Let $f(x)$ be bounded and continuous for $x \in \Bbb R^n$ and satisfy $\int |f(y)| dy < \infty.$ Prove that there exists a solution $u(x,t)$ of $$ \begin{array}{l} u_t=\Delta u, \text{ for $x \in \Bbb R^n$ and $t>0$,} \\ u=f(x), \text{ for $x \in \Bbb R^n$ and $t=0$,} \end{array} $$ such that $\lim_{t \rightarrow \infty}u(x,t)=0$ How can I prove this?
I think it we know $f$ is bounded and continuous, then use the fundamental solution of heat equation. We have if we consider $t$ goes to infinite the solution will decreases exponentially to 0. Is this right? If right, what $\int |f(y)| dy < \infty$ is do? Since I didn't use it in mine proof.
If not, how can I prove that? And where I am wrong?
Since $f$ is bounded and continuous, one may use the fundamental solution to the heat equation to write the solution to the IVP as $$ u(x,t)=(4\pi t)^{-n/2}\int_{\mathbb{R}^n} e^{-\frac{(x-y)^2}{4t}}f(y)\,dy. \tag{1} $$ Therefore, \begin{align} 0\leq |u(x,t)|&\leq (4\pi t)^{-n/2}\int_{\mathbb{R}^n} e^{-\frac{(x-y)^2}{4t}}|f(y)|\,dy \\ &\leq (4\pi t)^{-n/2}\int_{\mathbb{R}^n}|f(y)|\,dy. \tag{2} \end{align} Since $\int_{\mathbb{R}^n}|f(y)|\,dy<\infty$, it follows that $$ \lim_{t\to\infty}(4\pi t)^{-n/2}\int_{\mathbb{R}^n}|f(y)|\,dy=0. \tag{3} $$ Finally, from $(2)$ and $(3)$ one may conclude that $\lim_{t\to\infty}u(x,t)=0.\quad\square$