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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 Alas, I don't have the material to check the details right now. – D. Thomine Jun 24 '12 at 14:08
up vote 4 down vote accepted

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.

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