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I have a bounded operator $T$ from $L^p$ to itself for $1 \leqslant p \leqslant \infty$. Furthermore, on $L^2$ we have that $T$ is self-adjoint.

Now I wish to relate $\|(Tf)g\|_{L^1}$ to $\|f(Tg)\|_{L^1}$ (equal up to a constant perhaps). What properties should I need for $T$ for this to hold?

The question is not really well-defined, but I don't know what property I should look for in my operator.

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    Some thoughts: since you know $T$ is bounded on every $L^p$, I think it's enough to consider $f$ and $g$ continuous with compact support (the general case whould then follow by approximation, perhaps with the case $f\in L^1$, $g\in L^\infty$ needing separate treatment). Secondly, is your operator $T$ given formally by convolution with some kind of kernel function?2012-02-08
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    I'm also not quite sure how your title relates to the question you ask. It certainly follows from your assumption that $T$ can be identified with $T^*, perhaps after throwing in complex conjugation, but I don't see how that relates to your question about norms2012-02-08
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    @YemonChoi The title is because of to the "analogy" with an inner product. Do you have a suggestion for a better title? I do have an integral kernel for $T$, but is not a convolution.2012-02-08
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    But you are asking about the norms of two functions, rather than the identity $\langle Tf, g\rangle = \langle f, Tg \rangle$2012-02-08
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    Hence the "some kind of" :-).2012-02-08
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    What is the operator $T$? Without knowing it there is absolutely nothing you can say. For convenience sake let the underlying space be the circle. Let $Tf = \frac{1}{2\pi}\int f$ the constant function. Then $\langle Tf,g\rangle = \langle f,Tg\rangle = \int f \int g$. But If you let $f = 1 + \lambda \sin \theta$ and $g = 1$, $\|Tf g\| = 2\pi$ but $\| f Tg\| = \|f\|$ which can be arbitrarily large by taking $\lambda \nearrow \infty$.2012-02-08

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