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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

An Isomorphic Teen
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  • Just for fun, one can prove what is known as James’ theorem: $X$ is reflexive if and only if every element of $X’$ attains its norm. As a corollary your claim would follow. – Son Gohan Apr 03 '22 at 13:34
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    Interesting! Ill have to look into that at some point. I would rather have this proven directly at this point though – An Isomorphic Teen Apr 03 '22 at 13:36
  • This may help. (Maybe there is a simpler proof for your specific space.) – David Mitra Apr 03 '22 at 16:15
  • Thank you for that. Just from first glance it appears to be a bit over my head in regards to what I'm looking for. If you think of something simpler for my case please do share. – An Isomorphic Teen Apr 03 '22 at 16:23
  • @AnIsomorphicTeen I think you have to use the fact that $L^{q}$ is the dual space of $L^p$, where $q$ is the conjugate exponent to $p$. Now the problem reduces to "given $f \in L^q$, find $g \in L^p$ with $|g|_p = 1$ such that $|\int gf| = |f|_q$. – Mason Apr 03 '22 at 16:53
  • Interesting. Do you think the Riez representation theorem for the dual of $L^p$ can come into play at all? – An Isomorphic Teen Apr 03 '22 at 17:02
  • If you put $\ f=e^{i\theta}|f|\ $, with $\ \theta:X\rightarrow(-\pi,\pi]\ $, and $$ g=e^{-i\theta}\left|\frac{f}{|f|_q}\right|^\frac{1}{p-1} $$ then $\ g\in L^p\ $ with $\ |g|_p=1\ $ and $\ \int fg =|f|_q\ $. – lonza leggiera Apr 05 '22 at 14:40
  • Does this work for any $f$? – An Isomorphic Teen Apr 05 '22 at 15:06
  • Okay thank you. If you could formalize this as a canonical answer so that I may accept the answer and give the bounty that would be very helpful. – An Isomorphic Teen Apr 05 '22 at 15:12
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    It'll be a little while before I'll be able to do that but I'll do it in a day or so if nobody else beats me to it. – lonza leggiera Apr 05 '22 at 15:29

1 Answers1

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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\|\ $.

lonza leggiera
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