Let $(X, \mathfrak{M}, \mu)$ be a $\sigma$-finite measure space, let $1 < p < \infty$, and let $T$ be a bounded linear functional
on $L^p(X, \mu).$
I am trying to prove to myself that $T$ attains its norm.
The only thing that is coming to mind that we would need to show that for any $x \in X$, $$||Tx|| \leq ||T|| \: ||x|| \text{ and } ||Tx|| \geq ||T|| \: ||x||$$
Help would be appreciated.
I would like to use Mason's suggestion to reduce the problem down to:
Given $f \in L^q,$ find $g \in L^p$ with $||g||_p = 1$ such that $|\int gf| = ||f||_q$
Your thoughts would be appreciated
Best Answer
Since $\ L^q\ $ is the dual of $\ L^p\ $, where $\ q=\frac{p}{p-1}\ $, then there exists $\ f_T\in L^q\ $ such that $\ Tx=\int_\limits{X}f_Tx\,d\mu\ $ for all $\ x\in L^p\ $. By the Hölder inequality, therefore, $\ |Tx|\le\|f_T\|_q\,\|x\|_p\ $, and it follows that $\ \|T\| \le\|f_T\|_q\ $.
Now let $\ u=\frac{f_T}{|f_T|}\ $ and $$ g=u^{-1}\left|\frac{|f_T|}{ \|f_T\|_q}\right|^\frac{1}{p-1}\ . $$ Then \begin{align} \int_X|g|^p\,d\mu&=\int_X\frac{|f_T|^q}{\|f_T\|_q^q}\,d\mu\\ &=1\ , \end{align} so $\ g\ $ lies in the unit ball of $\ L^p\ $. But \begin{align} Tg&=\int_Xf_Tg\,d\mu\\ &=\int_X\frac{|f_T|^{1+\frac{1}{p-1}}}{\|f_T\|_q^ \frac{1}{p-1}}\,d\mu\\ &=\frac{\|f_T\|_q^q}{\|f_T\|_q^ \frac{1}{p-1}}\\ &=\|f_T\|_q\ . \end{align} It follows that $\ \|T\|= \|f_T\|_q $ and $\ Tg=\|T\|\ $.