$f\in L^{1}_{\text{loc}}(\mathbb{R}^{n})$ is such that $\int_{\mathbb{R}^{n}}f(x)g(x)dx \geq 0 $ for all $g\in C_{c}(\mathbb{R}^{n},\mathbb{R}^{+})$

Question

Assume that $f\in L^{1}_{\text{loc}}(\mathbb{R}^{n},\mathbb{R})$ is such that $$\int_{\mathbb{R}^{n}}f(x)g(x)dx \geq 0 $$ for all $g\in C_{c}(\mathbb{R}^{n},\mathbb{R}^{+}).$ Here $C_{c}(\mathbb{R}^{n},\mathbb{R}^{+})$ denotes the set of all continuous functions on $\mathbb{R}^{n}$ taking non-negative real values and which have compact support. I think it follows that $f(x)\geq 0$ for a.e. $x\in\mathbb{R}^{n}$. However I only managed to prove this implication in the case where $f$ is (essentially) bounded. How to go from this simpler case to the general case ?

I need $f$ to be essentially bounded because I can only take $g\in C_{c}(\mathbb{R}^{n},\mathbb{R}^{+})$ to be an approximation in the $L^1$ norm of the indicator function of a chosen Borel subset, and then I need to use Hölder's inequality — Sqrt, Nov 13 '21 at 14:48
I assume you have not been convolution? approximate identity? — Arctic Char, Nov 13 '21 at 15:28
Well, I could deduce that $f*g_{\epsilon}\geq 0$ for every $\epsilon>0$, where $(g_{\epsilon})_{\epsilon>0}$ is a sequence of mollifiers, but I would need something more to deduce that this sequence of mollifications converges pointwise a.e. to $f$, which is only locally integrable — Sqrt, Nov 13 '21 at 15:39

Arctic Char · Accepted Answer · 2021-11-13T16:14:57.183

Fix a family of mollifier $(g_\epsilon)_\epsilon>0$. For any $R>0$ let $$f_R = f \chi_{B_{R+1}} \ \ ,$$ where $B_{R+1}$ is the ball of radius $R+1$ in $\mathbb R^n$. Then $f_R * g_\epsilon$ converges in $L^1 (\mathbb R^n)$ to $f _R$. Restricting to $B_R$, $f_R * g_\epsilon$ converges in $L^1 (B_R)$ to $f _R=f$.

Also, if $\epsilon <1$, then the condition implies that $$f_R * g_\epsilon (x) \ge 0, \ \ \forall x\in B_R.$$

Together with the convergence of $f_R * g_\epsilon \to f$ in $L^1(B_R)$, there is a sequence $\epsilon _n \to 0$ so that

$$f_R * g_{\epsilon_n} \to f$$ a.e. in $B_R$ (see here). Thus $f \ge 0$ in $B_R$. Since $R>0$ is arbitrary, choosing $R = K$. Using

$$ \{ x\in \mathbb R^n : f(x) < 0\} = \bigcup_{K\in \mathbb N} \{ x\in B_K : f(x) < 0\},$$

one also has $f \ge 0$ a.e. in $\mathbb R^n$.

$f\in L^{1}_{\text{loc}}(\mathbb{R}^{n})$ is such that $\int_{\mathbb{R}^{n}}f(x)g(x)dx \geq 0 $ for all $g\in C_{c}(\mathbb{R}^{n},\mathbb{R}^{+})$

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