# Alternate proof of integral equality using MVT

Here is the original problem:

Let $$f:[0,1]\to \mathbb{R}$$ be a positive continuous function. Show that for every positive integer $$n$$ there is a unique $$a_n\in (0,1]$$ with $$\int _0^{a_n} f(x)\,dx = \frac{1}{n} \int _0^1 f(x)\,dx.$$ Further, compute $$\lim_{n\to\infty} n a_n.$$

My work:

Let $$F(x) = \int _0^x f(t)\,dt.$$ We have $$F(0)=0$$ and by the FTC $$F$$ is continuously differentiable, and strictly increasing since $$f$$ is positive. Note also that $$F(1) = \int _0^1 f(x)\,dx = M<\infty.$$ Since $$F$$ is differentiable (hence continuous), by the Intermediate Value Theorem there is some point $$a_2\in (0,1)$$ such that $$F(a_2) = M/2;$$ further, since $$F$$ is increasing, $$a_2$$ is unique. The generalization to $$a_n$$ is immediate.

Since $$F$$ is strictly increasing and differentiable, it has a well-defined inverse on $$[0,1].$$ Then the limit is $$\lim_{n\to \infty} n a_n = \lim _{n\to \infty} n\cdot F^{-1}\left(\frac{M}{n}\right)$$Make the substitution $$u=M/n$$: $$= M\lim _{u\to0^+} \frac{F^{-1}(u)}{u}.$$Since $$F(0)=0,$$ $$F^{-1}(0)=0$$ and since $$F$$ is differentiable, we may use L'Hopital's Rule: $$= M\lim_{u\to 0^+} \frac{d}{du}{F^{-1}(u)} = \lim _{u\to 0^+} \frac{M}{F'(F^{-1}(u))}= \frac{M}{f(F(0))}=\frac{1}{f(0)}\int_0^1 f(x)\,dx.$$

The problem provided the following hint:

First show the existence of $$\{a_n\}.$$ Then show that $$a_n\to 0$$ and $$\int _0^{a_n} f(x)\,dx=a_n f(\xi_n)$$ for some $$0<\xi_n

I think the idea is to use the hint and the First MVT for Integrals to show that $$\lim_{n\to\infty}na_n =\frac{1}{f(0)}\int_0^1 f(x)\,dx$$ but I'd be curious to see how and if there are other methods of proof as well.

• The hint gives you $$na_n = \frac{1}{f(\xi_n)}\int_0^1 f(x)\,dx\,.$$ – Daniel Fischer Oct 29 at 18:29
• Hmm, that's disappointing: I like my way a lot more! ;) – Integrand Oct 29 at 18:30

For the first part, you don't need MVT. The IVT already suffices: let $$I=\int_0^1f(x)dx$$ and $$g(a)=\int_0^af(x)dx$$. Since $$f$$ is positive and continuous, $$g$$ is strictly increasing and continuous. Also, $$g(0)=0<\frac1nI\le g(1)$$. Therefore, by IVT, there exists some $$a_n\in[0,1]$$ such that $$g(a_n)=\frac1nI$$. The value of $$a_n$$ cannot be zero because $$g(0). It is also unique because $$g$$ is strictly increasing.
For the second part, as Daniel Fischer commented, the hint was likely intended to mean that $$na_n=\frac{1}{\xi_n}\int_0^1f(x)dx$$.
Alternatively, note that $$(a_n)$$ is a decreasing sequence that is bounded below. Therefore it converges and $$g\left(\lim_{n\to\infty}a_n\right)=\lim_{n\to\infty}g(a_n)=\lim_{n\to\infty}\frac1nI=0$$. Since $$g>0$$ on $$(0,1]$$, $$\lim_{n\to\infty}a_n$$ must be zero. Therefore, by the first FTC, \begin{aligned} I=\lim_{n\to\infty}I &=\lim_{n\to\infty}na_n\frac{g(a_n)}{a_n}\\ &=\lim_{n\to\infty}na_n\cdot\lim_{n\to\infty}\frac{g(a_n)-g(0)}{a_n-0}\\ &=\lim_{n\to\infty}na_n\cdot\lim_{a\to0}\frac{g(a)-g(0)}{a-0}\\ &=\lim_{n\to\infty}na_ng'(0)\\ &=\lim_{n\to\infty}na_nf(0) \end{aligned} and hence $$\lim_{n\to\infty}na_n=\frac{1}{f(0)}I$$.