If $f,g\in\mathcal C^1[0,1],\,f$ monotone, and $g(x)>g(1)=g(0)$ on $(0,1)$, then $\int_0^1 f(x)g'(x)\,dx=0$ if and only if $f$ is constant The Problem: Let $f,g$ be continuously differentiable on $[0,1],\,f$ monotone, and $g(x)>g(0)=g(1)$ on $(0,1).$ Prove that
$$\int_0^1 f(x)g'(x)\,dx=0\quad\text{if and only if }f\text{ is constant.}$$
My Thoughts: I first try the easy direction. So suppose that $f$ is constant, hence there is some $c\in\mathbb R$ such that $f(x)=c$ for all $x\in[0,1]$. Then the fundamental theorem of calculus implies that
$$\int_0^1 f(x)g'(x)\,dx=\int_0^1 cg'(x)\,dx=c[g(1)-g(0)]=0.$$
However, I am having difficulty with the other direction. I tried applying integration by parts in the following way
$$0=\int_0^1 f(x)g'(x)\,dx=g(0)\left[f(1)-f(0)\right]-\int_0^1 f'(x)g(x)\,dx.$$
Then the Mean Value Theorem implies that there is some $d\in(0,1)$ such that
$$f'(d)=\frac{1}{g(0)}\int_0^1 f'(x)g(x)\,dx.$$
I did the above with the idea of showing that $f(1)=f(0)$, which would yield the conclusion. But, I have been stuck for a long time at this point.
Could anyone please give me a hint on how to get going from the point I am at, or if the above is not a correct path, just a small hint on how to start on the right path?
Thank you for your time, and really appreciate all feedback.
 A: You're almost there. Since $f$ is monotone then either $f'\geq 0$ or $f'\leq 0$ on all of $[0,1]$. Let us say the latter is true (otherwise change $f$ by $-f$). Then we have $-f'(x)g(x)\geq f'(x)g(0)$ on $[0,1]$, and
\begin{align*}
0&=g(0)[f(1)-f(0)]-\int_0^1 f'(x)g(x)dx\\
&\geq g(0)[f(1)-f(0)]-\int_0^1f'(x)g(0)dx\\
&=0\end{align*}
so the inequality in the middle is actually an equality. This means that
$$\int_0^1f'(x)g(x)dx=\int_0^1f'(x)g(0)dx$$
or equivalently
$$\int_0^1 f'(x)(g(x)-g(0))dx=0.$$
The function $x\mapsto f'(x)(g(x)-g(0))$ is non-positive on $[0,1]$ and has integral $0$, so it must be $0$ on $[0,1]$. Since $g(x)\neq g(0)$ on $(0,1)$ then $f'=0$ on $(0,1)$, and by the Mean Value Theorem $f$ is constant.
A: You need to use the hypothesis $g(x)>g(0)$. Assuming $f$ is nondecreasing, one has $f'(x)g(x)\geq f'(x)g(0)$ :
$$\int_0^1 f'(x) g(x) \, \mathrm{d}x \geq \int_0^1 f'(x)g(0) \,\mathrm{d}x = g(0) [f(1)-f(0)]
$$
Now the above inequality is actually an equality when $f'(x) g(x) = f'(x) g(0)$ for every $x\in (0,1)$. But since $g(x)>g(0)$, that means $f'(x)=0$ for every $x$.
Using this in combination with your integration by parts yields the result.
A: hint
Assume $ f $ increasing and put
$$F(x)=f(x)-f(0)$$
and$$\; G(x)=g(x)-g(0)$$
then
$$\int_0^1F(x)G'(x)dx=0$$
$$=\Bigl[F(x)G(x)\Bigr]_0^1-\int_0^1F'(x)G(x)dx$$
$$= 0-\int_0^1F'(x)G(x)dx$$
Now, observe that $$\forall x\in[0,1]\;\; F'(x)G(x)\ge 0$$
