We define the short time Fourier transform as follows: $$V_{g}f(x,w)=\int_{\mathbb R} f(t)g(t-x)e^{-2\pi itw} dt, (x,w \in \mathbb R).$$
(We may assume that $f$ and $g$ nice functions so that every make sense)
My Question is: How $V_{g}f(x,w)$ and $V_{g}(-w, x)$ are related ? Is it true that $\mid V_{g}f(x,w)\mid= \mid V_{g}(-w,x)\mid ?$
Suppose that $g$ has compact support, say $[-1,1]$. Then $$\begin{align} \phantom{-}V_{g}f(x,w)&=\int_{x-1}^{x+1}f(t)\,g(t-x)\,e^{-2\pi itw}\,dt,\\ V_{g}f(-x,w)&=\int_{-x-1}^{-x+1}f(t)\,g(t+x)\,e^{-2\pi itw}\,dt. \end{align} $$ If $|x|\ge1$, $V_{g}f(x,w)$ and $V_{g}f(-x,w)$ take into account values of $f$ on disjoint intervals, so that in general there is no relation between the two.