I'm working on some problems involving Fourier transforms and convolution problems and there is one problem I cannot solve. In my situation we have a complex valued function $f(ix)$, with $x\in\mathbb{R}$ and two functions from which we know the following: $|g_2(ix)|\le|g_1(ix)|=1$.
The convolution problem is the following, I have two functions (spectra in my case) that are defined as $$ U_1(i\omega) = \int_{-\infty}^\infty f(ix)g_1(i(\omega-x))\mathrm{d}x $$ $$ U_2(i\omega) = \int_{-\infty}^\infty f(ix)g_2(i(\omega-x))\mathrm{d}x $$ assuming these exists. I want to show that $|U_2(i\omega)|\le|U_1(i\omega)|$ for all $\omega$. Intuitively, this is true, since the magnitude of $g_1$ is equal to $1$ and $g_2$ is less or equal to $1$. Also, when I perform some simulation experiments with random functions $f$, I obtain that this relationship holds, however I cannot prove it. sadface
What I have been able to show is this: $$|U_1(i\omega)| = \int_{-\infty}^\infty f(ix)g_1(i(\omega-x))\mathrm{d}x \le \int_{-\infty}^\infty |f(ix)g_1(i(\omega-x))|\mathrm{d}x =\int_{-\infty}^\infty |f(i(\omega-x))|\mathrm{d}x$$ $$|U_2(i\omega)| = \int_{-\infty}^\infty f(ix)g_2(i(\omega-x))\mathrm{d}x \le \int_{-\infty}^\infty |f(ix)g_2(i(\omega-x))|\mathrm{d}x \le\int_{-\infty}^\infty |f(i(\omega-x))|\mathrm{d}x$$ Hence, the upperbound on $|U_2|$ is less or equal than the upperbound on $|U_1|$, however, this does not guarantee that $|U_2(i\omega)|\le|U_1(i\omega)|$ for all $\omega$.
Do you maybe have any idea how to approach this further?
Edit: According to some comments here below, I can extend this question with; How could I formulate extra (nontrivial) assumptions on $f$, $g_1$ and $g_2$, such that I can prove this relationship. (It is quite common for example that the considered functions are rational functions with polynomials as numerator and denominator)
