Sobolev Inequality in 1D Reading through Brezis and I am having trouble understand the proof of the following theorem:
There exists a constant $C$ such that $\|u\|_{\infty}\leq C\|u\|_{W^{1,\,p}}$ $\forall u\in W^{1,\,p}$, $1\leq p\leq\infty$.
Proof
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,
$$w'=G'(v)=p|v|^{p-1}v.$$
This is where I encounter my first problem: Differentiating $G(v)$ with the chain rule you have,
$$G'(v)=(p-1)|v|^{p-1}\frac{v}{|v|}+|v|^{p-1}$$
In order to get the result Brezis has quoted you have to assume $\frac{v}{|v|}=1$. How can you make this assumption?
Proof Continued
Thus, for $x\in\mathbb{R}$, we have
$$G(v(x))=\int_{-\infty}^{x}p|v(t)|^{p-1}v'(t)dt,$$
and by H$\ddot{\text{o}}$lder's inequality,
$$|v(x)|^{p}\leq p\|v\|_{p}^{p-1}\|v'\|_{p}$$
Here is my second problem: How do you use H$\ddot{\text{o}}$lder's inequality to arrive at the above inequality?
 A: I can't tell if you've miscopied or misread, but your statement about $w'$ is wrong.  What Brezis actually writes (p. 213, section 8.2) is
$$ w' = G'(v)v' = p|v|^{p-1}v' $$
Now, $w' = G'(v)v'$ is just the chain rule like you say, and for $G'(v)$ we calculate:
$$ G'(v) = (p-1)|v|^{p-2}\cdot \mathop{sgn} v\cdot v + |v|^{p-1}$$
which simplifies to $(p-1)|v|^{p-1}$ ready for multiplication by $v'$.
For the second part, we recall Holder's inequality first (theorem 4.6 on p.92):
$$\int |fg| \leq \| f\|_p \|g\|_{p'} \qquad \mbox{where } \frac{1}{p}+\frac{1}{p'} =1  $$
and observe that 
$$ |v(x)|^p = |G(v(x))| \leq \int \left| p\cdot |v|^{p-1} v'\right| \leq p\|v\|^{p-1}_{p'} \|v'\|_{p}$$
Finally we remember that $1/p+1/{p'} =1$ so that $p' = (p-1)/p$, which allows us to move from $\|\cdot\|_{p'}$ to $\|\cdot\|_p$ since $\|v\|^{p-1}_{p'} = (\int |v|^{(p-1)\cdot p'})^{1/p'} = (\int |v|^{(p-1)\cdot (p-1)/p})^{p/(p-1)} = (\int (|v|^{p-1})^p)^{1/p}$ (cancelling and rearranging, relying on the fact that $|v|$ is non-negative always), and this is just $\|v\|^{p-1}_p$.
