= Solution
The indicated <integration by substitution> gives
$$
u(x,t)=\frac1{\sqrt\pi}\int_{\mathbb R}
\varphi(x+2\sqrt t\,\eta)e^{-\eta^2}\,d\eta.
$$
Pointwise, the argument of $\varphi$ tends to $x$. Since the <Gaussian integral> is $\int e^{-\eta^2}d\eta=\sqrt\pi$, the expected limit is $\varphi(x)$. To justify it for unbounded data, use the <approximate identity> property rather than assuming a uniform bound on $\varphi$.
In the original variable, normalization gives
$$
u(x,t)-\varphi(x)=\int_{\mathbb R}K(x-\xi,t)\bigl(\varphi(\xi)-\varphi(x)\bigr)\,d\xi.
$$
For $|\xi-x|<\delta$, <continuity> makes the difference smaller than any prescribed $\varepsilon$ if $\delta$ is small enough, and the kernel mass is at most one. On the complement the Gaussian mass tends to zero; the data-weighted tail tends to zero by <heat-kernel convergence for Gaussian-integrable initial data>. This proves
$$
\boxed{\lim_{t\downarrow0}u(x,t)=\varphi(x)}.
$$
For <heat-kernel convergence from a conditionally convergent Gaussian integral>, the same tail conclusion follows using the bounded primitive of $(\varphi(\xi)-\varphi(x))e^{-\xi^2/(4T)}$. The factor $q=K(x-\xi,t)e^{\xi^2/(4T)}$ has its maximum at $\xi=xT/(T-t)$, approaching $x$. For sufficiently small $t$, it is monotone on each of the two tails outside $|\xi-x|<\delta$, and its values at the tail endpoints tend to zero exponentially. <Integration by parts> bounds each tail by a constant times its endpoint value, proving the same limit without an absolute-convergence assumption. The approach $t\downarrow0$ is from positive time.
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