= Heat-kernel convergence for Gaussian-integrable initial data
{title2=$K_t*\varphi\to\varphi$}
Suppose $\varphi$ is continuous at $x$ and, for some $T>0$,
$$
\int_{\mathbb R}|\varphi(\xi)|e^{-\xi^2/(4T)}\,d\xi<\infty.
$$
For $0<t<T$, its <convolution> with the <heat kernel> $K_t(s)=(4\pi t)^{-1/2}e^{-s^2/(4t)}$ is smooth and solves the <heat equation>. On every <compact set> in $\mathbb R\times(0,T)$, spatial or time <derivatives> of the <heat kernel> are bounded by the displayed integrable weight times a constant: their <polynomial> factors are absorbed by the strictly narrower Gaussian. This justifies <differentiation under the integral sign>.
Moreover, $\int K_t=1$, so
$$
(K_t*\varphi)(x)-\varphi(x)=\int K_t(x-\xi)\bigl(\varphi(\xi)-\varphi(x)\bigr)\,d\xi.
$$
Inside $|\xi-x|<\delta$, <continuity> makes the difference uniformly small. Outside that interval, for $t\leq T/4$, the inequality $\xi^2\leq2(\xi-x)^2+2x^2$ gives
$$
K_t(x-\xi)e^{\xi^2/(4T)}
\leq\frac{e^{x^2/(2T)}}{\sqrt{4\pi t}}e^{-\delta^2/(8t)}.
$$
The right side tends to zero, proving that the weighted tail contribution vanishes; the constant term $\varphi(x)$ is controlled by the vanishing Gaussian tail mass. Thus $(K_t*\varphi)(x)\to\varphi(x)$ as $t\downarrow0$. In particular, continuous data of <polynomial growth> satisfy this hypothesis, even when they are unbounded. This extends the <Gaussian approximate identity> argument beyond bounded data.
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