the solution to heat equation in convolution form

Question

Let $g\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})$. Let $u$ be defined as the function $$u(t,x)=\int_{\mathbb{R}}p_t(x-y)g(y)\,dy$$, where $$p_t(x)=\frac{1}{\sqrt{4\pi t}}e^{-\frac{|x|^2}{4t}},\quad t>0, x\in\mathbb{R}$$

It can be easily proven that $$p_{s+t}=p_s\star p_t(x),\quad s,t>0,x\in\mathbb{R}$$ Let us take for granted $u$ is well defined.

I wish to show the translation invariant property, i.e. $$u(s+t,g)=u(t,u(s,g)).$$

holds for any $s,t\ge 0$ and any $g\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})$.

Thanks for any helps.

Answer 1

Use the associativity of convolution: $$ u(s+t,g)=p_{s+t}\star g=(p_t\star p_s)\star g=p_t\star (p_s\star g)=p_t\star u(s,g)=u(t,u(s,g)). $$