$\int_{0}^{1}f(x)g(x)=0 \implies f(x)=0 \ \forall x \in [0,1]$ 
Let $f:[0,1]\to \mathbb{R}$ be a continuous function. If $\int_{0}^{1}f(x)g(x)=0$ for all continuous functions $g(x)$, then $f(x)=0 \ \forall x \in [0,1]$. I would like to know if my proof holds, please.

As the statement holds for every continuous function $g(x)$ on $[0,1]$, then it holds in particular for $g(x)=f(x)$. So, we have to show that $\int_{0}^{1}f^2(x)=0 \implies f(x)=0 \ \forall x \in [0,1]$.
First, as $f$ is continuous on $[0,1]$, $f^2(x)$ is continuous on $[0,1]$ as well. Therefore, $f^2(x)$ has a primitve $F(x)$ (which is differentiable on $[0,1]$). Thus,
$F(1)-F(0)=0\implies F(1)=F(0)$ and $F'(x)=f^2(x)\ge 0$.
As $F'(x)\ge 0$, then $F(x)$ is increasing. We will show now that $F(x)$ is constant.
Consider $0\le x\le 1$. As $F$ is increasing, then $F(0)\le F(x)\le F(1)=F(0)$. So, $F(x)=F(1)=F(0)=cste \ \forall x\in [0,1]$ .
We conclude that $F'(x)=cste'=0=f^2(x) \implies f(x)=0$ as wanted.
 A: that proof looks like it works :)
However, it is overcomplicating things a bit. Let see how we can use the definition of continuity directly. (and this will also be a more helpful proof method for similar problems in analysis)
Suppose there exists an $a$ such that $f(a)$ is not $0$. Then $f^2(a)$ is also non zero. Let's write $f^2(a) = y$. ($y>0$)
The definition of continuity at $a$ states that for all $\epsilon$ there exists a $\delta$ such that if $|x-a|<\delta$ then $|f(x)-f(a)|<\delta$
Now we set $\epsilon = \frac{y}{2}$. We then get a $\delta$ such that for all $x$ in $(a-\delta, a+\delta)$ we also have $f^2(x) > \frac{y}{2}$
Then the integral $\int_{a-\delta}^{a+\delta}f^2(x)dx> 2\delta\times\frac{y}{2}=\delta y >0$
Then we must have also that $\int_a^b{f(x)^2dx}>0$, which contradicts our hypothesis.
A: Just a suggestion : I guess you know that if $f\geq 0$ is continuous, then $\int_0^1 f=0\implies f=0$. Using that :
$$\forall g\in \mathcal C[0,1], \int_0^1 fg=0\implies \int_0^1 f^2=0\implies f=0.$$
A: Your proof looks correct to me.
Here is an alternative way to prove
$$\int_0^1 f = 0 , \quad f \ge 0\implies f = 0.$$
Assume to the contrary that there is $x \in (0,1)$ with $f(x) \ne 0$. By continuity, there is a neighborhood $I = [x-\epsilon, x+\epsilon]\subseteq (0,1)$ with $f(y) \ne 0$ when $y \in I$. Let $m$ be the minimum of $f$ on $I$. Then
$$0 = \int_0^1 f \ge \int_{x-\epsilon}^{x+\epsilon} f \ge m2 \epsilon > 0$$
which is a contradiction.
