= Solution
Because $f$ is <continuous>, $g(x,y)=e^{-x^2}f(e^{-x^2/2}y)$ is continuous on $\mathbb R^2$, so $|g|$ is a nonnegative measurable function. Apply <Tonelli's theorem> before assuming integrability. For each fixed $x$, set $z=e^{-x^2/2}y$, so $dy=e^{x^2/2}\,dz$. <Lebesgue substitution on an interval> gives
$$
\int_{\mathbb R}|g(x,y)|\,dy
=e^{-x^2/2}\int_{\mathbb R}|f(z)|\,dz.
$$
Therefore the <Gaussian integral> yields
$$
\boxed{\iint_{\mathbb R^2}|g(x,y)|\,dx\,dy
=\sqrt{2\pi}\int_{\mathbb R}|f(z)|\,dz<\infty.}
$$
This proves that $g$ is <Lebesgue integrable> on $\mathbb R^2$. <Fubini's theorem> now permits changing the order for $g$ itself. The same substitution without absolute values gives the stronger exact identity
$$
\int_{\mathbb R}\int_{\mathbb R}g(x,y)\,dx\,dy
=\sqrt{2\pi}\int_{\mathbb R}f(z)\,dz.
$$
Taking absolute values and using the <triangle inequality for an integral> proves
$$
\boxed{\left|\int_{\mathbb R}\int_{\mathbb R}g(x,y)\,dx\,dy\right|
\leq\sqrt{2\pi}\int_{\mathbb R}|f(z)|\,dz.}
$$
The constant is sharp: any nonzero nonnegative continuous integrable $f$ gives equality. This is a <weighted dilation kernel> with $a(x)=e^{-x^2}$ and $b(x)=e^{-x^2/2}$. The inner $x$ integral in the requested order is guaranteed to be integrable for almost every $y$, which is all the iterated integral requires.
Back to article page
