If $ \int fg = 0 $ for all compactly supported continuous g, then f = 0 a.e.?

Question

I was wondering whether the following statement is true and, if so, how it can be shown:

If $ f \in L^{1}_{Loc}(\mathbb{R}^n) $ and if for all compactly supported continuous functions $ g: \mathbb{R}^n \to \mathbb{C} $ we have that the Lebesgue integral of $ f $ multiplied by $ g $ equals zero, i.e. $$ \int_{\mathbb{R}^n} f(x)g(x) \mathrm{d}x = 0 , $$ then $ f(x) = 0 $ almost everywhere.

I would be very grateful for any answers or hints!

N.B. I am aware that this question has already been addressed. In If $f\in L^1_{loc}(\mathbb{R})$ and $\int f\varphi=0$ for all $\varphi$ continuous with compact support, then $f=0$ a.e., I am not quite sure about how to create a sequence of compactly supported continuous functions such that $ \varphi_n\to \frac{f}{|f|+1} $. This particular question may have its answer in If $f\in L^1(\mathbb{R})$ is such that $\int_{\mathbb{R}}f\phi=0$ for all continuous compactly supported $\phi$, then $f\equiv 0$., however here I am unsure about the meaning of a "regularizing sequence"; why does $ \phi_n\ast f\to f $ in $L^1$ sense if $ \phi_n(x) = n\phi(nx) $, where $ \phi\in \mathcal C^\infty_c(\Bbb R) $ with $ \phi\ge 0 $ and $ \int_{\Bbb R}\phi(x)dx=1 $?

Once again, any answer would be much appreciated!

Answer 1

First it is well known that, $C^\infty_c$ is dense in $L^p$ for $1\le p<\infty.$

But if $f\in L^p$ so is $\frac{f}{|f|+1}$ because

$$\frac{|f|}{|f|+1} \le |f|$$ therefore there is automatically a sequence $\phi_n$ in $C^\infty_c$ converging $L^p$ to $\frac{f}{|f|+1}$ this solve your first question

Now note that $$ \int_\mathbb R\phi_n(y) dy= \int_\mathbb R \phi(y) dy = 1$$ then, for every $x\in \mathbb R$,

\begin{split} f*\phi_n(x) -f(x) &=&f*\phi_n(x) -f(x)\int_\mathbb R \phi_n(y) dy \\ &= &\int_\mathbb R (f(x-y)-f(x))\phi_n(y) dy\\ &=& n\int_\mathbb R (f(x-y)-f(x))\phi(ny) dy\\ &=& \int_\mathbb R (f(x-\frac{1}{n}y)-f(x))\phi(y) dy \end{split}

Hence assuming $f\in L^1(\mathbb R)$ and using Fubini, we have,

\begin{split} \|f*\phi_n -f\|_1 &=&\int_\mathbb R \int_\mathbb R(f(x-\frac{1}{n}y)-f(x))\phi(y)dxdy\\ &=& \int_\mathbb R \|f(.-\frac{1}{n}y)-f(.)\|_1\phi(y) dy\\ \end{split}

• $\|f(\cdot-\frac{1}{n}y)-f(\cdot)\|_1|\phi(y)|\le 2\|f(\cdot)\|_1\phi(y) \in L^1(\mathbb R)$ for almost every $y$.
• And We know the following $\|f(\cdot-\frac{1}{n}y)-f(\cdot)\|_1\to 0$ as $n\to \infty$ for very $f\in L^1(\mathbb R)$ Therefore by convergence dominated theorem we get $$ \|f*\phi_n -f\|_1 \to 0 \qquad , n\to \infty$$

i,e $f*\phi_n \to f$ in $L^1$.

Answer 2

For any ball $B(a,r),$ there is a sequence $g_k$ of continuous functions with support in $B(a,r),$ with $|g_k|\le 1$ everywhere, such that $g_k(x) \to \text { sgn }(f(x))$ pointwise a.e. in $B(a,r).$ Thus

$$\tag 1 \int_{B(a,r)} |f| = \int_{B(a,r)} f\cdot \text { sgn }(f) = \lim \int_{B(a,r)} f\cdot g_k =\lim 0 =0.$$

The second equality in $(1)$ follows from the dominated convergence theorem. From $(1)$ we see $f=0$ a.e. on $B(a,r).$ Since $B(a,r)$ was an arbitrary ball, we have $f=0$ a.e. in $\mathbb R^n.$