If an integral of a product of a function and an arbtrary bump function around $t$ is $0$, $f(t)=0$? In my physics textbook, it is stated that if $\displaystyle \int_{t-\delta}^{t+\delta} b(x)f(x)dx=0$ for all bump function $b$ whose support $supp(b)\subset [t-\delta,t+\delta]$ and $f$ is a continuous function, then $f(t)=0$.
Is it true? If it is true, could you give me information about the proof? I'd also like to know that if $\displaystyle \int_{t-\delta}^{t+\delta} \int_{t-\delta}^{t+\delta} b(x,y)f(x,y)dxdy=0$ for all bunmp function $b$ whose support $supp(b)\subset [t-\delta,t+\delta]\times [t-\delta,t+\delta]$ and f is a countinuous function, then $f(t,t)=0$?
 A: I'll discuss the 1-d case only. The 2-d case is identical in spirit.
Suppose for the sake of contradiction that $f(t) > 0$. ($f(t)<0$ can be handled similarly.) Then there is a small neighborhood $(t-\epsilon, t+\epsilon)$ around $t$ such that $f(\tau) > 0$ for all $\tau\in(t-\epsilon,t+\epsilon)$. Moreover, we can choose $\epsilon>0$ to be as small as we like while maintaining this condition.
In particular, let's take $0 < \epsilon < \delta$. We next pick a bump function $b$ supported in $(t-\epsilon,t+\epsilon)$ with $b(\tau) > 0$ for all $\tau\in(t-\epsilon,t+\epsilon)$. (If you want to know how to pick such a bump function, pick your favorite bump function which is everywhere positive, then rescale and translate.) Then we have
$$
0 = \int_{t-\delta}^{t+\delta} b(\tau)f(\tau)d\tau \geq \int_{t-\epsilon}^{t+\epsilon} b(\tau)f(\tau)d\tau > 0
$$
where the first equality comes from the hypothesis on $f$ and the last inequality is because $b(\tau)f(\tau)>0$ on $(t-\epsilon,t+\epsilon)$. This is a contradiction, and thus we conclude that $f(t) = 0$.
