If Double Integral of a(x, y)h(x, y) vanishes for any C^2 h(x, y) then C^0 a(x, y) vanishes

Theorem
Let $\alpha\left(x, y\right)$, $h\left(x, y\right)$ be functions in $\R$.

Suppose $\alpha \in C^0$ in a closed region $R$.

Suppose $h\in C^2$ in $R$ and equals 0 on $\Gamma$, the boundary of $R$.

Suppose $\displaystyle\int\int_R\alpha(x,y)h(x,y)\mathrm{d}{x}\mathrm{d}{y}=0$

Then $\alpha\left(x, y\right)$ vanishes everywhere in $R$.

Proof
Suppose, the function $\alpha\left(x, y\right)$ is nonzero at some point in $R$.

Then $\alpha\left(x, y\right)$ is also nonzero in some disk $D$ such that $\left(x-x_0\right)^2+\left(y-y_0\right)^2\le\epsilon^2$.

Suppose

$\displaystyle h\left(x, y\right)= \operatorname{sgn}\left(\alpha\left(x, y\right)\right)\left[\epsilon^2-\left(x-x_0\right)^2+\left(y-y_0\right)^2\right]^3$

in this disk and 0 elsewhere.

Thus $h\left(x, y\right)$ satisfies conditions of the theorem.

However

$\displaystyle\int\int_R\alpha(x,y)h(x,y)\mathrm{d}{x}\mathrm{d}{y}=\int\int_D\left\vert\alpha\left(x, y\right)\right\vert\left[\epsilon^2-\left(x-x_0\right)^2+\left(y-y_0\right)^2\right]^3\ge 0$

Hence, by Proof by Contradiction, the theorem is proved.