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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 the inequality in (5.) below. Could you explain how to find the best constant $C_k$?

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 $$ \| u \|_{H^1} \le \frac{C}{\alpha} \|f\|_{L^1} \le \frac{C}{\alpha} \|f\|_{L^\infty}, $$ which implies $$ \|u\|_{L^\infty} \le \frac{C^2}{\alpha} \|f\|_{L^\infty}. $$

Akira
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2 Answers2

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For the second part, it is easy to see $u$ is a weak solution of the following BVP $$ -u''+ku=f \text{ in } I, u(0)=u(1)=0. \tag{1} $$ A general solution to the homogeneous equation is given as $$ u_h(t)=c_1e^{\sqrt kt}+c_2e^{-\sqrt kt}. $$ Choose $u_1(0)=0$ and one picks $$ u_1=e^{\sqrt kt}-e^{-\sqrt kt}. $$ Choose $u_2(1)=0$ and one picks $$ u_2=e^{\sqrt kt}-e^{\sqrt k(2-t)}. $$ So $$ W(u_1,u_2)(\xi)=2\sqrt k(e^{2\sqrt k}-1). $$ So the Green function is \begin{eqnarray*} G(t,\xi)&=&\left\{\begin{array}{ll} \frac{u_1(\xi)u_2(t)}{p(t)W(u_1,u_2)(\xi)}, 0\le\xi\le t,\\ \frac{u_1(t)u_2(\xi)}{p(t)W(u_1,u_2)(\xi)}, t\le\xi\le 1, \end{array}\right.\\ &=&\left\{\begin{array}{ll} \frac{1}{2\sqrt k(e^{2\sqrt k}-1)}(e^{\sqrt k\xi}-e^{-\sqrt k\xi})(e^{\sqrt k(2-t)}-e^{\sqrt kt}), 0\le\xi\le t,\\ \frac{1}{2\sqrt k(e^{2\sqrt k}-1)}(e^{\sqrt kt}-e^{-\sqrt kt})(e^{\sqrt k(2-\xi)}-e^{\sqrt k\xi}), t\le\xi\le 1. \end{array}\right. \end{eqnarray*} and hence (1) has the explicit solution $$ u=\int_0^1G(t,\xi)f(\xi)d\xi. $$ Thus \begin{eqnarray*} |u| &\le& \|f\|_{L^\infty}\int_0^1|G(t,\xi)|d\xi\\ &=&\frac{\|f\|_{L^\infty}}{2\sqrt k(e^{2\sqrt k}-1)}\bigg[(e^{\sqrt k(2-t)}-e^{\sqrt kt})\int_0^t(e^{\sqrt k\xi}-e^{-\sqrt k\xi})d\xi\\ &&+(e^{\sqrt kt}-e^{-\sqrt kt})\int_t^1(e^{\sqrt k(2-\xi)}-e^{\sqrt k\xi})d\xi\bigg]\\ &=&M_k(t)\|f\|_{L^\infty}. \end{eqnarray*} Here $$ M_k(t)=\frac{-e^{-\sqrt{k} (t-1)}-e^{\sqrt{k} t}+e^{\sqrt{k}}+1}{k(e^{\sqrt{k}} +1)} $$ which attains the maximum $$ M_k(\frac12) =\frac{\left(e^{\frac{\sqrt{k}}{2}}-1\right)^2}{k(e^{\sqrt{k}}+1)}$$ at $t=\frac12$. So $$ \|u\|_{L^\infty}\le M_k(\frac12) \|f\|_{L^\infty}. $$ Note if $f$ is constant, then (1) has a unique solution $$ u(t)=M_k(t)f $$ and hence $$ \|u\|_{L^\infty}= M_k(\frac12) \|f\|_{L^\infty}. $$ Therefore the best constant is $M_k(\frac12)$.

xpaul
  • 47,821
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I have a short answer for this equation without using the Green function. Let $v$ and $w$ be solutions of the following BVPs $$ -v''+kv=-\|f\|_{L^\infty} \text{ in } I, v(0)=v(1)=0 $$ $$ -w''+kw=\|f\|_{L^\infty} \text{ in } I, w(0)=w(1)=0 $$ respectively. Let $U=u-w$. It is easy to see $U$ satisfy $$ -U''+kU\le0, U(0)=U(1)=0. \tag1$$ Multiplying both sides of (1) by $U^+$ gives $$ \int_I\bigg[((U^+)')^2+k(U^+)^2\bigg]\le0 $$ which implies $U^+=0$ or $u\le w$. Similarly $u\ge v$ and hence $$ v\le u\le w. $$ It is easy to see $$ v=-M_k(t)\|f\|_{L^\infty}, w=M_k(t)\|f\|_{L^\infty}$$ from which one has $$ |u|\le M_k(t)\|f\|_{L^\infty}. $$ So $$ \|u\|_{L^\infty}\le M_k(\frac12)\|f\|_{L^\infty}. $$ Note if $f$ is constant, then (1) has a unique solution $$ u(t)=M_k(t)f $$ and hence $$ \|u\|_{L^\infty}= M_k(\frac12) \|f\|_{L^\infty}. $$ Therefore the best constant is $M_k(\frac12)$.

xpaul
  • 47,821