the short time Fourier transform is obtained by the formula:
$$Sf(u,\epsilon)=\int_\mathbb{R}f(t)g(t-u)e^{-i\epsilon t}dt$$
where $f,g \in L^2(\mathbb{R})$ are the signal and window respectively:
and the reconstruction formula is given by:
$$f(t)=\frac{1}{2\pi}\int_\mathbb{R}\int_\mathbb{R}Sf(u,\epsilon)g(t-u)e^{i\epsilon t}d\epsilon du$$.
My problem is: I currently don't see why the right hand side of the inversion formula even exists. What you may use is that $Sf \in L^2(\mathbb{R}^2)$ and $g \in L^2(\mathbb{R})$, too. It smells like Cauchy Schwarz, but I just don't see it.
If anything is unclear, please let me know.
I'm basically going to paraphrase from Mallat's wavelet tour text where you can find this proof. You can fill in the gaps, make it more rigorous, change assumptions etc...
I'm going to assume that the window function, $g(t)$, is real, symmetric, and has unit norm, i.e. $g(t) \in \mathbb{R}$, $g(t) = g(-t)$, and $||g||_2 = 1$.
Using Parseval's identity on the integral over $u$, we have:
$$ \int_{\mathbb{R}} Sf(u,\epsilon) g(t-u) du = \frac{1}{2\pi} \int_{\mathbb{R}} \widehat{Sf}(w,\epsilon) \hat{g}(w) e^{iwt} dw $$
Next, note that:
$$ Sf(u,\epsilon) = e^{-i\epsilon u}(f*g_{\epsilon})(u) $$
where $g_{\epsilon}(u) = g(u) e^{i\epsilon u}$, so that:
$$ \widehat{Sf}(w,\epsilon) = \hat{f}(w+\epsilon) \hat{g}(w) $$
Finally, we have by substituting back in: $$ \frac{1}{2\pi}\int_{\mathbb{R}}\int_{\mathbb{R}} Sf(u,\epsilon) g(t-u) e^{i\epsilon t} du d\epsilon = \frac{1}{(2\pi)^2} \int_{\mathbb{R}} \hat{g}(w)^2 \int_{\mathbb{R}} \hat{f}(w+\epsilon) e^{i(\epsilon +w)t} d\epsilon du = f(t) $$