A functional equation on continuous and monotone functions

Question

Let $f: [0,1] \to \mathbb{R}$ be a continous increasing function with $\int_0^1 f(x)dx=0$. Prove that if there exists a nonconstant continuous decreasing function $\phi:[0,1]\to \mathbb{R}$ such that $\int_0^1 f(x) \phi(x) dx=0$, then $f(x)=0, \forall x\in [0,1]$.
I began by assuming that there is some $x_0 \in [0,1]$ such that $f(x_0)\ne 0$. I considered the case when $f(x_0)>0$. Then we would have that $f(x)\ge f(x_0)>0, \forall x\in [x_0, 1]$.
From the MVT for integrals we know that $\exists c\in [0,1]$ such that $f(c)=0$. Obviously, this $c$ is now going to be in $[0,x_0)$ and since $f$ is increasing we would have that $f(x)\le f(c)=0, \forall x\in [0,c]$.
This doesn't really give us much. I also tried to consider $\alpha= \min \{x| f(x)>0\}$, but I couldn't make any further progress.

Answer 1

If $f$ is constant then $f=0$ from the first equation, so suppose $f$ is non constant.

We must have $f(0) <0, f(1)>0$, so choose some $x^*$ such that $f(x^*) = 0$.

Then let $g(x) = \phi(x)-\phi(x^*)$. Note that $\int g f = 0$ and $gf \le 0$ hence we have $gf = 0$.

In particular, we must have $g(0) = g(1) = 0$ and so $\phi$ must be a constant which is a contradiction.