If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$.

Question

I am wondering about a proof of the fact that If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$. I am familiar with the proof given here http://www.academia.edu/740170/An_Introduction_to_Weak_Derivatives on page 4, but I am wondering if there is another proof using density.

Basically something like take a sequence of $\phi_n\in C_c({\mathbb{R}}$) such that $\phi_n\to f$ in $L^1$, then $\displaystyle0=\int_\mathbb{R} f\phi_n\to\int_{\mathbb{R}}f^2$, hence $f=0$ a.e. But I need to justify this last limit. If $f$ is bounded it is fine, since $\displaystyle|\int_\mathbb{R} f\phi_n-f^2|\le\int_\mathbb{R} |f||\phi_n-f|\le\|f\|_{\infty}\int_\mathbb{R}|\phi_n-f|\to 0$. I am having trouble doing it without the assumption that $f$ is bounded though.

Just use dominated convergence theorem - You might also need that if $\varphi_n \to f$ in $L^1$, then there is a subsequence that convergence a.e. — Prahlad Vaidyanathan, Dec 05 '13 at 06:41
The referred article basically proofs the theorem if $f$ is a simple function using the density argument. — Vobo, Dec 06 '13 at 16:57
one can give a one line proof of this using the Lebesgue differentation theorem — Alexander Grothendieck, Dec 07 '13 at 12:23

TZakrevskiy · Accepted Answer · 2013-12-15T00:06:36.800

Take $\phi\in \mathcal C^\infty_c(\Bbb R)$, such that $\phi\ge 0$ and $\int_{\Bbb R}\phi(x)dx=1$. Let's call $\phi_n(x) = n\phi(nx)$.

On the one hand, this is a regularising sequence. Therefore, $\phi_n\ast f\to f$ in the sense of $L^1$ (the proof of this fact, however, relies on density of $\mathcal C _c$ in $L^1$.)

On the other hand, since $\phi_n(y-x)\in \mathcal C^\infty_c(\Bbb R)$ for a fixed $y$, we can write by our hypothesis $$0=\int_{\Bbb R}\phi_n(y-x)f(x)dx.$$

Thus, $f=0$ in the sense $L^1$.

If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$.

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