In Brezis book, in the chapter of the Sobolev spaces, there is a proof for the inequality \begin{gather} \| u \|_{L^{\infty}(I)} \leq C \| u \|_{W^{1.p}(I)} \end{gather} for every $u \in W^{1,p}(I)$ and $1 \leq p \leq \infty$ where he sets $I =\mathbb{R}$. Then he proceeds by setting $G(s) = |s|^{p-1}s$ and taking the function $w = G(v) \in C^1_c(\mathbb{R})$. He computes \begin{gather} w^\prime = G^\prime(v)v^\prime = p |v|^{p-1}v^\prime \end{gather} and thus for $x \in \mathbb{R}$ \begin{gather} G(v(x)) = \int_{-\infty}^x p |v|^{p-1}v^\prime(t)\ \mathrm{d}t \end{gather} He says that by Hölder's inequality \begin{gather} |v(x)|^p \leq p \|v\|^{p-1}_p \|v^\prime\|_p \end{gather} From which he concludes that \begin{gather} \| v \|_{\infty} \leq C \|v\|_{W^{1,p}} \end{gather} and then by density he proceeds to prove the claim.
I quite don't get how he passes through the last two steps, after defining the function $G(v(x))$. Can you please explain me the procedure in a bit more detail? Also, do you have any alternative prove? Particularly, I don't understand the motivation behind the definition of $G(s)$.
Thanks!
Maybe it is best to first consider $p=1$. In this case $G$ is simply the identity and the proof simplifies to $$ \lvert v(x)\rvert=\left\lvert\int_{-\infty}^x v'(t)\,dt\right\rvert\leq \int_{-\infty}^x\lvert v'(t)\rvert\,dt\leq \lVert v'\rVert_{L^1}. $$ Now we can take the supremum over $x\in \mathbb{R}$ and obtain $\lVert v\rVert_{L^\infty}\leq \lVert v'\rVert_{L^1}\leq \lVert v\rVert_{W^{1,1}}$.
The (rough) idea in the general case is to run the same proof, but apply the fundamental theorem of calculus to $\lvert v\rvert^p$ instead of $v$. The function $G$ is really just a replacement for $\lvert\cdot\rvert^p$ that repairs the non-differentiablity at $0$.
Let's now come to the steps you don't understand. Taking absolute values on both sides, your third equality reads $$ \lvert v(x)\rvert^p=p\left\lvert \int_{-\infty}^x \lvert v\rvert^{p-1}v'\,dt\right\rvert. $$ Next we apply Hölder's inequality with exponents $p/(p-1)$ and $p$ to the factors $\lvert v\rvert^{p-1}$ and $v'$ in the integral: $$ \left\lvert\int_{-\infty}^x \lvert v\rvert^{p-1}v'\,dt\right\rvert\leq \left(\int_{-\infty}^x (\lvert v\rvert^{p-1})^{p/(p-1)}\,dt\right)^{(p-1)/p}\left(\int_{-\infty}^x \lvert v'\rvert^p\,dt\right)^{1/p}\leq \lVert v\rVert_{L^p}^{p-1}\lVert v'\rVert_{L^p}. $$ Thus $$ \lVert v\rVert_{L^\infty}\leq p^{1/p}\lVert v\rVert_{L^p}^{(p-1)/p}\lVert v'\rVert_{L^p}^{1/p}. $$ In the last step we apply Young's inequality with exponents $p/(p-1)$ and $p$ to get $$ \lVert v\rVert_{L^\infty}\leq C(\lVert v\rVert_{L^p}+\lVert v'\rVert_{L^p})=C\lVert v\rVert_{W^{1,p}}. $$ Here all the fancy factors with $p$ where absorded in the constant $C$.