I was learning the sobolev inequality in one dimensional case ,we have the following $\|u\|_{L^\infty(I)} \le C \|u\|_{W^{1,p}}(I)$
I can read the proof ,but I don't understand the motivation of the proof,let me present the proof below:
We start by proving for $I=\mathbb{R}$; the general case then follows from this by the extension theorem . 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 $$ G(v(x))=\int_{-\infty}^{x} p|v(t)|^{p-1} v^{\prime}(t) d t $$ and by Hölder's inequality $$ |v(x)|^{p} \leq p\|v\|_{p}^{p-1}\left\|v^{\prime}\right\|_{p}, $$ then apply the interpolation $a^\theta b^{1-\theta} \le \theta a + (1- \theta)b$ gets the result for $C^1_c$ function,finally use the density arguement gets the result for $W^{1,p}$
Using the function $|v|^{p-1} v$ looks a bit magical for me,is there some general technique behind the proof?
I can give an alternative proof. I use $\|u\|_{W^{1, p}} = \|u\|_{L^p} + \|u'\|_{L^p}$. By a density argument, it should suffice to establish a bound $$\| u \|_{L^{\infty}} \leq C\|u\|_{W^{1, p}} \text{ for }u \in C_c^{\infty}(\mathbb{R}).$$ Since $\| u\|_{L^{\infty}} = |u(x_0)|$ for some $x_0$, we can use translation invariance of the $W^{1, p}$ norm to reduce to proving a bound $$|u(0)| \leq C\|u\|_{W^{1, p}} \text{ for }u \in C_c^{\infty}(\mathbb{R}).$$ Now fix $\phi \in C_c^{\infty}(B_1(0))$ with $\phi(0) = 1$. It suffices to prove a bound $$|u(0)| \leq C\|(\phi u)'\|_{L^p}\text{ for }u \in C_c^{\infty}(\mathbb{R}).$$ So it suffices to prove a bound $$|u(0)| \leq C\|u'\|_{L^p}\text{ for }u \in C_c^{\infty}(B_1(0)).$$ But this is straightforward from $-u(0) = u(1) - u(0) = \int_{0}^{1}u'(x)\,dx$.