Counter example for a norm relationship

Question

Function $w(x)$, $x\in[0,1]$ is four times differentiable, and its forth derivative is absolutely continuous. In addition, it satisfies the boundary conditions $w(0)=w(1)=w''(0)=w''(1)$. I am wondering if $\| w\|_2<C$ gives $\| w\|_p<K_C$ and vice versa.

Answer 1

The differentiability of $w$ and the boundary conditions do not really matter here. First, using Hölder's inequality, it is easy to check that $$ \Vert w\Vert_{L^p([0,1])}\leq \Vert w\Vert_{L^q([0,1])} $$ for any $1\leq p\leq q$. If $p>q$, then it is known that $w\in L^p$, since it is continuous on a compact, thus $L^\infty$, but there is no control of the $L^p$ norm in terms of the $L^q$ norm. Indeed, choose any $\rho\in C^\infty(\mathbb R)$ with compact support in $[0,1]$ and such that $\Vert \rho\Vert_q=1$. Then, for $\epsilon>0$, define $$ \rho_\epsilon:x\mapsto \epsilon^{-1/q}\rho(x/\epsilon). $$ It is easy to check that $(\rho_\epsilon)_{0<\epsilon\leq1}$ verifies the conditions you asked for, and $\Vert \rho_\epsilon\Vert_q=1$, but $\Vert \rho_\epsilon\Vert_p=\epsilon^{1/p-1/q}\Vert \rho\Vert_p\to+\infty$ as $\epsilon \to 0$.