Is it in $L^1$?

Question

Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$, and $d: \Omega \longrightarrow \mathbb{R}$ be a measureable function. Is it true that if $$ \int_\Omega df < \infty \ , $$ for every $f \in C_c(\Omega)$ -- compactly supported continuous functions -- then $d \in L^1(\Omega)$?

Notes:

$\Omega$ can have many connected components.

If we allow all "test" functions $f \in L^\infty (\Omega)$ then we are basically viewing $f$ as an element in the dual space.

If need be, you can add the assumption that $d$ is constant on each connected component of $\Omega$. The reason for this weird condition comes from the fact that my $d$ is actually the Brouwer degree for some function.

Thanks in advance.

Thanks for quick refutation of the claim from several people. Now I add the assumption that there is a universal constant $C$ such that $$ \int_\Omega df < C \ \|f\|_{L^\infty}. $$ Does that change anything?

score 2 · Answer 1 · answered Jul 24 '18 at 20:03

The statement is false. Your condition that $\int_\Omega df < \infty$ really only guarantees that $d$ is well behaved away from the boundary. You can still have that $d$ blows up near $\partial \Omega$ and $d\notin L^1(\Omega)$. For instance, let $\Omega = (0,1)$ and $d(x) = \frac 1 x$. For any $f \in C_c(\Omega)$, we have that $\int_\Omega d f <\infty$. However, $$ \int_\Omega |d| = \int_0^1\frac 1 x\, dx = \infty.$$

Elle Najt · Accepted Answer · 2018-07-24T21:00:26.840

1

If $d$ is constant on each connected component of $\Omega$, then since $\Omega$ is bounded, $d \in L^1(\Omega)$ is true as long as $\Omega$ only has finitely many components. ( An upper bound is $|\Omega|max(|d|)$.)

If $\Omega$ is allowed to have infinitely many components you can construct an example similar to Trevors. That is, take $\Omega = [0,1] \setminus \{1/2, 1/3, 1/4, \ldots \}$ and have $d (x) = n$ for$x \in (1/(n+1), 1/n)$.

Regarding revised question:

The revised version is true, I think. You can take $f_n (x) = sgn( d(x))1_{A_n}$, where $A_n$ is an increasing sequence of compact sets whose union is $\Omega$. Then $||f_n||_{\infty} = 1$, so your condition has $\int d f_n \leq C$.

But $\int df_n = \int_{A_n} |d|$, so in fact $\int_{A_n} |d| \leq C$. Now, the monotone convergence theorem implies that $\int_{A_n} |d| \to \int_{\Omega} |d|$, and together with the inequality this shows that $\int_{\Omega} |d| \leq C$.

edited Jul 24 '18 at 21:00

answered Jul 24 '18 at 20:35

Elle Najt

20,740

Thanks, that wraps it all up! – Behnam Esmayli Jul 24 '18 at 20:49
Does the revised version hold? (See question again.) – Behnam Esmayli Jul 24 '18 at 20:56
@BehnamEsmayli I think so, see my updated answer. – Elle Najt Jul 24 '18 at 21:00
Wow, nice! Looks convincing. The reason for that universal boundedness is the area formula that $\int df = \int f\circ u \det u $ with the latter having a bound coming from a bound on $\det u$. Here $u$ is a map from a region in $R^n$ to $R^n$. Thanks a lot. – Behnam Esmayli Jul 24 '18 at 21:04
@BehnamEsmayli I don't really understand the context of the problem (but I think you mean det Jac(u) ? ) ... maybe you could add the context to your question? – Elle Najt Jul 24 '18 at 21:13
You're right. No that will be a whole new thing. I'd rather keep it simple. – Behnam Esmayli Jul 24 '18 at 21:13

Is it in $L^1$?

2 Answers2