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This problem arose while I was trying to solve exercise 6.41 in Folland's real analysis.

Suppose $1<p\leq \infty$ and let $q$ be the conjugate exponent of $p$ (i.e. $p^{-1}+q^{-1}=1$). If $T$ is a bounded operator in $L^p$ such that $$ \int (Tf) g = \int f(Tg) \qquad \forall f,g\in L^P\cap L^q $$ then can one show that $T$ is linear?

I was able to show that for any functions $f_1, f_2$ on $L^p\cap L^q$ and any $g\in L^p \cap L^q$ we have
$$ \int (T[f_1+f_2]) g = \int (Tf_1 + Tf_2)g $$ I am not sure how to proceed from here. Any hints are appreciated.

My ultimate goal is to extend $T$ to a bounded operator on $L^r$ where $r$ is some number between $p$ and $q$. If there is another approach which does not require to show that $T$ is linear that would also solve my problem as well.

Thanks in advance.

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It seems to me like you are pretty much done. Indeed, if you can show that $$\int (T[f_1 + \alpha f_2])g = \int (Tf_1 + \alpha Tf_2)g$$ for all $g \in L^p \cap L^q$, then by boundedness of $T$ and density of $L^p \cap L^q$ in $L^q$, you have $$\int (T[f_1 + \alpha f_2])g = \int (Tf_1 + \alpha Tf_2)g, \,\,\, \forall g \in L^q.$$ But then $h \in L^p$ defined by $h = T[f_1 + \alpha f_2] - Tf_1 - \alpha Tf_2$ satisfies $$\int hg = 0, \,\,\, \forall g \in L^q$$ and so $$\| h \|_p = \sup_{g \in L^q, \|g \|_q \le 1} \left \lvert \int hg \right\rvert = 0$$ which shows that $h = 0$ and thus $$T[f_1 + \alpha f_2] = Tf_1 + \alpha Tf_2,$$ so $T$ is linear.

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  • $\begingroup$ So I have to assume that $T$ is a bounded operator from $L^p$ to $L^p$? (and not just from $L^p$). Otherwise, I don't know how to go from $g\in L^q\cap L^p$ to $g\in L^q$ $\endgroup$ – Quoka Mar 6 '18 at 20:34
  • $\begingroup$ I'm not exactly sure I understand what you're asking. I thought you had already assumed that $T$ was bounded from $L^p \to L^p$; otherwise what does "$T$ is a bounded operator in $L^p$" mean? $\endgroup$ – User8128 Mar 6 '18 at 20:41
  • $\begingroup$ The author never specified. I think this assumption is indeed required, I was wondering if maybe I was missing something.Thanks! $\endgroup$ – Quoka Mar 6 '18 at 20:45

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