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$f$ is positive continuous function on $[0,1]$. Define $$\int_{0}^{a_n} f(x) dx = \frac{1}{n} \int_{0}^1 f(x) dx$$ where $a_n>0$. Find $ \lim_{n\to \infty} n a_n$.

It is clear that $lim_{n\to \infty} a_n =0$ because $f(x)$ is positive.I tried to use Weierstrass approximation of continuous function by polynomials but could no quite get the right way.I do not see a way to bring down this equation $\int_{0}^{a_n} f(x) dx = \frac{1}{n} \int_{0}^1 f(x) dx$ to $n a_n$.

This is a qualifying problem of real analysis. Small hint works for me.

Thanks.

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HINT: Call $F$ a primitive of $f$; thus your relation yelds to $$ F(a_n)-F(0)=\frac1n[F(1)-F(0)] $$

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  • $\begingroup$ I have tried that one. But let me see one more time. $\endgroup$ – mp100 Jul 14 '15 at 23:27
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HINT:

Find the limit

$$\lim_{n\to \infty}\left(\frac{1}{a_n}\int_{0}^{a_n}\,f(x)\,dx\right)$$

SPOILER ALERT: SCROLL TO SEE ANSWER

$$\begin{align}\lim_{n\to \infty}n\,a_n&=\lim_{n\to \infty}n\,a_n\frac{\int_0^{a_n}f(x)dx}{\int_0^{a_n}f(x)dx}\\\\&=\lim_{n\to \infty}\frac{n\,\int_0^{a_n}f(x)dx}{\frac{1}{a_n}\int_0^{a_n}f(x)dx}\\\\&=\lim_{n\to \infty}\frac{\int_0^{1}f(x)dx}{\frac{1}{a_n}\int_0^{a_n}f(x)dx}\\\\&=\frac{1}{f(0)}\int_0^{1}f(x)dx\end{align}$$

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    $\begingroup$ A $n$ got lost :) $\endgroup$ – user251257 Jul 14 '15 at 23:40
  • $\begingroup$ @user251257 Thanks! A pitfall of using a "smart phone" to post. $\endgroup$ – Mark Viola Jul 15 '15 at 0:29
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Let $u = nx \to du = ndx \to \dfrac{1}{n}\displaystyle \int_{0}^1f(x)dx= \displaystyle \int_{0}^{a_n} f(x)dx = \displaystyle \int_{0}^{na_n} f\left(\dfrac{u}{n}\right)\left(\dfrac{1}{n}du\right)=\dfrac{1}{n}\displaystyle \int_{0}^{na_n} f\left(\dfrac{u}{n}\right)du\Rightarrow \displaystyle \int_{0}^1 f(x)dx = \displaystyle \int_{0}^{na_n} f\left(\dfrac{u}{n}\right)du$. Now let $L = \displaystyle \lim_{n\to\infty} na_n$, and observe due to continuity of $f$ at $x = 0$, $\displaystyle \lim_{n\to \infty} f\left(\dfrac{u}{n}\right) = f(0)$, we have: $\displaystyle \int_{0}^1 f(x)dx = \displaystyle \int_{0}^L f(0)du = Lf(0) \to L = \dfrac{\displaystyle \int_{0}^1 f(x)dx}{f(0)}$

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  • $\begingroup$ For the proof to work, you need to know that $L$ exists and $L<\infty$. Little bit wonky. $\endgroup$ – user251257 Jul 14 '15 at 23:43
  • $\begingroup$ Mean value theorem on primitive of f makes theorem much more easy $\endgroup$ – mp100 Jul 15 '15 at 2:28
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Let $F(x)=\int_0^x f(y) dy$. Then $F$ is a strictly increasing function satisfying $F(a_n)=\frac{1}{n} F(1)$. This means that $\lim_{n \to \infty} a_n = 0$. Hence $F(a_n)=f(0) a_n + o(a_n)$ (which is what the FTC gives) is a useful approximation. So

$$f(0) a_n + o(a_n) = \frac{1}{n} F(1) \Rightarrow n a_n = \frac{F(1)}{f(0)} + o(n a_n).$$

Move the error term to the other side:

$$(1+o(1)) n a_n = \frac{F(1)}{f(0)}.$$

Take limits on both sides:

$$\lim_{n \to \infty} n a_n = \frac{F(1)}{f(0)}.$$

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Hint: You might want to consider the limit of $$ \frac{1}{a_n} \int_0^{a_n} f(x) \mathrm d x = \frac{1}{a_n n} \int_0^1 f(x) \mathrm d x $$ and use $a_n\to 0$ for $n\to\infty$.

Since $f$ is positive and continuous, we have $\frac{1}{a_n} \int_0^{a_n} f(x) \mathrm d x > 0$ and by fundamental theorem of calculus we have $$ \lim_{n\to\infty} a_n n = \frac{\int_0^1 f(x) \mathrm d x}{\lim_{n\to\infty}\frac{1}{a_n} \int_0^{a_n} f(x) \mathrm d x} = \frac{\int_0^1 f(x) \mathrm d x}{f(0)}. $$

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  • $\begingroup$ This one works. Thanks $\endgroup$ – mp100 Jul 15 '15 at 1:01
  • $\begingroup$ No wonder it got a(n) anonymous and unexplained downvote. $\endgroup$ – Gary. Jul 15 '15 at 2:21
  • $\begingroup$ @Gary.: What do you mean? $\endgroup$ – user251257 Jul 15 '15 at 2:23
  • $\begingroup$ @user251257: Just a sarcastic comment on anonymous downvotes/downvoters. I wish I was registered and could upvote your, nice answer. I will do so when I register. $\endgroup$ – Gary. Jul 15 '15 at 2:25
  • $\begingroup$ @user251257: I have received a few anonymous downvotes myself. Instead of asking you to clarify, they automatically assume you're wrong and put you on the defensive. $\endgroup$ – Gary. Jul 15 '15 at 2:28

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