Let $f\in \mathcal{S}(\mathbb R)$ with $\hat{f}$ has a compact support. For $r>0,$ put $f_{r}(x)= r^{-1}f(x/r), (x\in \mathbb R).$
We note that, $\int_{\mathbb R} |f_{r}(x)| dx = r^{-1} \int_{\mathbb R} |f(x/r)| dx = r^{-1}\int_{\mathbb R} |f(y)| r dy= \int_{\mathbb R} |f(x)| dx;$
and, $\hat{f_{r}} (\xi)= \int_{\mathbb R} f_{r}(x) e^{-2\pi i \xi \cdot x} dx= \hat{f}(r\xi); (\xi \in \mathbb R).$
Fix $y\in \mathbb R;$ define $f_{r}^{y}(t):= f_{r}(t-y), (t\in \mathbb{R} ).$
We note that, $\|f_{r}^{y}-f_{r}\|_{L^{1}(\mathbb R)} = \int_{\mathbb R} r^{-1} | f((x-y)/r)- f(x/r)| dx = \int_{\mathbb R} |f(z-r^{-1}y)- f(z)| dz \to 0 $ as $r\to \infty.$ (I guess, this can be justified by dominated convergence theorem; but my little confusion is, how should I choose dominating function, how to take $g\in L^{1}(\mathbb R)$ such that $|f_{r}(z)| \leq g(z), z\in \mathbb R ;$ please let point out to me this here; )
We define the short-time Fourier transform, of $f$ with respect to non zero window function $g\in \mathcal{S}(\mathbb R)$(Schwartz space), as follows: $$V_{g}f(x, w)=\int_{\mathbb R} f(t) g(t-x) e^{-2\pi i w\cdot t} dt= \widehat{fg^{x}}(w); (x,w)\in \mathbb R^{2}.$$
My Question is: Suppose $V_{g}f_{r}, V_{g}f^{y}_{r}\in L^{1}(\mathbb R \times \mathbb R),$ and fix $y\in \mathbb R.$ Can we expect $\|V_{g}f^{y}_{r}- V_{g}f_{r}\|_{L^{1}(\mathbb R \times \mathbb R)} \to 0$ as $r\to \infty,$ that is, $\int_{\mathbb R^{2}} |\int_{\mathbb R} (f_{r}(t-y)- f_{r}(t)) g(t-x) e^{-2\pi i w\cdot t} dt|dx dw \to 0 $ as $ r\to \infty $ ?