= Solution
The <heat kernel> is $H_t(x)=(4\pi t)^{-1/2}e^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is \b[$u(t,x)=(H_t*g)(x)$]. Differentiating under the integral gives the <heat equation>, and the approximate-identity property gives the initial value.
For a <tempered distribution>, its derivative is defined by $\langle T',\psi\rangle=-\langle T,\psi'\rangle$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE=\delta_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E''=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E''=E-\delta_0$, and
$$
\boxed{(-D^2+1)E=\delta_0.}
$$
For an ansatz $u=e^tv$, the forced equation becomes $(1-D^2)v=\phi$. With Fourier convention $\widehat v(k)=\int e^{-ikx}v(x)dx$, its unique solution in the <Schwartz space> is
$$
\boxed{\widehat v(k)=\frac{\widehat\phi(k)}{1+k^2},\qquad v=E*\phi.}
$$
The smooth multiplier and its derivatives have <polynomial> bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's <Fourier transform> by $1+k^2$ gives zero, forcing the difference to vanish.
Subtract this particular solution and evolve the remaining initial data by the <heat kernel>:
$$
\boxed{u(t,x)=e^tv(x)+H_t*(f-v)(x).}
$$
For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot)\to v$ uniformly. If additionally $f-v$ is integrable, its uniform <norm> is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.
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