Weak derivative: Showing a function is equal to zero a.e.

Question

I am very beginner in the theory of weak derivative. I am trying to fix the following problem:

Suppose that $f\in{L_{loc}^{1}}$ and $\int_{a}^{b}f(x)\phi({x})dx=0$ for all $\phi\in{C_{0}^{\infty}}$ then $f(x)=0$ $a.e.$ on $(a,b)\subset{\mathbb{R}}$.

I will appreciate your ideas.

Answer 1

One useful technique you can use here is regularization by convolution. For instance, a typical regularization result (that is relevant here) is the following:

Regularization: Let $\phi\in C^\infty_0$ be nonnegative, with $\int \phi\,dx = 1$. For each $t>0$, define $\phi_t = t^{-1}\phi(x/t)$. Then for any $f\in L^1_{loc}$, the convolutions $f*\phi_t$ are smooth functions and $f*\phi_t\to f$ in $L^1_{loc}$ as $t\to 0$.

Now let's apply this result to your problem. Choose a $\phi$ as in the theorem. By definition, $$(f*\phi_t)(x) = \int f(y)\phi_t(x - y)\,dy.$$ But the function $y\mapsto \varphi_t(x - y)$ is a compactly supported smooth function, so by your assumption on $f$, the integral is $0$, i.e., $(f*\phi_t)(x) = 0$ for all $x$. That is to say $f*\phi_t \equiv 0$. As $t$ was arbitrary, if you let $t\to 0$ the regularization theorem says that $0 \equiv (f*\phi_t)\to f$ in $L^1_{loc}$. Thus $f = 0$ in $L^1_{loc}$.