About proof of fundamental lemma of calculus of variation I have found several proofs of fundamental lemma of calculus of variation for $C^1$ functions, but could someone help me to show this lemma with the following hypothesis?:
Let be $g:[x,y]\rightarrow{\mathbb{R}}$ a continuous function such that $\int_x^yg(t)h(t)=0$  $\forall{h\in{C^\infty_c(x,y)}}$, show that $g=0$.
Thanks.
 A: Preparation:
Consider the function
$$
h(x)=\begin{cases} exp\left(\frac1{x^2-1}\right) & |x|<1\\0 & otherwise\end{cases}
$$
Then $h\in C_c^\infty(\mathbb R)$, $h(x)>0$ for $|x|<1$ and $supp(h)=[-1,1]$.
For $x_0\in\mathbb R$ and $\varepsilon>0$ you can consider $h_{x_0,\varepsilon}(x):=h\left(\frac{x-x_0}\varepsilon\right)$. 
Then we get $h_{x_0,\varepsilon}\in C_c^\infty(\mathbb R)$, $h_{x_0,\varepsilon}(x)>0$ for $|x-x_0|<\varepsilon$ and $supp(h_{x_0,\varepsilon})=[x_0-\varepsilon,x_0+\varepsilon]$.
Proof of the lemma:
Assume $g\not\equiv 0$. Then there exists $x_0\in(x,y)$ such that $g(x_0)\neq 0$. Wlog $g(x_0)>0$. By continuity of $g$, there exists $\varepsilon>0$ such that $g(\xi)>0$ for all $\xi\in[x_0-\varepsilon,x_0+\varepsilon]\subset (x,y)$.
Now, we choose $h_{x_0,\varepsilon}\in C_c^\infty(x,y)$. 
We deduce $gh_{x_0,\varepsilon}$ is continuous with $(gh_{x_0,\varepsilon})(\xi)> 0$ for $|\xi-x_0|<\varepsilon$ and $supp(gh_{x_0,\varepsilon})=[x_0-\varepsilon,x_0+\varepsilon]$.
Finally, we conclude
$$
\int_x^y gh_{x_0,\varepsilon}=\int_{x_0-\varepsilon}^{x_0+\varepsilon}\underbrace{gh_{x_0,\varepsilon}}_{\geq 0}\geq \int_{x_0-\frac12\varepsilon}^{x_0+\frac12\varepsilon}\underbrace{gh_{x_0,\varepsilon}}_{> 0}>0.
$$
But this is a contradiction. Hence $g\equiv 0$.
