Integral = 0 implies function = 0?

Question

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 $C^{\infty}$...

Answer 1

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?

Answer 2

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 $$ \int_{-\infty}^\infty\int_{-\infty}^\infty|f(x,t)|^2\,\mathrm{d}x\,\mathrm{d}t=0 $$