Problem involving integration and mean value theorem Let $f:[0,1]\rightarrow \mathbb{R}$ be a continuous function and $\int_0 ^1 f(t)dt =0$. Show that there is a $c\in (0,1)$ such that
$(a)$ $f(c)=\int_0 ^c f(t)dt$.
$(b)$ if $f(0)=0$, then $cf(c)=(1-c)\int_0 ^c f(t)dt$.
$(c)$ $\int_0 ^c tf(t)dt=0$.
Now, using the functions $h(t)=e^{-t}\int_0 ^t f(u)du$ and $h(t)=\frac{e^{t}}{t}\int_0 ^t f(u)du$ and using MVT, $(a)$ and $(b)$ can be proved. But I can not see how to find a solution for $(c)$. How should I proceed?
Thank you.
 A: Okay, after a couple of almost sleepless nights, I think I'm there. I'll be happy if someone finds a mistake.
Let $G(x)=\int_0 ^x tf(t)dt=x\int_0 ^x f(t)dt -\int_0 ^x \Big( \int_0 ^u f(t)dt\Big)du=x^2\frac{d}{dx}\Big(\frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big)du\Big)$, $x\in(0,1)$.
If possible, let $G$ be never zero in $(0,1)$. Continuity of $G$ suggests that we can assume $G(x)>0$ for all $x\in (0,1)$ (we can always replace $f$ by $-f$). Since $G>0$, we have $\int_0 ^x f(t)dt >\frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big) du$ for all $x\in (0,1)$.
Also we have $\frac{d}{dx}\Big(\frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big)du\Big)>0$ for all $x$. So for $x,y\in (0,1)$ with $x>y$ we have
$$\frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big)du>\frac{1}{y}\int_0 ^y \Big(\int_0 ^u f(t)dt\Big)du.$$
Now, let us choose $x_1,x,y,y_1\in (0,1)$ such that $x_1 >x>y>y_1$. So using the inequalities above, we find that
$$\int_0 ^{x_1} f(t)dt>\frac{1}{x_1}\int_0 ^{x_1} \Big(\int_0 ^u f(t)dt\Big)du\\>\frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big)du\\>\frac{1}{y}\int_0 ^y \Big(\int_0 ^u f(t)dt\Big)du\\
>\frac{1}{y_1}\int_0 ^{y_1} \Big(\int_0 ^u f(t)dt\Big)du.$$
Now, we keep these arbitrarily chosen $x$ and $y$ fixed and we let $x_1 \rightarrow 1$ and $y_1 \rightarrow 0$ (recall that we have chosen $x_1,x,y,y_1\in (0,1)$ such that $x_1 >x>y>y_1$). This gives (for the expression involving $y_1$, apply L'Hospital or the Mean Value Theorem)
$$0\geq \frac{1}{x}\int_0 ^x \Big(\int_0 ^u f(t)dt\Big)du>\frac{1}{y}\int_0 ^y \Big(\int_0 ^u f(t)dt\Big)du\geq 0.$$
This is obviously a contradiction. Hence, there must be a point $c\in (0,1)$ such that $\int_0 ^c tf(t) dt=0$
