on all of $[0,\infty)$.

Let $\epsilon>0$. Set $\delta={\epsilon\over M}$. Choose $N$ so that for any $n>N$ and any $x>\delta$, $L-\epsilon\le f(nx)\le L+\epsilon$.

Now $$\int_0^2 f(nx) =\int_0^\delta f(nx) +\int_\delta^2 f(nx) .$$ For $n>N$ $$(2-\delta)(L-\epsilon)\le\int_\delta^2 f(nx) \le (2-\delta)(L+\epsilon).$$

It follows that $$\int_0^\delta f(nx) + (2-\delta)(L-\epsilon) \le \int_0^2 f(nx) \le\int_0^\delta f(nx) + (2-\delta)(L+\epsilon);$$ whence $$-\delta M+ (2-\delta)(L-\epsilon) \le \int_0^2 f(nx) \le\delta\cdot M+ (2-\delta)(L+\epsilon).$$ From this, we have $$-\epsilon+ (2-\delta)(L-\epsilon) \le \int_0^2 f(nx) \le\epsilon+ (2-\delta)(L+\epsilon),$$ for all n>N. But then $$2L+{\epsilon^2\over M}-3\epsilon-{\epsilon\over M}L \le \int_0^2 f(nx) \le 2L-{\epsilon^2\over M}+3\epsilon-{\epsilon\over M}L ,$$for all n>N.

Since $L$ and $M$ are fixed, the result follows.

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