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Let $I$ be the open interval $(0, 1)$ and $k >0$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.23

  1. Given $f \in L^1(I)$, prove that there exists a unique $u \in H_0^1(I)$ satisfying $$ (1) \quad \int_I u' v'+k \int_I u v=\int_I f v \quad \forall v \in H_0^1(I) . $$
  2. Show that $u \in W^{2,1}(I)$.
  3. Prove that $$ \|u\|_{L^1} \leq \frac{1}{k}\|f\|_{L^1} . $$
  4. Assume now that $f \in L^p(I)$ with $1<p<\infty$. Show that there exists a constant $\delta>0$ independent of $k$ and $p$, such that $$ \|u\|_{L^p} \leq \frac{1}{k+\delta / (p p')}\|f\|_{L^p}, $$ where $p'$ is the Hölder conjugate of $p$.
  5. Prove that if $f \in L^{\infty}(I)$, then $$ \|u\|_{L^{\infty}} \leq C_k\|f\|_{L^{\infty}}, $$ and find the best constant $C_k$.

I have proved $(4.)$ in case $p \in (2, \infty)$. Could you elaborate on how to prove it for $p \in (1, 2)$?

  1. We fix $\gamma \in C^1(\mathbb R, \mathbb R)$ with $\gamma' \ge 0, \gamma(0)=0$, $\gamma (t) =1$ for $t\ge 1$ and $\gamma (t) =-1$ for $t\le -1$. For $n \in \mathbb N$, let $v_n := \gamma (n u)$. Then $v_n \in H_0^1(I), v_n' =nu'\gamma'(nu)$ with $v_n (x) \to \operatorname{sgn} (u(x))$ as $n \to \infty$. Plugging $v_n$ in $(1)$, we get $$ \begin{align*} \int_I [n|u'|^2\gamma'(nu)+kuv_n] &= \int f v_n \\ \implies k\int_I uv_n &\le \int f v_n \\ \implies k\int_I uv_n &\le \int |f v_n|. \\ \end{align*} $$

The claim then follows by taking the limit $n \to \infty$ in the above inequality.

4.

  • $p \in [2, \infty)$.

Let $\gamma(t) = |t|^{p-1} \operatorname{sgn} (t)$ for $t \in \mathbb R$. Then $\gamma (0)=0$ and $\gamma'(t) = (p-1)|t|^{p-2}$. It follows from $p\ge 2$ that $\gamma'$ is continuous and thus $\gamma \in C^1 (\mathbb R)$. Let $v:=\gamma(u)=|u|^{p-1} \operatorname{sgn} (u)$. By Corollary 8.11 (in the same book), $v \in H^1_0 (I)$ with $v'=\gamma'(u)u'=(p-1)u'|u|^{p-2}$. Plugging $v$ in $(1)$, we get $$ \begin{align*} \int_I [(p-1)|u'|^2 |u|^{p-2} +k|u|^p] &= \int f |u|^{p-1} \operatorname{sgn} (u). \end{align*} $$

Let $w := |u|^{p/2} \operatorname{sgn} (u)$. Then $w' = \frac{p}{2}u'|u|^{(p-2)/2}$. Clearly, $w \in H^1_0(I)$. By Poincaré's inequality, there is a constant $C >0$ such that $\int_I |w|^2 \le C \int_I |w'|^2$ and thus $\int_I |u|^p \le \frac{Cp^2}{4} \int_I |u'|^2 |u|^{p-2}$. Then $$ (p-1)\int_I |u'|^2 |u|^{p-2} \ge \frac{4(p-1)}{Cp^2} \int_I |u|^p = \frac{4}{Cpp'} \int_I |u|^p. $$

By Hölder's inequality, $$ \int f |u|^{p-1} \operatorname{sgn} (u) \le \| f \|_{L^p} \| u \|_{L^p}^{p/p'}. $$

It follows that $$ \left (k + \frac{4}{Cpp'} \right ) \| u \|_{L^p}^{p} \le \| f \|_{L^p} \| u \|_{L^p}^{p/p'}. $$

  • $p \in (1, 2)$.
Akira
  • 17,367

2 Answers2

2

We can use a duality argument.

Let $p\in(1,2]$, then $p'\in[2,\infty)$. By duality, we have $$\|u\|_{L^p}=\sup \left\{\left|\int_Iu\varphi \right|:\varphi\in L^{p'}(I), \|\varphi\|_{L^{p'}}\leq 1\right\}.\tag{$*$}$$ Now we take $\varphi\in L^{p'}(I)\subset L^1(I)$ with $\|\varphi\|_{L^{p'}}\le1$. Let $v\in H_0^1(I)$ be the unique solution of $$\int_Iv'w'+k\int_Ivw=\int_I\varphi w,\qquad \forall w\in H_0^1(I).\tag{$**$}$$ Then we have $$\|v\|_{L^{p'}}\leq \frac{1}{k+\delta / (p’p)}\|\varphi\|_{L^{p'}}\leq \frac{1}{k+\delta / (p p')}$$ since $p'\in[2,\infty)$. Taking $w=u$ in $(**)$ gives that \begin{align*} \int_Iu\varphi =\int_I u'v'+k\int_Iuv=\int_I fv, \end{align*} hence $$\left|\int_Iu\varphi \right|\leq \|f\|_{L^p}\|v\|_{L^{p'}}\leq \frac{1}{k+\delta / (p p')}\|f\|_{L^p}.$$ Now the desired result follows from $(*)$.

Feng
  • 13,705
0

We consider the linear map $T:L^1(I) \to H^1(I), f \mapsto u$. It follows from $(1)$ that $\int_I [ |u'|^2 + k|u|^2 ] = \int_I f u$. By Hölder's inequality, $\alpha \| u \|_{H^1}^2 \le \|f\|_{L^1} \|u\|_{L^\infty}$ where $\alpha := \min \{1, k\}$. By Theorem 8.8 (in the same book), there is a constant $C>0$ such that $\|u\|_{L^\infty} \le C \| u \|_{H^1}$. Then $\| u \|_{H^1} \le \frac{C}{\alpha} \|f\|_{L^1}$. Then $T$ is bounded.

For $p \in (1, \infty)$, let $p'$ be its Hölder conjugate. Consider the bounded linear operators $$ \begin{align*} \varphi : L^p(I) \to L^p(I), f \mapsto Tf, \\ \psi : L^{p'}(I) \to L^{p'}(I), g \mapsto Tg. \end{align*} $$

It follows from $(1)$ that $$ \begin{align*} \int_I (Tf)' (Tg)'+k \int_I (Tf) (Tg) &= \int_I f (Tg), \\ \int_I (Tg)' (Tf)'+k \int_I (Tg) (Tf) &= \int_I g (Tf). \end{align*} $$

It follows that $\int_I (Tf)g= \int_I f(Tg)$, which implies $\varphi$ and $\psi$ are adjoint of each other. Now we fix $p \in (1, 2)$. Then $p' \in (2, \infty)$. We have $$ \begin{align*} \|Tf\|_{L^p} &= \sup_{\|g\|_{L^{p'}} \le 1} \int_I (Tf)g \\ &= \sup_{\|g\|_{L^{p'}} \le 1} \int_I f(Tg) \\ &\le \|f\|_{L^p} \sup_{\|g\|_{L^{p'}} \le 1} \|Tg\|_{L^{p'}} \\ &\le \|f\|_{L^p} \sup_{\|g\|_{L^{p'}} \le 1} \left ( \frac{1}{k+\delta / (p p')} \|g\|_{L^{p'}} \right ) \\ &\le \frac{1}{k+\delta / (p p')} \|f\|_{L^p}. \end{align*} $$

Akira
  • 17,367