Exercise 8.23 Brezis book

Question

Let us begin by stating the problem

Let $I = \left( {0,1} \right)$ and fix a constant $k>0$.

1. Given $f \in {L^1}\left( I \right)$, prove that there exists a unique $u \in H_0^1\left( I \right)$ satisfying $$\int_0^1 {\left( {{u^\prime }{v^\prime } + kuv} \right)} = \int_0^1 {fv} ,\,\forall v \in H_0^1\left( I \right).$$
2. Show that $u \in {W^{2,1}}\left( I \right)$.
3. Prove that $${\left\| u \right\|_{{L^1}}} \leqslant \frac{1}{k}{\left\| f \right\|_{{L^1}}}$$.
4. Assume now that $f \in {L^p}\left( I \right)$ with $1 < p < \infty $. Show that there exists a constant $\delta > 0$ independent of $k$ and $p$, such that $${\left\| u \right\|_{{L^p}}} \leqslant \frac{1}{{k + \frac{\delta }{{p{p^\prime }}}}}{\left\| f \right\|_p}.$$
5. Prove that if $f \in {L^\infty }\left( I \right)$ then $${\left\| u \right\|_{{L^\infty }}} \leqslant {C_k}{\left\| f \right\|_{{L^\infty }}},$$ and find the best constant $C_k$.

I finished questions 1, 2, and 3. For question 4, the author gave a hint as follows. If $p \geq 2$, choose $v = {\left| u \right|^{p - 2}}u$. And when $1 < p < 2$, use duality.

For the case $p \geqslant 2$, if we choose $v = {\left| u \right|^{p - 2}}u$, then we have ${v^\prime } = \left( {p - 1} \right){\left| u \right|^{p - 2}}{u^\prime }$. Hence, we obtain $$\left( {p - 1} \right)\int_0^1 {{{\left| u \right|}^{p - 2}}{{\left( {{u^\prime }} \right)}^2}} + k\left\| u \right\|_{{L^p}}^p = \int_0^1 {f{{\left| u \right|}^{p - 2}}u} .$$ By adopting Holder's inequality, we can easily discover ${\left\| f \right\|_{{L^p}}}\left\| u \right\|_{{L^p}}^{p - 1} \geqslant k\left\| u \right\|_{{L^p}}^p$. And this fact implies ${\left\| u \right\|_{{L^p}}} \leqslant \frac{1}{k}{\left\| f \right\|_{{L^p}}}$. So in this case, we CAN NOT choose $\delta > 0$ which fulfills the inequality in question 4.

In the case $1<p<2$, the author hinted duality method. Unfortunately, I do not know this method.

I hope someone can help me in clarifying these doubts.

Thank you in advance.

Answer 1

We can do the following: $$ \int_0^1 |u|^{p-2} (u')^2 = \frac{4(p-1)}{p^2} \int_0^1 ( (|u|^{p/2})')^2 dx \ge c \frac{4(p-1)}{p^2} \int_0^1 |u|^p, $$ where in the last step I used Poincare inequality applied to $|u|^{p/2}$. Using this in your calculations gives $$ \left( k + c \frac{4(p-1)}{p^2} \right) \|u\|_{L^p}^2 \le \|f\|_{L^p} \|u\|_{L^p}^{p-1}. $$ Note that $$ c\frac{4(p-1)}{p^2} = \frac{4c}{pp'}, $$ so the claim is true with $\delta = 4c$, and $c$ is from Poincare inequality. Since the domain is the unit interval, $c=\frac1{\pi^2}$.