Assume $\int_{D}fgdx=0\forall g\in C_{c}^{\infty}\left(D\right)$. Then $f = 0$ a.e. on D.

I want to prove this result: Let $D$ be an open subset of $\mathbb{R}^n$, $p \in[1,\infty)$ and $f$ be in $L^p(D)$. Assume $\int_{D}fgdx=0\forall g\in C_{c}^{\infty}\left(D\right)$. Then $f = 0$ a.e. on $D$.

I found a corollary of the book Haim Brezis - Functional Analysis, Sobolev Spaces and Partial Differential Equations that is a nearly result. It's corollary $4.24$. But the assumption is $f \in L^1_{loc}(D)$. Can I apply the proof of corollary $4.24$ for my problem?

• Since smooth compactly supported functions are dense in $L^q(D)$ for $D$ open and $1 < q < \infty$, this follows immediately from $L^p-L^q$ duality for $p > 1$. The $L^1_{loc}$ case is one you genuinely need to use approximate identities and the differentiation theorem for. – Chris Janjigian Jan 27 '14 at 0:55
• Can you explain more precise? Or notice the theorems which you used. – chuyenvien94 Jan 27 '14 at 1:05
• The dual space of $L^p(D)$ is always $L^q(D)$ where $\frac{1}{p} + \frac{1}{q} = 1$ so long as $1 < p < \infty$. On an open domain, the set of $C_c^\infty(D)$ functions is dense in $L^q(D)$ for $1 \leq q < \infty$. You can basically prove this by convolving with an approximate identity. There is a tiny amount of subtlety in the argument, but not much. It follows from this density that $\int fg = 0$ for all $g$ in $L^q(D)$. From the variational characterization of the norm $\|f \|_p = \sup_{\|g\|_q = 1} \int fg$, this implies that $\|f\|_p = 0$. – Chris Janjigian Jan 27 '14 at 15:50