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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

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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 $\epsilon \to 0$.

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    Would the limit of the function $\rho$ not a member of $C^\infty$? – 2017-02-01
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    The sequence $\rho_\epsilon$ has no limit in $L^p$ indeed, the purpose of this is just to show that there can not be a constant $K$ depending only on $C$ such that $\Vert w\Vert_2 < C \implies \Vert w\Vert_p < K$, as $\Vert w\Vert_p$ can be as big as we want it to be. – 2017-02-01
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    True; that's clear; but, I have restricted my functions to be in a much smaller subspace. I mean both $L_p$ and $L_q$ norms are defined for all members of this subspace. – 2017-02-01
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    I hear you, but the set of $C^\infty$ functions with compact support in $[0,1]$ is a subset of the subspace you are considering. For all $0<\epsilon\leq 1$, the functions $\rho_\epsilon$ definitely are in the required subspace. – 2017-02-01
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    So having said this; I am guessing the function $f: L^2\to L^2 , \; w\mapsto w^3$ is not even locally Lipschitz function since we cannot write $\|w\|_6 \le K_C \|w\|_2, \, \forall \|w\|_2\le C$; true? – 2017-02-02
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    That is true, since in any neighbourhood of any point in $L^2$, following the reasoning, it is possible to find a function with as big an $L^6$ norm as you want it to be. That is also true in the particular subspace you desvribed. – 2017-02-06