I'm trying to understand the proof of the Inverse Fourier Transform theorem in Stéphane Mallat's "A wavelet tour of signal processing". Near the end of the proof, we have:
$ \lim_{\epsilon \rightarrow 0} \int_{-\infty}^{\infty} |I_\epsilon(t) - f(t)| dt = \lim_{\epsilon \rightarrow 0} \int \int g_\epsilon(t-u) |f(u) - f(t)| du dt = 0$
where
$ I_\epsilon(t) = \int_{-\infty}^\infty g_\epsilon(t-u) f(u) du$
and $g_\epsilon$ is a squeezed Gausian that satisfies $\lim_{\epsilon \rightarrow 0} \int g_\epsilon(t-u) h(t) dt = \int \delta(t-u) h(t) dt = h(u)$
I don't understand the first equality in the first equation. It seems like the $g_\epsilon$ is taken out of the absolute value and I don't know how to justify this. Any pointers?
More than likely, the equality would be inequality, with equality only in the limit. Such arguments almost always involve assuming that the approximate delta function satisfies $$ \int_{-\infty}^{\infty}g_{\epsilon}(t-u)du = 1. $$ Therefore, one can move any constant (with respect to the variable of integration) inside $$ \int_{-\infty}^{\infty}g_{\epsilon}(t-u)f(u)du-f(t) = \int_{-\infty}^{\infty}g_{\epsilon}(t-u)(f(u)-f(t))du $$ Then $$ \left|\int_{-\infty}^{\infty}g_{\epsilon}(t-u)f(u)du-f(t)\right| \le \int_{-\infty}^{\infty}g_{\epsilon}(t-u)|f(u)-f(t)|du. $$ Finally, $$ \int_{-\infty}^{\infty}\left|\int_{-\infty}^{\infty}g_{\epsilon}(t-u)f(u)du-f(t)\right|dt \le \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g_{\epsilon}(t-u)|f(u)-f(t)|dudt $$ The limits of the two expressions are the same, but only because the limit on the right is 0, and the limit on the left is sandwiched between 0 and the limit on the right, which is 0. It's a sloppy way to state such a limit, because the existence of the limit on the left is being tacitly proved. A $\limsup$ would have been better, with a conclusion that the limit must actually exist and be 0.