= Solution
Write the kernel as $K(s,t)=(4\pi t)^{-1/2}e^{-s^2/(4t)}$, with $t>0$. Its <partial derivatives> are
$$
K_t=\left(-\frac1{2t}+\frac{s^2}{4t^2}\right)K,
\qquad K_{ss}=\left(-\frac1{2t}+\frac{s^2}{4t^2}\right)K.
$$
Thus the <heat kernel> itself satisfies $K_t=K_{ss}$. The proposed <convolution> consequently gives
$$
u_t(x,t)=\int_{\mathbb R}\varphi(\xi)K_t(x-\xi,t)\,d\xi
=\int_{\mathbb R}\varphi(\xi)K_{ss}(x-\xi,t)\,d\xi
=u_{xx}(x,t).
$$
This verifies the <heat equation>. The parameter is positive time; the real integral formula is not defined at negative time.
The passage of <derivatives> through the <integral> needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in <heat-kernel convergence for Gaussian-integrable initial data>; it holds, in particular, for data of <polynomial growth>. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so <differentiation under the integral sign> applies.
Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi)$ be a primitive of $\varphi(\xi)e^{-\xi^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi,t)=K(x-\xi,t)e^{\xi^2/(4T)}$. This <function>, its spatial and time <derivatives>, and their $\xi$-derivatives decay as Gaussians when $t<T$. <Integration by parts> rewrites the <convolution> as $-\int Hq_\xi\,d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent <integral>; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.
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