Integral = 0 implies function = 0? If $f : \mathbb{R}^2 \to \mathbb{R}$ is continuous and such that
$$
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \! f(x,t) \phi(x,t) \, \mathrm{d}x \, \mathrm{d}t = 0
$$
for all $\phi \in C_c^{\infty}(\mathbb{R}^2)$ (that is, for all compactly supported smooth functions), then is it true that $f$ must be identically equal to $0$ ?
I thought taking, in particular, $\phi_n \in C_c^{\infty}(\mathbb{R}^2)$ to be equal to $0$ outside the disk of radius $n$ centered at the origin, to $f$ where $f \geq 0$ and to $-f$ where $f < 0$ would permit us to prove this, but unfortunately it seems like such $\phi_n$ need not be $C^{\infty}$...
 A: We can assume that $f$ has compact support. If not, take a $C_c^\infty$ partition of unity $\{\psi_n\}$ and show that $\psi_n(x)f(x)=0$ as below. Then $f=\sum\limits_n\psi_n(x)f(x)=0$.
Given an $\epsilon\gt0$, we can find a $\phi\in C_c^\infty$ so that
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)-\phi(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t\le\epsilon
$$
The hypothesis then implies
$$
\begin{align}
&\int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t\\
&\le\int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)|^2+|\phi(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)-\phi(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t\\[6pt]
&\le\epsilon
\end{align}
$$
Since $\epsilon$ is arbitrary, we have
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t=0
$$
A: A nonvanishing continuous function has a nonzero value in a disk $B(x, R)$ of radius $R$ around some $x.$ Now, take $\phi$ a partition of unity function (aka bump function) whose support lies in $B(x, R)$ What happens?
