$\int_{\mathbb R^{2}} |\int_{\mathbb R} (f(t-y)- f(t)) g(t-x) e^{-2\pi i w\cdot t} dt|dx dw \to 0 $ as $y\to 0$?

Question

Fact: It is well-known that translation is continuous in the $L^{1}$ norm, that is, if $f\in L^{1}(\mathbb R)$ then $\lim_{y\to 0} \|f_{y}-f\|_{L^{1}(\mathbb R)}=0;$ (where, $f_{y}(x)= f(x-y)$, $x\in \mathbb R$ ).

(Bit roughly speaking, we consider the analogous case for the Fourier transform) Suppose $f, \hat{f} \in L^{1}(\mathbb R).$

Consider, $\widehat{(f_{y}-f)} (\xi)= \int_{\mathbb R} (f_{y}-f)(x) e^{-2\pi i \xi \cdot x} dx= \int_{\mathbb R} (f(y-x)-f(x)) e^{-2\pi i \xi \cdot x} dx = (e^{-2\pi i y \cdot \xi }-1) \hat{f}(\xi); (\xi \in \mathbb R);$ therefore, by dominated convergence theorem, it follows that, $\|\widehat{f_{y}- f}\|_{L^{1}(\mathbb R)}\to 0$ as $y\to 0.$(Please correct me, If I am wrong 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 $f, \hat{f} \in L^{1}(\mathbb R),$ and $\|V_{g}f\|_{L^{1}(\mathbb R^{2})} <\infty.$ Can we expect $\|V_{g}f_{y}- V_{g}f\|_{L^{1}(\mathbb R \times \mathbb R)} \to 0$ as $y\to 0,$ that is, $\int_{\mathbb R^{2}} |\int_{\mathbb R} (f(t-y)- f(t)) g(t-x) e^{-2\pi i w\cdot t} dt|dx dw \to 0 $ as $y\to 0$ ?

Thanks,

Answer 1

For $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$ and $y\in\mathbb{R}^{d}$, let us write $\left(M_{y}f\right)\left(x\right):=e^{2\pi i\left\langle x,y\right\rangle }\cdot f\left(x\right)$. It is then easy to see $\left\Vert M_{y}f\right\Vert _{1}=\left\Vert f\right\Vert _{1}$ for $f\in L^{1}(\mathbb{R}^{d})$ [in fact, this remains valid for every $p\in\left(0,\infty\right]$ instead of only $p=1$].

Now a direct calculation yields \begin{eqnarray*} \left(V_{g}f_{y}\right)\left(x,w\right) & = & \int_{\mathbb{R}}f\left(t-y\right)\cdot g\left(t-x\right)\cdot e^{-2\pi iwt}\, dt\\ & \overset{s=t-y}{=} & \int_{\mathbb{R}}f\left(s\right)\cdot g\left(s+y-x\right)\cdot e^{-2\pi iw\left(s+y\right)}\, ds\\ & = & e^{-2\pi iwy}\cdot\int_{\mathbb{R}}f\left(s\right)\cdot g\left(s-\left(x-y\right)\right)\cdot e^{-2\pi iws}\, ds\\ & = & e^{-2\pi iwy}\cdot\left(V_{g}f\right)\left(x-y,w\right)\\ & = & e^{-2\pi iwy}\cdot\left(\left(V_{g}f\right)_{\left(y,0\right)}\right)\left(x,w\right)\\ & = & \left(M_{\left(0,-y\right)}\left(\left(V_{g}f\right)_{\left(y,0\right)}\right)\right)\left(x,w\right). \end{eqnarray*} Using this and the isometry of $M_{\left(0,-y\right)}$ that we noted above, we derive \begin{eqnarray*} \left\Vert V_{g}f_{y}-V_{g}f\right\Vert _{1} & \leq & \left\Vert M_{\left(0,-y\right)}\left(\left(V_{g}f\right)_{\left(y,0\right)}\right)-M_{\left(0,-y\right)}\left(V_{g}f\right)\right\Vert _{1}+\left\Vert M_{\left(0,-y\right)}\left(V_{g}f\right)-V_{g}f\right\Vert _{1}\\ & = & \left\Vert \left(V_{g}f\right)_{\left(y,0\right)}-V_{g}f\right\Vert _{1}+\left\Vert M_{\left(0,-y\right)}\left(V_{g}f\right)-V_{g}f\right\Vert _{1}\\ & \xrightarrow[y\rightarrow0]{} & 0. \end{eqnarray*} Here, the first term vanishes because (as you noted yourself), $\left\Vert f_{y}-f\right\Vert _{1}\xrightarrow[y\rightarrow0]{}0$ and the second term vanishes basically as in the other part of your prove, because for each sequence $\left(y_{n}\right)_{n}$ with $y_{n}\rightarrow0$ you have $M_{\left(0,-y_{n}\right)}\left(V_{g}f\right)-V_{g}f\rightarrow0$ pointwise, which together with the majorization $$ \left|M_{\left(0,-y\right)}\left(V_{g}f\right)-V_{g}f\right|\leq2\left|V_{g}f\right|\in L^{1}(\mathbb{R}^{2}) $$ yields $L^{1}$-convergence.