To prove : $\lim_{t \rightarrow \infty} g_t \ast f = 0$.

Question

Let $g_t(x) = e^{-x^2/2t^2}/t\sqrt{2\pi}$. To prove that for $f \in C_0(\mathbb{R})$, $$ \lim_{t \rightarrow \infty} g_t \ast f = 0.$$ Is the question valid ? Any hints on how to solve it ?

Answer 1

The family $g_t$ is a well-known convolution kernel. Most often, one considers the limit $t\to 0$, when $g_t$ converges (in the sense of distributions) to the Dirac measure, but the limit $t\to +\infty$ is also interesting (up to insignificant normalisation, $g_t$ is the heat kernel).

When looking at the limiting behaviour for convolutions with a parametrised kernel family, it is often convenient to move the parameter from the kernel to the convolution partner, here too:

$$\begin{align} (g_t\ast f)(x) &= \int_{-\infty}^\infty g_t(y)f(x-y)\,dy\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{y^2}{2t^2}} f(x-y)\frac{dy}{t}\tag{$y = tz$}\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-z^2/2} f(x-tz)\,dz. \end{align}$$

Now we have one very well-behaved factor in the integrand that is independent of the parameter $t$, and we need to look at how $f(x-tz)$ behaves for $t\to \infty$. When doing that, remember Lebesgue.