Suppose we have an integral operator $A$ such that $$Af(x) = \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}e^{-\frac{(x-y)^2}{2}}f(x) \, dy$$

This operator is bounded and $\|A\|=1$ (see Norm of the integral operator in $L^2(\mathbb{R})$.). So in order to prove $A$ to be self-adjoint it's suffice to show that it is symmetric, i.e. $$\forall f,g\in L^2(\mathbb{R}):\,\, \langle Af, g \rangle = \langle f,Ag \rangle$$

We can prove it using unitary Fourier transform $F$ and its property that $\langle f,g \rangle = \langle Ff, Fg \rangle$ for any $f,g\in L^2(\mathbb{R})$:

$$ \langle Af, g \rangle = \langle FAf, Fg \rangle = \langle \exp\left[-\frac{x^2}{2}\right](Ff), Fg \rangle = $$ $$ = \langle Ff, \exp\left[-\frac{x^2}{2}\right](Fg) \rangle = \langle Ff, FAg \rangle = \langle f, Ag \rangle $$ So operator $A$ really is self-adjoint.

The question is how one proves this fact without using Fourier transform?

My attempts:

$$\langle Af, g \rangle = \int\limits_\mathbb{R}\left( \frac{1}{\sqrt{2\pi}} \int\limits_{\mathbb{R}} e^{\frac{-(x-y)^2}{2}}f(y) \, dy \right)\overline{g(x)} \, dx.$$ Move the constant outside of the outer integral and the actor $\overline{g(x)}$ inside of the inner integral:

$$\langle Af, g \rangle = \frac{1}{\sqrt{2\pi}} \int\limits_\mathbb{R} \int\limits_{\mathbb{R}} e^{\frac{-(x-y)^2}{2}}f(y)\overline{g(x)} \,dy\,dx$$ It's to justify the following interchange of integrals:

$$\langle Af, g \rangle = \frac{1}{\sqrt{2\pi}} \int\limits_\mathbb{R} \int\limits_{\mathbb{R}} e^{\frac{-(x-y)^2}{2}}f(y)\overline{g(x)} \, dx\,dy$$ Move real-valued function and real constant under conjugation line and function $f(y)$ that doesn't depend on $x$ outside of the inner integral:

$$\langle Af, g \rangle = \int\limits_\mathbb{R} f(y)\int\limits_{\mathbb{R}} \overline{\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-y)^2}{2}}g(x)} \, dx\,dy$$ Do conjugation before integration:

$$\langle Af, g \rangle = \int\limits_\mathbb{R} f(y)\left(\overline{\int\limits_{\mathbb{R}} \frac{1}{\sqrt{2\pi}}e^{\frac{-(x-y)^2}{2}}g(x) \, dx}\right) \, dy = \langle f, Ag \rangle$$

So if we justify the interchange of integrals, the result will stand.


You can justify the interchange of orders of integration by letting $$ G(x,y) = e^{-(x-y)^{2}/2} $$ and noticing that $G(x,y)f(x)\overline{g(y)}$ is jointly measurable in $x,y$, and is bounded by \begin{align} |G(x,y)f(x)\overline{g(y)}| & = |G(x,y)^{1/2}f(x)||G(x,y)^{1/2}\overline{g(y)}| \\ & \le \frac{1}{2}G(x,y)|f(x)|^{2}+\frac{1}{2}G(x,y)|g(y)|^{2}. \end{align} So the expression on the left is absolutely integrable on $\mathbb{R}\times\mathbb{R}$ because the expression on the far right is absolutely integrable. Fubini's Theorem now applies to justify the interchange of orders of integration.

  • $\begingroup$ That's awesome, thank you! Fubini's Theorem... I should've studied better. $\endgroup$ – Glinka Aug 2 '15 at 19:23

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