1
$\begingroup$

I'm struggling with the following question from Haim Brezis' book:


Let $q\in[1,\infty)$ and $r\in (1,\infty]$. Show that there exists a constant $C=C(q,r)$ such that:

$\Vert u\Vert_{L^\infty(0,1)} \leq C\cdot \Vert u \Vert_{W^{1,r}(0,1)}^a \cdot \Vert u \Vert_{L^{q}(0,1)}^{1-a} \quad \text{for all} \quad W^{1,r}(0,1)$

where $a\in (0,1)$ is determined by:

$a\Big( \frac{1}{q} +1- \frac{1}{r} \Big)= \frac{1}{q}$

And it gives the following hint:

When $u(0)=0$ write $G\big( u(x) \big)= \int_0^x G' \big( u(t) \big)u'(t)dt$ for $G(t)=\vert t \vert^{\alpha-1} \cdot t$ and $\alpha =\frac{1}{a}$. When $u(0)\neq 0$, use the previous case for $\zeta u$ where $\zeta \in C^1[0,1]$ satisfying $\zeta(0)=0$ and $\zeta(t)=1$ for $t\in [\frac{1}{2},1]$.


My attempts thus far:

I know since $r>1$ that $u$ has a version which is in $C[0,1]$, and that if $r<\infty$ then it is also $(1-\frac{1}{r})$- Holder continuous. Also since $L^\infty(0,1)$ is embedded continuously into $C[0,1]$, there exists $C_1>0$ such that:

$ \Vert u \Vert_{L^\infty(0,1)} \leq C_1 \cdot \Vert u \Vert_{W^{1,r}(0,1)} \quad \text{for all} \quad u\in W^{1,r}(0,1) $

And because of this it now only remains to show that:

$\Vert u \Vert_{W^{1,r}(0,1)}^{1-a}\leq C_2 \cdot \Vert u \Vert_{L^q(0,1)}^{(1-a)} \quad \text{for all} \quad u\in W^{1,r}(0,1) $

for some constant $C_2>0$. But I have not been able to understand where the hint comes into it at all. Also it seems to me that $G(t)$ is simply $t^\alpha$. I would appreciate any hints or help.

$\endgroup$
  • 1
    $\begingroup$ Did you try to apply Hoelder inequality to the hinted term about $G(u(x))$? Then you have a chance to bring the $W^{1,r}$-norm into the game. $\endgroup$ – daw Mar 13 '18 at 9:43
  • $\begingroup$ Just a further question, am I wrong in saying that $G'(t)=\alpha t^{\alpha-1}$? $\endgroup$ – Keen-ameteur Mar 13 '18 at 10:11
  • 1
    $\begingroup$ The derivative is $G'(t) = \alpha |t|^{\alpha-1}$. $\endgroup$ – daw Mar 13 '18 at 10:15
  • $\begingroup$ I was able to solve it, thanks $\endgroup$ – Keen-ameteur Mar 13 '18 at 15:35
1
$\begingroup$

I'll write my answer in case some one else will also want the solution for this question:

Let $u\in W^{1,r}(0,1)$. Every $v\in W^{(0,1)}$ satisfies that:

$v\in C[0,1]$ and $v(x)-v(y)=\int_y^x v'(t)dt$.

By a theorem stating that if $G(0)=0$ and $\vert G' \vert \leq M$ then:

$G\circ u \in W^{1,r}(0,1)$ and $\big( G\circ u \big)'(x)=G'\big( u(x)\big) \cdot u'(x)$, if $u(0)=0$ we can write therefore:

$G\circ u(x)= G\circ u(x)- G\circ u(0)= \int_0^x\big( G\circ u \big)'(t)dt$

Also $\vert G(u)\vert=\vert u\vert ^\alpha$, and thus:

$\vert u(x)\vert^\alpha= \Big \vert \int_0^x\big( G\circ u \big)'(t)dt \Big \vert \leq \int_0^1 \Big\vert G'\big( u(t) \big) u'(t) \Big\vert dt= \alpha \cdot \int_0^1 \Big\vert \big( u(t) \big)^{\alpha-1} \cdot u'(t) \Big\vert dt $

By Holder's inequality for $r$ and it's conjugate we obtain:

$\int_0^1 \Big\vert \big( u(t) \big)^{\alpha-1} \cdot u'(t) \Big\vert dt \leq \Bigg( \int_0^1 \vert u'(t)\vert^rdt \Bigg)^{\frac{1}{r}} \cdot \Bigg( \int_0^1 \vert u(t)\vert^{(\alpha-1) \frac{r}{r-1}} dt \Bigg)^{1-\frac{1}{r}}$

Which gives us that:

$\vert u(x) \vert^\alpha \leq \alpha \Bigg( \int_0^1 \vert u'(t)\vert^rdt \Bigg)^{\frac{1}{r}} \cdot \Bigg( \int_0^1 \vert u(t)\vert^{(\alpha-1) \frac{r}{r-1}} dt \Bigg)^{1-\frac{1}{r}}$ for all $x\in (0,1)$

Simple calculations give us that:

$1-a=\frac{q}{\alpha} \cdot (1-\frac{1}{r}) \quad$ and $\quad (\alpha-1) \cdot \frac{r}{r-1}=q$, and therefore taking the $\alpha$-th root from both sides gives us that:

$\text{(*)} \quad \vert u(x) \vert \leq \alpha^{\frac{1}{\alpha}} \Vert u' \Vert_{L^r(0,1)}^a \cdot \Vert u \Vert_{L^q(0,1)}^{(1-a)}$ for all $x\in (0,1)$.

Remembering that $ \Vert u' \Vert_{L^r(0,1)} \leq \Vert u \Vert_{W^{1,r}(0,1)} $, this solves the case when $u(0)=0$. When $u(0)\neq 0$ using (*) for $\zeta u$ or $\zeta(x) \cdot u(1-x)$ shows that:

$\vert u(x) \vert \leq C \Vert u \Vert_{W^{1,r}(0,1)}^a \cdot \Vert u \Vert_{L^q(0,1)}^{(1-a)}$ for all $x\in (0,1)$

And since $u$ is continuous this shows what is needed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.