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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': $f\in L^1_{\text{loc}}(\Omega)$.
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    $\begingroup$ 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 $f\in C^1_0$ ? $\endgroup$ Sep 12, 2020 at 19:57
  • $\begingroup$ @PietroMajer Your first definition. Thank you for pointing that out. I have edited the post to make it more clear. $\endgroup$
    – Ben Ciotti
    Sep 12, 2020 at 21:02

1 Answer 1

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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, $\int_\Omega fg\ge \sum_{j\ge1}\frac1j=+\infty$.

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    $\begingroup$ Note that no assumption on $\partial\Omega$ are needed $\endgroup$ Sep 12, 2020 at 23:23
  • $\begingroup$ 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 $\Omega$. $\endgroup$
    – Ben Ciotti
    Sep 13, 2020 at 0:44
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    $\begingroup$ Isn't this the same as saying $f$ is $ L^1_{loc}(\Omega)$? $\endgroup$ Sep 13, 2020 at 6:06
  • $\begingroup$ @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 $\endgroup$ Sep 13, 2020 at 8:52
  • $\begingroup$ Ok, so in that case, could the result be shown? Perhaps using the additional assumptions? $\endgroup$
    – Ben Ciotti
    Sep 14, 2020 at 0:17

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