The following looks quite similar to Poincare's inequality:

Let $\displaystyle{ 1 \leq p < \infty}$ and $\displaystyle{ U \subset \mathbb R^n}$ open and such that $\displaystyle{ U \subset \mathbb R^{n-1} \times (0,L) }$ with $L>0$. Show that for $\displaystyle{ u \in C_c ^ {\infty} (U) \cap W^{1,p} (U) }$ the following inequality hold:

$$ \int_U |u|^p \leq \frac{L^p}{p} \int_U | \nabla u|^p $$.

Here it is some thoughts I did, although I didn't manage to prove till the end.

Let $ x =(x' ,x_n) $ where $ x' = (x_1, \cdots, x_{n-1} )$. Now integrating with respact to the last variable we get:

$\displaystyle{ |u(x' ,t)|=\left | \int_0^{x_n} \partial_n u (x' , t) dt \right | \leq \int_0^L | \partial_n u (x' , t) |dt = \| 1 \cdot | \partial_n u (x' , t) \|_{L^1 ([0,l])} }$

and now form Holder's inequality (where $q$ is the conjugate exponent of $p$) we get that this last is

$$ \leq L^{1/q} \left(\int_0^L | \partial_n u (x' , t) |^pdt \right)^{1/p} $$

Integrating now with respect to $x_n$ we get that

$\displaystyle{ \int_0^L |u(x' ,t)|^p dx_n \leq L^{p/q +1 } \left(\int_0^L | \partial_n u (x' , t) |^pdt \right) = L^p \int_0^L |\partial_n u (x' , t) |^p dt}$

Integrating now with respect to $x'$ we have:

$\displaystyle{ \int_U |u(x)|^p dx \leq L^p \int_U | \partial_n u (x' , t) |^p dx \leq L^p \int_U |\nabla u(x)|^p dx }$

I can't see how this $p$ in the denominator of the fraction in the constant appears.

Any ideas?


1 Answer 1


We have $$|u(x',x_n)|\leqslant \int_0^{x_n}|\partial_n u(x',t)|\mathrm dt\leqslant \lVert \partial_n u(x',\cdot)\rVert_px_n^{1-1/p}.$$ The integration with respect to $x_n$ produces the constant $\frac 1p$.

Indeed, we have $$|u(x',x_n)|^p\leqslant\lVert \partial_n u(x',\cdot)\rVert_p^p\int_0^Lx_n^{p-1}\mathrm dx_n,$$ and the last integral is $[s^p/p]_{s=0}^{s=L}=\frac{L^p}p$.

Now we conclude as in the opening post integrating with respect to the other variables.

  • $\begingroup$ Dear Davide Giraudo, I would like to ask if it is possible to write some more details, about the integration, because I have been stuck. Thank you! $\endgroup$
    – passenger
    Nov 18, 2013 at 14:49
  • $\begingroup$ I've edited. I hope it's clearer now. $\endgroup$ Nov 18, 2013 at 15:04
  • $\begingroup$ O.k yes it is clear! I didn't think to raise them to the $p$ power. One last question: in the norm of $\partial_n$ why do you put the dot? Could I put a variable $t$ ? $\endgroup$
    – passenger
    Nov 18, 2013 at 15:08
  • 1
    $\begingroup$ Why not. I wrote it in order to point out that we take the norm with respect to the last variable. In general, when $u$ depends of one variable, we write $\lVert u\rvert_p$ rather than $\lVert u(t)\rVert_p$. $\endgroup$ Nov 18, 2013 at 15:10
  • 2
    $\begingroup$ O.k I thought about it, but I wasn't sure! Thank you very much for your time and your explanations! $\endgroup$
    – passenger
    Nov 18, 2013 at 15:12

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