1
$\begingroup$

Let $X=(\mathbb R^2, \|.\|_3)$ be a real normed space, where $\|(x_1,x_2)\|_3=[|x_1|^3+|x_2|^3]^{1/3}$. How to find the norm of bounded linear functional $ax+by$?

I tried this way: $|ax+by|\leq |a||x|+|b||y|\leq \max\{|a|,|b|\}(|x|+|y|)\leq\max\{|a|,|b|\}([|x|^3+|y|^3]^{1/3}+|y|)\leq \max\{|a|,|b|\}([|x|^3+|y|^3]^{1/3}+[|x|^3+|y|^3]^{1/3})= 2\max\{|a|,|b|\}([|x|^3+|y|^3]^{1/3})=2\max\{|a|,|b|\} \|(x,y)\|_3 $. This implies $ax+by$ is bounded linear functional. Now how should I find the norm of this functional?

$\endgroup$
  • $\begingroup$ try to make all your inequalities equalities $\endgroup$ – mathworker21 Dec 10 '18 at 7:24
  • $\begingroup$ @mathworker21 I could not do. Please give some more hints. $\endgroup$ – Infinity Dec 10 '18 at 7:28
1
$\begingroup$

Hints: use the inequality $|ax+by| \leq (|x|^{3}+|y|^{3} )^{1/3} (|a|^{3/2}+|b|^{3/2} )^{2/3}$ (This is Holder's Inequality) to see that the norm of the functional is less than or equal to $(|a|^{3/2}+|b|^{3/2} )^{2/3}$. To show that the the norm is exactly equal to this number take $x=\pm t|a|^{1/2},y=\pm t|b|^{1/2}$ where $t =\frac 1 {\|(|a|^{1/2},|b|^{1/2})\}\|}$. (In the first term $\pm$ stands for $1$ if $a >0$, $-1$ otherwise. Similarly choose $\pm$ sign in the second term based on the sign of $b$).

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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