Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.