# How is $\Vert x \Vert_3$ constrained by $\Vert x \Vert_2$ and $\Vert x \Vert_4$ for $x\in \ell^2$?

It seems we can find some $x\in \ell^2$ with $\Vert x \Vert_2=1$ that has $\Vert x \Vert_4=a$ for any $0<a\le 1$.

But can we find an $x$ with $\Vert x \Vert_2=1,\Vert x \Vert_3=b,\Vert x \Vert_4=a$ for every choice of $0<a<b<1$?

(This was inspired by this longstanding MO question that made me curious about the flexibility of the three norms.)

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Since $1/3 = 1/3*1/2 + 2/3*1/4$, I would expect something like $\|x\|_3 \leq \|x\|_2^{1/3} \|x\|_4^{2/3}$, or in other words $b \leq a^{2/3}$, to be necessary (see en.wikipedia.org/wiki/Interpolation_space). Alas, I don't have the material to check the details right now. – D. Thomine Jun 24 '12 at 14:08

This is extended version of D. Thomine's comment. It is known that for a given real numbers $\{\theta_k:k\in\{1,\ldots,n\}\}$$\subset(0,1), such that \sum_{k=1}^n \theta_k=1 and real numbers \{p_k:k\in\{1,\ldots,n\}\}\subset\mathbb{R}_+ we have the following generalized Hölder inequality:$$ \Vert x\Vert_{p}\leq\prod\limits_{k=1}^n\Vert x\Vert_{p_k}^{\theta_k} $$where p^{-1}=\sum_{k=1}^n\theta_k p_k^{-1}. In your particular case we take n=2, \theta_1=1/3, \theta_2=2/3, p_1=2 and p_2=4. Then we get p=1/3 and$$ \Vert x\Vert_3\leq\Vert x\Vert_2^{1/3}\Vert x\Vert_4^{2/3}$$which is equivalent to$b\leq a^{2/3}$. Thus in general we can't find$x\in\ell_2\$ satisfying all conditions mentioned above.