Integral of a product of two functions Let $\Omega$ be an open subset of $\mathbb{R^2}$ and $f$ a continuos function in $\Omega$.
Is it true that, if $$\int_{\Omega} f(x)g(x) dx = 0$$ $\forall g \in C_c^{\infty}(\Omega) $ then $f=0$ in $\Omega$?
 A: Suppose $f\neq 0$, i.e., there is $y\in \Omega $ s.t. $f(y)\neq 0$ (suppose WLOG that $f(y)>0$). By continuity, there is a ball $B(y,r)\subset \Omega $ s.t. $f(x)>0$ for all $x\in \overline{B(y,r)}$. Let $r'>r$ s.t. $B(y,r)\subset B(y,r')\subset \Omega $. Take $g\in \mathcal C^\infty $ s.t. 
$$g(x)=\begin{cases}1&x\in \overline{B(y,r)}\\ 0&x\in B(y,r')^c\end{cases}$$
and $0\leq g(x)\leq 1$ if $x\in B(y,r')\setminus \overline{B(y,r)}$. Such function exist. Then, $$\int_\Omega fg\geq \int_{B(y,r)}f>0,$$
which is a contradiction. 
A: Under all these assumptions I would say that the answer is yes. 
Assume towards contradiction that $f(x)\not \equiv0$, then there exists $x\in \Omega $ such that $f(x_0)\neq 0$, because $f$ is continuous. Assume without loss of generality that $f(x_0)=\epsilon>0$. Then there exists $\delta>0$ such that $f(x)>\frac{\epsilon}{2}$ for all $\Vert x-x_0\Vert<\delta$.
There exists $h_\delta\in C_c^\infty$ satisfying that $h_\delta \vert_{B(x_0,\delta/2)}=1$, $h_\delta (x)=0$ when $\Vert x-x_0\Vert \geq \delta$ and $\vert h_\delta \vert\leq 1$.
You would get that $\int f \cdot h_\delta \geq \frac{\epsilon}{2}\cdot \vert B(x_0,\delta/2)\vert>0$
and this is a contradiction. I can see that @Surb has already written an answer, but since it took me a while to write this, I'll leave it.
