Is this some kind of adjoint? 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.
 A: 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.
A: 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. 
