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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? – user16299 Feb 8 '12 at 22:15
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 norms – user16299 Feb 8 '12 at 22:17
@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. – Jonas Teuwen Feb 8 '12 at 22:20
But you are asking about the norms of two functions, rather than the identity $\langle Tf, g\rangle = \langle f, Tg \rangle$ – user16299 Feb 8 '12 at 22:25
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 $\lambda \nearrow \infty$. – Willie Wong Feb 8 '12 at 23:38

Consider e.g. $Tf(x) = \int_0^1 f(y)\ dy$ on $L^p[0,1]$. This is about as nice an operator as you would hope to find: in $L^2$ it's the orthogonal projection on the constant function $1$. Then $\|(Tf) g\|_1 = |\langle f, 1 \rangle| \|g\|_1$ while $\|f (Tg)\|_1 = \|f\|_1 |\langle g,1\rangle|$. For example, if $f(x) = 1$ and $g(x) = e^{2 \pi i n}$ where $n$ is a nonzero integer, $\|(Tf)g\|_1 = \|g\|_1 = 1$ while $\|f(Tg)\|_1 = 0$. I don't know what kind of relation between these you could hope to find.

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Hah, same counterexample that I came up with. – Willie Wong Feb 8 '12 at 23:56
Great. Thanks. I definitely need to think more about it. – Jonas Teuwen Feb 9 '12 at 6:45

To expand on my comment, consider $T:L^2\to L^2$ bounded self-adjoint operator, a necessary condition is that $\not\exists f\in \ker T$ and $g\in (\ker T)^\perp$ such that $f Tg \not\equiv 0$. Otherwise $\|Tf g\| = 0$ is not comparable to $\|f Tg\| \neq 0$.

This is, of course, not sufficient, since if you let $\widehat{Tf} = e^{-\lfloor\xi\rfloor}\hat{f}(\xi)$. This operator is bounded on $L^2$, and has no kernel. But if you let $\hat{f}$ be the ball of radius 1 and $\hat{g}_n$ be the annulus of inner radius $n$ and outer radius $n+1$, you will still have no uniform constant comparing $\|fTg\|$ with $\|Tfg\|$.

A sufficient condition, OTOH, at least for functions in $L^2$, you can get via the spectral theorem. If $T$ is a bounded linear operator, consider the measure space $(X,\Sigma,\mu)$ and the isomorphism $\Phi$ from $L^2$ to $L^2_\mu(X)$ such that $T$ is represented by $f\mathrm{d}\mu$. If $\Phi$ is such that whenever $u,v\in L^2_\mu(X)$ have disjoint essential support then so do $\Phi^{-1}u,\Phi^{-1}v$, we can conclude that (if I am not mistaken) that $\|Tf g\| = \|f Tg\|$. But this is a very artificial and strong condition. And I don't know if this is at all easy to verify in applications.

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The operator $T$ is quite "complex" it it built out of a partial derivative, an unbounded operator and its semigroup (but is bounded itself). I will try to figure out if I have more useful conditions. – Jonas Teuwen Feb 9 '12 at 6:47

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