I am studying Sobolev spaces, and my professors give us the following exercise, but I do not understand how is possible conclude it.
Let $1\leq p< \infty$ and $v\in C_c^\infty(\mathbb{R})$. If $$ |v(x)|^p\leq p || v ||_p^{p-1} || v' ||_p ,$$ then there existe $C>0$, such that $$ || v ||_\infty \leq C ||v||_{W^{1,p}}.$$
How is possible proof it ?
Recall Young's inequality: for $\alpha, \beta \in [0,1]$ with $\alpha + \beta = 1$, one has $a^{\alpha}b^{\beta} \leq \alpha a + \beta b$ for any non-negative real numbers $a$ and $b$. Take in this case $\alpha = (p-1)/p$, $\beta = 1/p$, $a = ||v||^{p}_{p}$, $b = ||v'||^{p}_{p}$.
From the condition and the above inequality, you have that: \begin{equation} |v(x)|^p \leq p||v||^{p \times \frac{p-1}{p}}_{p} ||v'||^{p \times \frac{1}{p}}_{p} \leq p \left( \frac{p-1}{p} ||v||_{p}^{p} + \frac{1}{p} ||v'||_{p}^{p} \right) \leq C(p)||v||_{W^{1,p}}^{p}, \end{equation} where $C(p) = p \max\{ (p-1)/p, 1/p\}$. By continuity and compactness, take the $p$-th root, then the supremum over all $x$ to conclude.