Prove the existence of a point $c$. Problem
Let $f$ be a continuous function, $f:[0,1]\to\mathbb{R}$ with $\int_{0}^1 (2x-1)f(x) dx = 0$.
Prove that there exists a point c between $(0, 1)$ such that $\int_{0}^c (x-c)(f(x)-f(c)) dx = 0$.
This problem was given at a regional competition in Romania for $12$th grade students.
Attempt
From Rolle's Theorem we have that there exists q such that $f(q) = (2q - 1)f(q) = 0$, where we have $f(q)=0$.
My other attempt was: fix $p$ to $0.5$ and separate according to the Mean Value Theorem for Integrals $\int_{0}^1 (2x-1)f(x) dx = 0$ into $-f(c1) * \int_{0}^p (1-2x) dx$ $+ $ $f(c2) * \int_{p}^1 (2x-1) dx = 0$ where we get that $f(c1)=f(c2)$
 A: Meaning no disrespect to the existing answer, there is a much simpler solution.
Consider the continuously differentiable function $\Lambda:[0,1]\to\Bbb R$ given by: $$c\mapsto\int_0^c(2xc-c^2)f(x)\,\mathrm{d}x$$
$\Lambda(0)=0=\Lambda(1)$ by assumption. By “Rolle’s” theorem, $\Lambda’(c)=0$ for some distinguished $c$ in $(0,1)$. For this $c$, we then know (Leibniz integral rule): $$0=(2c^2-c^2)f(c)(1)+\int_0^c\partial_c[(2xc-c^2)f(x)]\,\mathrm{d}x=2\int_0^c(x-c)(f(x)-f(c))\,\mathrm{d}x$$As desired.
A: First, let's take a look at our conclusion:
$\int_0^c(x-c)(f(x)-f(c))dx = 0 \Leftrightarrow \int_0^cxf(x)dx - \int_0^cxf(c)dx - \int_0^c cf(x)dx + \int_0^c cf(c)dx = 0 \Leftrightarrow \int_0^cxf(x)dx - f(c)\int_0^cxdx - c\int_0^cf(x)dx + c^2f(c) = 0 \Leftrightarrow \int_0^cxf(x)dx - c\int_0^cf(x)dx + \frac{c^2f(c)}{2} = 0$
This is essentially what we need to prove. Therefore, consider $F : [0, 1] \rightarrow \mathbb{R}, F(x) = \int_0^x f(t)dt$. One can immediately see that $F$ is differentiable, and that $F'(x) = f(x), \forall x \in [0, 1]$.
$\int_0^x tf(t)dt = \int_0^xtF'(t)dt = xF(x) - \int_0^xF(t)dt$
At this step, look at what we need to prove. Subtract $xF(x)$ and add $\frac{x^2f(x)}{2}$. Hence, we obtain:
$\int_0^xtf(t)dt - xF(x) + \frac{x^2f(x)}{2} = \frac{x^2f(x)}{2} - \int_0^xF(t)dt$
One can easily see that the left side of this equality is in fact equivalent to what we have to prove. Furthermore, proving the existence of $c \in (0, 1)$ for which the conclusion holds is in fact equivalent to proving that such a $c$ exists so that $\frac{c^2f(c)}{2} - \int_0^cF(t)dt = 0$.
At this step, consider $G, H : [0, 1] \rightarrow \mathbb{R}, G(x) = \int_0^xF(t)dt, H(x) = \frac{x^2f(x)}{2} - \int_0^xF(t)dt = \frac{x^2}{2}G''(x) - G(x)$. Note that $G$ is twice differentiable on $[0, 1]$. At this point in the proof, we need to prove that there exists a $c \in (0, 1)$ such that $H(c) = 0$.
Consider the function $g : [0, 1] \rightarrow \mathbb{R}, g(x) = \frac{x^2}{2}G'(x) - xG(x)$. The new defined $g$ is differentiable on $[0, 1]$, but what's more interesting is that $g'(x) = H(x), \forall x \in (0, 1)$. Now, $g(0) = 0$ for obvious reasons. We are left to prove that $g(1) = 0$.
$g(1) = \frac{G'(1)}{2} - G(1)$. From its definition, $G'(x) = F(x)$, therefore $G'(1) = F(1)$.
Recall the hypothesis: $\int_0^1(2x-1)f(x)dx = 0 \Leftrightarrow 2\int_0^1xf(x)dx = \int_0^1f(x)dx \Leftrightarrow \int_0^1xf(x)dx = \frac{1}{2}\int_0^1f(x)dx$
Therefore, $G(1) = \int_0^1F(t)dt = \int_0^1t'F(t)dt = F(1) - \int_0^1tf(t)dt = F(1) - \frac{1}{2}\int_0^1f(t)dt = F(1) - \frac{F(1)}{2} = \frac{F(1)}{2}$
The last equality comes directly from making use of the hypothesis. Therefore, $g(1) = 0$.
Since $g(0) = g(1) = 0, g$ is differentiable on $(0, 1)$ and continuous on $[0, 1]$, from Rolle's theorem it follows that there exists $c \in (0, 1)$ such that $g'(c) = 0 \Leftrightarrow H(c) = 0$, thus concluding our proof.
Note: As a general rule of thumb, whenever faced with such problems where an initial condition between $\int_0^1f(x)dx$ and $\int_0^1xf(x)dx$ is given, the idea is to follow the same approach until finding $H(x)$. Afterward, it's all a matter of finding a helpful function to apply Rolle's theorem.
