3
$\begingroup$

Let $\Omega \subseteq \mathbb{R}^n$ (open) and $u \in \mathcal{D}'$ be a distribution, that has a distributional derivative which is in $L^p(\Omega)$ (for some $p \geq 1$).

Show that $u \in L^p_{loc}(\Omega)$

...where $L^p(\Omega)$ is defined as in

https://en.wikipedia.org/wiki/Lp_space#Lp_spaces

and $L^p_{loc}(\Omega)$ is defined as in:

https://en.wikipedia.org/wiki/Locally_integrable_function#Generalization:_locally_p-integrable_functions


My thoughts:

1) For notational simplicity let's call $Du = D_i u$ and $\phi' = \partial_{x_i}\phi$ for some $i \in \lbrace 1,2 \dots n\rbrace$

Then by definition of distributional derivatives, we have that:

$\int_{\Omega} (Du) \phi dx=-\int_{\Omega} u \phi' dx, \forall \phi \in C^{\infty}(\Omega)$ with compact support

2) Let also $K \subset \Omega$ (compact). We need to show that $\int_K \vert u \vert^p dx < \infty$

3) A simple density argument of the test functions should help us go from $K$ to $\Omega$ so this shouldn't be a big deal

4) We can also use something like Holder's inequality (and the fact that $Du \in L^p$) to show something among the lines of:

$\vert \int_{\Omega} (Du) \phi dx \vert \leq \int_{\Omega} \vert (Du) \phi \vert dx \leq (\int_{\Omega} \vert Du \vert^p)^{\frac{1}{p}} (\int_{\Omega} \vert \phi \vert^q)^{\frac{1}{q}}< \infty$

5) Then from the definition of $Du$ we also have that:

$\vert \int_{\Omega} u \phi' dx \vert = \vert \int_{\Omega} (Du) \phi dx \vert$

6) But how do we find an inequality that involves $\int \vert u \vert ^p$ and $\int\vert u \vert$ (in this order) to fill in the chain of inequalities? (Assuming of course, that such a "chain" is the way to go in this problem)

7) Another idea I am toying with (see update below), is to use some sort of a density argument to go from - say - $\int\vert u \vert ^p$ to $\int\vert u \vert \phi_n$ for an appropriate sequence of $\phi_n$. Not sure if this is possible though.

Update:

Question: What density arguments are available for locally integrable functions? For instance, since $u \in L^1_{loc}$, is it possible to find a sequence $\phi_n$ in $C^{\infty}_0(\Omega)$ or even in $C^{\infty}(\Omega)$ such that:

$$\lim\limits_{n \rightarrow \infty}\phi_n = \vert u \vert^{p-1}$$

I think that would complete the chain of inequalities and thus the proof.

$\endgroup$

1 Answer 1

2
+50
$\begingroup$

I will outline the solution for the case $n=1$ and $p=2$, and if you can not generalize it to $p\ge1$ and $n\ge 1$, ask in comments, we will dive into the murky details. Without further ado:

Let $K$ be a non-trivial compact set, $\phi$ be a test function with support in $K$ and $u'\in L^2(K)\cap D'(\Bbb R)$.

We can say that by Cauchy-Bunyakovskiy-Schwarz inequality

$$|\langle u',\phi\rangle| = \left|\int_K u'(x)\phi(x)dx\right|\le \|u',L^2(K)\|\|\phi,L^2(K)\|$$ Then again, by Poincaré's inequality, for a constant $C$ that depends only on $K$, $$\|\phi,L^2(K)\|\le C \|\phi',L^2(K)\|.$$ This gives us $$|\langle u,\phi'\rangle| \le C\|u',L^2(K)\| \|\phi',L^2(K)\| .$$

Therefore, we can already say that $u$ is a continuous linear operator over the set $L^2(K)\cap \{f: \int_Kf=0\}$. We can extend it to the whole $L^2(K)$ by noticing that if $\psi$ is a test function supported in $K$ with $\int_k\psi=1$, then for any $\phi$ as above we have $$\langle u,\phi\rangle = \left\langle u,\phi-\psi\int_K\phi\right\rangle + \left(\int_K\phi\right)\langle u,\psi\rangle.$$ The test function $\phi-\psi\int_K\phi$ is a derivative of a test function (easy exercise), hence you can apply the estimations above; and the term $\left(\int_K\phi\right)\langle u,\psi\rangle$ is bounded by $C_1 \|\phi,L^2(K)\|$ (essentially, $C_1$ acts like an arbitrary constant when you find an antiderivative). Thus, $\langle u,\phi\rangle$ is bounded by $$C\|u',L^2(K)\| \left\| \phi-\psi\int_K\phi,L^2(K)\right\| + C_1 \|\phi,L^2(K)\|.$$ It easy to see that $u$ now can be seen as continuous linear functional over $L^2(K)$. Therefore, $u$ itself belongs to $L^2(K)$.

Notes for the generalizations: the Poincaré's inequality works for $p\in [1,\infty)$ and $n\ge 1$, so that part is ok. The representation theorem definitely works for $p>1$, some tinkering might be necessary to cover the case $p=1$.

$\endgroup$
5
  • $\begingroup$ Thank you very much for the detailed answer! I did not think to use Poincare's inequality! Please give me some time to go over the fine details. But for now, a few things: 1) Isn't $p=1$ trivial since $u$ is assumed to be already in $L^1_{loc}(\Omega)$ by definition (since u has a distributional derivative)? Maybe I am using a slightly different definition? 2) Why the assumption of $n=1$? Did you use it in any essential way during the proof? 3) Also, I haven't thought it thoroughly yet, but why the restriction over the set $\lbrace f : \int_K f = 0 \rbrace$ $\endgroup$
    – Pellenthor
    Apr 5, 2017 at 21:18
  • $\begingroup$ (1) Not really. All distributions have distributional derivatives, even those who are not themselves $L^1_{loc}$. But we can remedy that by saying that an integral of $L^1_{loc}$ function is continuous, hence locally integrable. (2) You are right, in what I wrote $n$ can be arbitrary. (3) We have shown that $u$ is is continuous on test functions which are the derivatives of test functions (third formula). And a test function is such a derivative if and only if its integral is equal to zero (recall the proof of "why all distributions have an antiderivative"). $\endgroup$ Apr 6, 2017 at 8:18
  • $\begingroup$ Thank you! I guess I only have a couple more questions 1) which test function has $\phi-\psi\int_K\phi$ as its derivative? and 2) For the general case, I assume we ll use Holder's inequality instead of CSB. Does that affect any other parts of the proof? $\endgroup$
    – Pellenthor
    Apr 6, 2017 at 15:51
  • $\begingroup$ 1) We do not really care. We can safely take $\tau (x) = \int_{-\infty}^x \left(\phi(s) - \psi(s)\left(\int_K\phi(y)dy\right) \right)ds$ for all it matters. 2) I don't think it does. $\endgroup$ Apr 6, 2017 at 15:55
  • $\begingroup$ So just use the FTC then. When you mentioned it was an "exercise", I thought there was more to it than that. Thanks! $\endgroup$
    – Pellenthor
    Apr 6, 2017 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .