# Showing integrability of a locally integrable function on a bounded domain under some additional assumptions

Suppose $$\Omega\subset \mathbb{R}^3$$ is a smooth and bounded domain, and $$f:\Omega\to[0,\infty]$$ is a given function which is finite almost everywhere and satisfies

• Assumption A: For all $$g\in C_0^1(\Omega)$$ we have the product $$fg\in L^1(\Omega)$$. (Here $$C_0^1(\Omega)$$ refers to functions which are continuously differentiable in $$\Omega$$ and extend continuously to $$0$$ on $$\partial\Omega$$).

Question 1: Can we show that $$f\in L^1(\Omega)$$?

Question 2: Does the answer to Question 1 change if we include some or all of the following assumptions:

• Assumption B: $$f$$ possesses a weak derivative which is finite almost everywhere in $$\Omega$$;

• Assumption C: There exists a nonnegative function $$f_0 \in H^2(\Omega)\cap C(\bar{\Omega})$$ such that $$f-f_0=0$$ (in the sense of trace) on $$\partial\Omega$$;

• Assumption D: There exists a nonnegative function $$h\in H^1(\Omega)$$ such that $$h$$ is nonzero almost everywhere in $$\Omega$$ and $$f=-\ln h$$ in $$\Omega$$.

Note: Assumption D more or less implies Assumption B. I wrote them separately in the hopes of formulating the problem as simply as possible.

Notation: Here $$H^k$$ is the standard Sobolev space notation for $$W^{k,2}$$.

9/14/20 Edit:
Question 1 has been answered in the affirmative. I additionally pose the following

Question 3: Answer Questions 1 and 2 in the case that Assumption A is replaced by

• Assumption A': $$f\in L^1_{\text{loc}}(\Omega)$$.
• In the definition of $C^1_0$, do you mean: (i) continuously differentiable on $\Omega$, continuous on $\overline \Omega$, vanishing on $\partial \Omega$, or (ii) : continuously differentiable on $\overline \Omega$, and vanishing on $\partial \Omega$. For instance : if $\Omega$ is the unit ball, and $f(x)=\sqrt{1-\|x\|^2}$, does $f\in C^1_0$ ? Sep 12, 2020 at 19:57
• @PietroMajer Your first definition. Thank you for pointing that out. I have edited the post to make it more clear. Sep 12, 2020 at 21:02

Let $$(g_k)_{k\ge0}$$ be a sequence of smooth functions such that $$g_k(x)=1$$ if $$\text{dist}(x,\partial\Omega)\ge 2^{-k}$$, $$g_k (x)=0$$ if $$\text{dist}(x,\partial\Omega)\le 2^{-k-1}$$ and $$0\le g_k\le 1$$ everywhere.
To prove the affirmative answer to Question 1 by contrapositive, let $$f\not\in L^1(\Omega)$$ be given: we want to find $$g\in C^1_0(\Omega)$$ such that $$fg\not\in L^1(\Omega)$$. We can assume $$fg_k\in L^1(\Omega)$$ for all $$k$$, otherwise we are done with $$g=g_k$$ for some $$k$$. Then $$\int_\Omega fg_k$$ is an increasing sequence of positive real numbers, that diverges to $$+\infty$$, for if it were bounded, $$f\in L^1(\Omega)$$ by Beppo Levi's theorem. So for some subsequence $$(g_{k_j})_j$$ we have $$\int_\Omega fg_{k_{j+1}}\ge \int_\Omega fg_{k_j}+1$$, that is $$\int_\Omega f(g_{k_{j+1}}-g_{k_j})\ge 1$$. For all $$j$$ the function $$g_{k_{j+1}}-g_{k_j}$$ is bounded between $$0$$ and $$1$$, and supported in the set $$\big\{ 2^{-k_{j+1}-1}\le \text {dist}(x,\partial\Omega)\le 2^{-k_j}\big\}$$.
But then $$g:=\sum _{j\ge1} \frac{g_{k_{j+1}}-g_{k_j} }j$$ is a locally finite sum of smooth functions, hence smooth in $$\Omega$$; clearly $$g(x)\to0$$ for $$x\to\partial\Omega$$, and, again by Beppo Levi's theorem, $$\int_\Omega fg\ge \sum_{j\ge1}\frac1j=+\infty$$.
• Note that no assumption on $\partial\Omega$ are needed Sep 12, 2020 at 23:23
• Very nice Pietro, thank you! May I ask a followup (which essentially comes down to the distinction you raised in your first comment)? Suppose I remove the assumption that $fg\in L^1$ for all $g\in C_0^1(\Omega)$, and replace it by the assumption that $fg\in L^1$ for all $g\in C_c^1(\Omega)$. Would the result still hold? Here $C^1_c(\Omega)$ is the set of continuously differentiable functions with compact support in $\Omega$. Sep 13, 2020 at 0:44
• Isn't this the same as saying $f$ is $L^1_{loc}(\Omega)$? Sep 13, 2020 at 6:06
• @BenCiotti, as a matter of fact, substituting $C^1_0(\Omega)$ with $C^1_c(\Omega)$ is equivalent to say that $f$ is a locally integrable function Sep 13, 2020 at 8:52