Brezis' Theorem 8.8: how to get the inequality $\|u\|_{L^{\infty}(I)} \leq C\|u\|_{W^{1, p}(I)}$?

Question

Let $I:=(a, b)$ be an open interval, possibly unbounded. I'm reading below theorem in Brezis' Functional Analysis:

Theorem 8.8. There exists a constant $C$ (depending only on $|I| \leq \infty)$ such that $$ (5) \quad \|u\|_{L^{\infty}(I)} \leq C\|u\|_{W^{1, p}(I)} \quad \forall u \in W^{1, p}(I), \quad \forall 1 \leq p \leq \infty . $$ In other words, $W^{1, p}(I) \subset L^{\infty}(I)$ with continuous injection for all $1 \leq p \leq \infty$. Further, if $I$ is bounded then

• $(6) \quad$ the injection $W^{1, p}(I) \subset C(\bar{I})$ is compact for all $1<p \leq \infty$,
• $(7) \quad$ the injection $W^{1,1}(I) \subset L^q(I)$ is compact for all $1 \leq q<\infty$.

The proof of $(5)$ is given as follows:

We start by proving (5) for $I=\mathbb{R}$; the general case then follows from this by the extension theorem (Theorem 8.6). Let $v \in C_c^1(\mathbb{R})$; if $1 \leq p<\infty$ set $G(s)=|s|^{p-1} s$. The function $w=G(v)$ belongs to $C_c^1(\mathbb{R})$ and $$ w^{\prime}=G^{\prime}(v) v^{\prime}=p|v|^{p-1} v^{\prime} . $$ Thus, for $x \in \mathbb{R}$, we have $$ \begin{align*} G(v(x))=\int_{-\infty}^x p|v(t)|^{p-1} v^{\prime}(t) d t, \tag{$*$}\label{*} \end{align*} $$ and by Hölder's inequality $$ \begin{align*} |v(x)|^p \leq p\|v\|_p^{p-1}\left\|v^{\prime}\right\|_p, \tag{$**$}\label{**} \end{align*} $$ from which we conclude that $$ (8) \quad \|v\|_{\infty} \leq C\|v\|_{W^{1, p}} \quad \forall v \in C_c^1(\mathbb{R}), $$ where $C$ is a universal constant (independent of $p$).

Argue now by density. Let $u \in W^{1, p}(\mathbb{R})$; there exists a sequence $\left(u_n\right) \subset C_c^1(\mathbb{R})$ such that $u_n \rightarrow u$ in $W^{1, p}(\mathbb{R})$ (by Theorem 8.7). Applying (8), we see that $\left(u_n\right)$ is a Cauchy sequence in $L^{\infty}(\mathbb{R})$. Thus $u_n \rightarrow u$ in $L^{\infty}(\mathbb{R})$ and we obtain (5).

My understanding let $p'$ be the Hölder conjugate of $p$. Then $(p-1)p'=p$ and thus $$ \begin{align*} \int_{-\infty}^x |v(t)|^{p-1} |v^{\prime}(t)| d t &\le \left (\int_{-\infty}^x |v(t)|^{(p-1)p'} dt \right)^{1/p'} \left (\int_{-\infty}^x |v^{\prime}(t)|^p \right)^{1/p} \\ &\le \|v\|_{p}^{p/p'} \| v'\|_p \\ &= \|v\|_{p}^{p-1} \| v'\|_p. \end{align*} $$

Then $|v(x)|^p \le p\|v\|_{p}^{p-1} \| v'\|_p$ and thus $\|v\|^p_\infty \le p\|v\|_{p}^{p-1} \| v'\|_p$.

Could you explain how to go from (\ref{**}) to $(8)$?

Answer 1

We have for all $x\in \mathbb R$ that $$ \vert v(x)\vert^p \leqslant p\| v\|_p^{p-1} \| v’\|_p \leqslant p \| v\|_{W^{1,p}}^p. $$ Now take $p$-th root of both sides then take the supremum over $x\in \mathbb R$ to get $$ \| v\|_\infty \leqslant p^{1/p} \| v\|_{W^{1,p}} . $$ Finally, use that $x \mapsto x^{1/x}$ is bounded from above to obtain the constant in (8) can be chosen independently of $p$.