= Solution
For the <heat kernel>
$$
K(x,t)=(4\pi t)^{-1/2}e^{-x^2/(4t)},
$$
direct <partial differentiation> gives
$$
K_t=\left(-\frac1{2t}+\frac{x^2}{4t^2}\right)K
=K_{xx}.
$$
Thus $K$ solves the <heat equation>. For bounded continuous $f$, <differentiation under the integral sign> gives
$$
u_t=\int_{-\infty}^{\infty}K_t(x-y,t)f(y)\,dy
=\int_{-\infty}^{\infty}K_{xx}(x-y,t)f(y)\,dy=u_{xx}.
$$
With $Y=(x-y)/\sqrt{4t}$,
$$
u(x,t)=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}
e^{-Y^2}f(x-2\sqrt t\,Y)\,dY.
$$
The <Gaussian integral> makes the weight have total mass one, and the <dominated convergence theorem> gives $u(x,t)\to f(x)$. This is the <Gaussian approximate identity>.
For the <viscous Burgers equation> with unit viscosity, the <Cole-Hopf transformation> $w=-2u_x/u$ reduces the equation to $u_t=u_{xx}$. To obtain initial value $g$, choose
$$
f(x)=\exp\left[-\frac12\int_0^x g(s)\,ds\right]
$$
(up to an irrelevant positive constant), and set $u=K_t*f$. Therefore
$$
\boxed{w(x,t)=-2\,\partial_x\log\!\left[
\int_{-\infty}^{\infty}K(x-y,t)f(y)\,dy
\right]}.
$$
Equivalently,
$$
w(x,t)=\frac1t
\frac{\int (x-y)K(x-y,t)f(y)\,dy}
{\int K(x-y,t)f(y)\,dy},
$$
and the approximate-identity limit gives $w(x,t)\to g(x)$.
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