Is it true that $2^p\|f\|^p_{L_p} >2\|f\|^p_{L_\infty}$?

Question

For any $C^1$ function defined in $(0,1)$, is it true that $$ 2^p\|f\|^p_{L_p} >2\|f\|^p_{L_\infty} $$ If it is true, how to prove it?

Answer 1

It is not true. Consider, for $\delta>0$, $$ f(t)=\begin{cases}\frac{(x-\delta)^2}{\delta^2},&\mbox{ if } x\in[0,\delta]\\ 0,&\mbox{ otherwise}\end{cases} $$ Then $\|f\|_\infty=\|f\|_\infty^p=1$, and $$ \|f\|_p^p=\frac1{\delta^2}\int_0^\delta(x-\delta)^{2p}\,dx=\frac1{\delta^2}\left.\frac{(x-\delta)^{2p+1}}{2p+1}\right|_0^\delta=\frac{\delta^{2p-1}}{2p+1}. $$ As $\delta$ can be chosen arbitrarily small, the inequality can be made to fail.