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
- Given $f \in L^1(I)$ prove that there exists a unique $u \in H_0^1(I)$ satisfying $$ (1) \quad \int_I u' v'+k \int_I u v=\int_I f v \quad \forall v \in H_0^1(I) . $$
- Show that $u \in W^{2,1}(I)$.
- Prove that $$ \|u\|_{L^1} \leq \frac{1}{k}\|f\|_{L^1} . $$
- 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 $p$.
- 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 $$ \|u\|_{L^\infty} \le \frac{C^2}{\alpha} \|f\|_{L^\infty}. $$
