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Consider a sufficiently fast decaying and smooth function $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ such that $$ \int_{\mathbb{R}} f(x)dx=0. $$ Now consider the Schrodinger semigroup $e^{it\partial_x^2}$, that is, the operator such that $$ \mathcal{F}\big[e^{it\partial_x^2}f\big](\xi)=e^{-it\xi^2}\mathcal{F}[f](\xi). $$ Is it true that $$ \int_{\mathbb{R}} e^{it\partial_x^2}\big(f(x-t)\big)dx=0? $$ I do believe that, if we put $f(x)$ instead of $f(x-t)$ in the last line it should be true since (sufficiently fast decaying) solutions to the equation $$ i\partial_t u+\partial_x^2u=0 $$ conserve the average $$ \int_{\mathbb{R}}u(t,x)dx=\int_{\mathbb{R}}u(0,x)dx. $$ So in that sense, I would like to do something like $$ \int_{\mathbb{R}} e^{it\partial_x^2}\big(f(x-t)\big)dx = \int_{\mathbb{R}} f(x-t)dx = \int_{\mathbb{R}} f(x)dx=0, $$ but I'm not 100% sure if that is allowed given the time dependence in $f(x-t)$. Does anybody know how to prove or disprove the above claim?

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Denote Fourier transform by $\hat{}$. $$ \hat{f}(0)=\int_{-\infty}^{\infty} f(x)dx=0 $$ Let $g_t(x)=f(x-t)$ and $h_t(x)=\left(e^{it\partial_x^2}g_t\right)(x)$. You're looking for the value of $$ \hat{h}(0)=\int_{-\infty}^{\infty} h(x)dx=0 $$ Now $$ \begin{align} \hat{g_t}(\xi)&=\int_{-\infty}^{\infty} g_t(x)e^{-ix\xi}dx\\ &=\int_{-\infty}^{\infty} f(x-t)e^{-ix\xi}dx\\ &=\int_{-\infty}^{\infty} f(x')e^{-i(x'+t)\xi}dx'\\ &=e^{-it\xi}\int_{-\infty}^{\infty} f(x')e^{-ix'\xi}dx'=e^{-it\xi}\hat{f}(\xi) \end{align} $$ and $$ \hat{h}(\xi)=e^{-it\xi^2}\hat{g_t}(\xi)=e^{-it\xi^2}e^{-it\xi}\hat{f}(\xi)=e^{-it(\xi^2+\xi)}\hat{f}(\xi) $$ so $$ \hat{h}(0)=e^{-it(0^2+0)}\hat{f}(0)=1\cdot 0=0 $$

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Yes, that's ok. For each fixed $t\in\mathbb R$, the operator $e^{it\partial_x^2}$ commutes with translations, as you can see via the PDE or via the Fourier transform, as you prefer. Therefore, $$ e^{it\partial_x^2}(f(\cdot - t))(x)=(e^{it\partial_x^2}f)(x -t).$$ Integrating in $x$, changing variable to $y=x-t$ (no problem here, since $t$ is fixed), you are back to the case you know.

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