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It is stated in Rosenberg's book "The Laplacian on a Riemannian Manifold" that a basic observation regarding the heat kernel $$ e(t,x,y) = \frac{1}{\sqrt{4\pi t}} \exp\{ - \frac{(x - y)^2}{4t}\} \qquad (t,x,y) \in (0,\infty) \times \mathbb R \times \mathbb{R} $$ is that for any function $f \in L^2(\mathbb R)$ we have $$ \lim_{t \to \infty} \frac{1}{\sqrt{4 \pi t}} \int_\mathbb R \exp\{ - \frac{(x - y)^2}{4t}\}\, f(y) \,dy = 0\,. $$

I would really like to be able to fill in the details for this computation, but I get stuck at the following point. So far I have assumed that $f \in L^2(\mathbb{R})$ is (essentially) bounded so that \begin{equation}\tag{1} \left|\frac{1}{\sqrt{4 \pi t}} \int_\mathbb R \exp\{ - \frac{(x - y)^2}{4t}\}\, f(y) \,dy\,\right| \le \frac{1}{\sqrt{4 \pi t}} \int_\mathbb R \exp\{ - \frac{(x - y)^2}{4t}\} \,dy \,\|f\|_\infty \end{equation} and then I can use the change of variable $u = (x - y)/\sqrt{4t}$ to compute $$ \int_\mathbb R \exp\{ - \frac{(x - y)^2}{4t}\} \,dy = \sqrt{4 t} \int_\mathbb R e^{-u^2} \,du = \sqrt{4\pi t}\,. $$ But then, substituting this into the right hand side of (1) shows that I have lost the $t$-dependence, hence the limit does not yield zero.

How can I achieve a better estimate?

Thanks very much for your help!

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  • $\begingroup$ Cauchy-Schwarz? $\endgroup$ – Giuseppe Negro Mar 14 '14 at 23:59
  • $\begingroup$ You can decompose $e(t,x,y) = \sum_{i=0}^{\infty} e^{\lambda_i t} g_i(x)g_i(y)$ where $g_i, \lambda_i$ is the spectrum of the Laplace-Beltrami operator. Your limit then follows from the negativity of the Laplace-Beltrami eigenvalues. $\endgroup$ – user7530 Mar 15 '14 at 0:17
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So far I have assumed that $f\in L^2(\mathbb R)$ is (essentially) bounded

Bad idea. For one thing, this is not true in general, for another, $L^\infty$ on a space of infinite measure is quite far from $L^2$.

Let's review the situation. The scaling in $$\phi_t(x) = \frac{1}{\sqrt{4\pi t}} \exp\left\{ - \frac{x^2}{4t}\right\} $$ is arranged so that the $L^1$ norm of $\phi_t$ is independent of $t$. Indeed, we can write $\phi$ for $\phi_1$; then $\phi_t(x) = t^{-1/2}\phi(t^{-1/2}x) $, which (by substitution) is seen to have the same integral as $\phi$.

From here we can already expect the $L^p$ norm for every $p>1$ to decay in long term, because $\phi_t$ gets more spread out. And indeed, the same substitution $x'=t^{-1/2}x$ shows that $$\int |\phi_t|^p = t^{(1-p)/2} \int |\phi|^p \to 0 \tag{1}$$

You only need $p=2$, actually. As Giuseppe Negro remarked, the Cauchy-Schwarz inequality finishes the problem: the integral is at most $\|f\|_{L^2} \|\phi_t\|_{L^2}$.

The extra generality in (1) shows that the same conclusion holds for $f\in L^q$, $1<q<\infty$ (using Hölder's inequality).

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