# Olympiad calculus problem

This problem is from a qualifying round in a Colombian math Olympiad, I thought some time about it but didn't make any progress. It is as follows.
Given a continuous function $f : [0,1] \to \mathbb{R}$ such that $$\int_0^1{f(x)\, dx} = 0$$ Prove that there exists $c \in (0,1)$ such that $$\int_0^c{xf(x) \, dx} = 0$$ I will appreciate any help with it.

• Where is the $c$ being used? Apr 24, 2011 at 0:56
• @moron: I had a typo, I already corrected it Apr 24, 2011 at 0:57
• @Theo Buehler: Bad misconception on my part. Thanks for the heads up. I realized I was thinking of analytic functions on compact sets, rather than continuous functions. Apr 24, 2011 at 1:58

## 4 Answers

[Major edit to make the second part of the proof a lemma]

Define:

$$F(x) = \int_0^x{f(t)\, dt}$$

Then $F(0)=F(1)=0$, and $F'(x)=f(x)$

Integrating by parts, we see that:

$$\int_0^t{xf(x) \, dx} = t F(t) - \int_0^t{F(x)\, dx}$$

To prove the theorem, we must find a $c \in (0,1)$ such that:

$$F(c) = {\frac{1}{c}}\int_0^c{F(x)\, dx}$$

This is shown with via the following:

Lemma: If $F$ is a continuous function on $[0,1]$ such that $F(0)=F(1)$ then there is a $c\in (0,1)$ such that: $$F(c) = {\frac{1}{c}}\int_0^c{F(x)\, dx}$$

Proof: Define:

$$G(t) = {\frac{1}{t}}\int_0^t{F(x)\, dx}$$

$G(t)$ is continuous and defined on $[0,1]$ (the value at $t=0$ is defined as the limit, and is just $F(0)$ by the fundamental theorem of calculus.)

$G(t)$ is the average of $F(x)$ for $x\in(0,t)$, so $G(t)$ must have as an upper bound the upper bound of $F$. That is, for all $t\in [0,1]$, $G(t)\leq \operatorname{max}_{x\in[0,1]} F(x).$

Since $F$ is continuous, there must be an $x_M \in [0,1]$ such that $F(x_M)= \operatorname{max}_{x\in[0,1]} F(x)$. Then, setting $t=x_M$, we see that:

$$G(x_M)\leq \operatorname{max}_{x\in[0,1]} F(x) = F(x_M)$$

Similarly, we have that $G(t)\geq \operatorname{min} F(x)$, and thus, when $F(x_m) = \operatorname{min}_{x\in[0,1]} F(x)$, we have $G(x_m)\geq F(x_m)$.

So the continuous function $H(x)=G(x)-F(x)$ has the property that $H(x_M)\leq 0$ and $H(x_m)\geq 0$. By the intermediate value theorem, there must be a $c$ between $x_m$ and $x_M$, inclusive, such that $H(c)=0$, and hence $F(c)=G(c)$.

If we can find $x_m$ and $x_M$ in $(0,1)$ - that is, not on the boundary - then we know that $c\in (0,1)$ and we are done.

If both $x_m$ and $x_M$ are on the boundary, then the maximum and minimum of $F$ are equal, and thus $F$ is constant, and hence we can choose any $c$.

So assume that $F(0)=F(1)$ is the minimum value for $F$, and that $F$ does not take the minimum value anywhere else on $[0,1]$. Then $F(x)>F(1)$ for all $x\in(0,1)$, so we know that $G(1)>F(1)$. So $H(1)>0$ and $H(x_M)\leq0$. Hence, there must be a $c$ in $[x_M,1]$ with $H(c)=0.$ But $c\neq 1$, so we know that $c \in [x_M,1)\subset(0,1)$.

• I did not understand the very last sentence. On which two points do you use the intermediate value theorem? Apr 24, 2011 at 9:14
• @user9325 Clarified my answer. Apr 24, 2011 at 9:30
• Very nice. (You added a little typo, though: $F(1)=0$). Apr 24, 2011 at 9:33
• @user9325 Thanks.. Fixed typo. Apr 24, 2011 at 9:41

This is a streamlined version of Thomas Andrews' proof:

Put $F(x):=\int_0^x f(t)dt$ and consider the auxiliary function $\phi(x)={1\over x}\int_0^x F(t)dt$. Then $\phi(0)=0$, $\ \phi(1)=\int_0^1 F(t)dt=:\alpha$, and by partial integration one obtains $$\phi'(x)=-{1\over x^2}\int_0^xF(t)dt +{1\over x}F(x)={1\over x^2}\int_0^x t f(t)dt\ .$$ The mean value theorem provides a $\xi\in(0,1)$ with $\phi'(\xi)=\alpha$. If $\alpha$ happens to be $0$ we are done. Otherwise we invoke $F(1)=0$ and conclude that $\phi'(1)=-\alpha$. It follows that there is a $\xi'\in(\xi,1)$ with $\phi'(\xi')=0$.

• That's very slick! I think it might be a case where the slickness obscures what's going on, but slick has its own aesthetic merits. Apr 27, 2011 at 15:59
• @Christian $\phi(0)$ is defined? Nov 21, 2019 at 20:11
• @Lucas: You have $|F(t)|\leq M t$ $(t\geq0)$ for some $M$ and therefore $\left|\int_0^x F(t)\>dt\right|\leq{M\over2} x^2$. This implies $\lim_{x\to0+}\phi(x)=0$. Nov 22, 2019 at 9:22

Let $$G(x) = \int_{0}^{x} t f(t) \, \mathrm{d}t$$, and suppose that $$G(c) \neq 0$$ for all $$c \in (0, 1)$$. Then by IVT, either

• (Case 1) $$G > 0$$ on $$(0, 1)$$, or
• (Case 2) $$G < 0$$ on $$(0, 1)$$.

However,

\begin{align*} 0 = \int_{0}^{1} f(x) \, \mathrm{d}x = \int_{0}^{1} \frac{G'(x)}{x} \, \mathrm{d}x = G(1) + \int_{0}^{1} \frac{G(x)}{x^2} \, \mathrm{d}x. \end{align*}

This is $$> 0$$ in Case 1 and $$< 0$$ in Case 2, a contradiction.

The problem has originally been asked in the 2006 Romanian National Olympiad.

Define a function $$\mathcal{G}(y)=y\int_0^{y}f(x)dx-\int_0^{y}xf(x)dx$$

Using chain rule and Leibnitz rule on differentiation under integral sign we get $$\mathcal{G}\prime(y)=\int_0^yf(x)dx$$

On inspection we get that $$\mathcal{G}\prime(1)=\mathcal{G}\prime(0)=0.$$

Now we will use Flett's variant of LMVT that for some $$c\in(0,1)$$ $$\quad\mathcal{G}\prime(c)=\frac{\mathcal{G}(c)-\mathcal{G}(0)}{c}$$

$$\implies c\int_0^cf(x)dx=c\int_0^{c}f(x)dx-\int_0^{c}xf(x)dx\implies\int_{0}^{c}xf(x)dx=0.$$