= Heat-kernel convergence from a conditionally convergent Gaussian integral
{title2=$\int_{\mathbb R}\varphi(\xi)e^{-\xi^2/(4T)}\,d\xi\text{ converges}$}
Let $\varphi$ be continuous and suppose that the displayed <improper integral> converges separately at both infinite endpoints for some $T>0$. Absolute convergence is not required. Then its <heat-kernel solution> exists, is smooth and solves the <heat equation> for $0<t<T$, and tends to $\varphi(x)$ as $t\downarrow0$ at each fixed $x$.
To prove the smoothing claim, take a primitive $H$ of $\varphi(\xi)e^{-\xi^2/(4T)}$. The assumed convergence makes $H$ bounded. For $q(x,\xi,t)=K_t(x-\xi)e^{\xi^2/(4T)}$, <integration by parts> gives
$$
\int\varphi(\xi)K_t(x-\xi)\,d\xi=-\int H(\xi)q_\xi(x,\xi,t)\,d\xi.
$$
The endpoint term vanishes. The coefficient of $\xi^2$ in the exponent of $q$ is $-(T-t)/(4tT)<0$. Its <derivatives> therefore have Gaussian decay on every compact subset of $\mathbb R\times(0,T)$, justifying <differentiation under the integral sign> in the absolutely convergent right-hand expression. Since $q_t=q_{xx}$, it satisfies the <heat equation>.
For the initial limit, apply the same construction to $\varphi(\xi)-\varphi(x)$ and split at $|\xi-x|=\delta$. <Continuity> controls the local part by its small oscillation times the unit kernel mass. On each remaining tail, $q$ is monotone for sufficiently small $t$: its maximum lies at $\xi=xT/(T-t)$, approaching $x$. If the relevant primitive has absolute bound $M$, <integration by parts> bounds that tail by $2M$ times the value of $q$ at its finite endpoint. For either endpoint $\xi=x\pm\delta$, that value tends to zero like $t^{-1/2}e^{-\delta^2/(4t)}$ times a fixed constant. Both tails vanish, proving the limit. Separate endpoint convergence matters here; merely assuming a symmetric <Cauchy principal value> does not provide the bounded primitive used in this proof.
Back to article page
