Spring 2015 - Problem 5

Fourier analysis

Let $u \in L^2\p{\R}$ and let us set

U\p{x, \xi} = \int e^{-\p{x+i\xi-y}^2/2} u\p{y} \,\diff{y}, \quad x, \xi \in \R.

Show that $U\p{x, \xi}$ is well-defined on $\R^2$ and that there exists a constant $C > 0$ such that for all $u \in L^2\p{\R}$ , we have

\iint \abs{U\p{x, \xi}}^2 e^{-\xi^2} \,\diff{x} \,\diff{\xi} = C \int \abs{u\p{y}}^2 \,\diff{y}.

First, observe that

\p{x + i\xi - y}^2 = \p{x - y}^2 + 2i\p{x - y}\xi - \xi^2.

Thus, its real part is $\p{x - y}^2 - \xi^2$ , which gives the estimate

\begin{aligned} \abs{U\p{x, \xi}} &\leq \int e^{-\p{x-y}^2/2} e^{-\xi^2/2} \abs{u\p{y}} \,\diff{y} \\ &\leq \p{\int e^{-\p{x-y}^2} e^{\xi^2} \,\diff{y}}^{1/2} \norm{u}_{L^2} && \p{\text{Cauchy-Schwarz}} \\ &= e^{\xi^2/2} \p{\int e^{-y^2} \,\diff{y}}^{1/2} \norm{u}_{L^2} && \p{y \mapsto x - y} \\ &< \infty, \end{aligned}

since $u \in L^2$ and $e^{-y^2} \in L^1$ . Thus, $U\p{x, \xi}$ exists everywhere on $\R^2$ . For the second part, observe that

\begin{aligned} \iint \abs{U\p{x, \xi}}^2 e^{-\xi^2} \,\diff{x} \,\diff{\xi} &= \iint \abs{\int e^{-\p{x-y}^2/2} e^{i\p{x-y}\xi} e^{\xi^2/2} u\p{y} \,\diff{y}}^2 e^{-\xi^2} \,\diff{x} \,\diff{\xi} \\ &= \iint \abs{\int e^{iy\xi} e^{-y^2/2} u\p{x - y} \,\diff{y}}^2 \,\diff{\xi} \,\diff{x} && \p{\text{Fubini-Tonelli}} \\ &= \iint \norm{\widehat{e^{-y^2/2} u\p{x - y}}}_{L^2}^2 \,\diff{x} \\ &= \int \norm{e^{-y^2/2} u\p{x - y}}_{L^2}^2 \,\diff{x} && \p{\text{Plancherel}} \\ &= \iint e^{-y^2} \abs{u\p{x - y}}^2 \,\diff{y} \,\diff{x} \\ &= \int e^{-y^2} \int \abs{u\p{x - y}}^2 \,\diff{x} \,\diff{y} && \p{\text{Fubini-Tonelli}} \\ &= \p{\int e^{-y^2} \,\diff{y}} \p{\int \abs{u\p{x}}^2 \,\diff{x} } && \p{x \mapsto x - y} \\ &= \sqrt{\pi} \int \abs{u\p{y}}^2 \,\diff{y}, \end{aligned}

so $C = \sqrt{\pi}$ .