I am considering an operator of the form $T_\mu = S_\mu^* S_\mu$ with $\mu,k \in \mathcal C^\infty(\mathbb R)$, which is a mix of a convolution- and multiplication operator (see also this and this question), $$ S_\mu(f)(x) = (k * (\mu f))(x) $$ By Fourier transformation, it can be written as $$ S_\mu(f)(x) = \int dp ~ e^{ipx} \widetilde k(p) ~ (\mu f)^\sim(p) = \int dp ~ e^{ipx} \widetilde k(p) (\widetilde \mu * \widetilde f)(p). $$ This operator is positive as it is of the form $S^* S$ and which can be seen directly $$ \langle f, T_\mu f \rangle = \int dp ~ \widetilde{k}^2(p) ~ |(\mu f)^\sim(p)|^2 \geq 0 $$ Question: Under what condition on functions $\mu,\nu \in \mathcal C^\infty$ is the difference $T_\mu - T_\nu$ a positive operator?
Attempts:
- One sufficient condition would be $|(\mu f)^\sim(p)|^2 > |(\nu f)^\sim(p)|^2$. However, it involves an arbitrary function $f$ and is therefore not easily proven for known $\mu,\nu$.
- For a pure multiplication operator, this requirement would be $|\mu(x)|^2 > |\nu(x)|^2$. For a pure convolution operator, it would be the same condition Fourier transformed. This is a mix of both cases (even though the convolution part is the same). But it is unclear to me how it changes the condition on $\mu,\nu$?
- Maybe $|\mu(x)|^2 > |\nu(x)|^2$ is a sufficient condition, even it is not necessary?
Let me know if more detail is needed. I am thankful for any suggestions!