# Norm of a bounded linear functional.

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?

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

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 $$b$$).