Let $\mathcal{S}(\mathbb{R})$ be the space of rapidly decreasing function on $\mathbb{R}$. Let $\Phi_t$ be defined as
$$\Phi_t(x)=\frac{1}{\sqrt{4\pi t}}e^{-\frac{|x|^2}{4t}},\quad t>0,\,\,x\in\mathbb{R}.$$
For any $f\in\mathcal{S}(\mathbb{R})$ we defined a function $$u(t,x)=\int_{\mathbb{R}}\Phi_t(x-y)f(y)\,dy,\quad t>0,x\in\mathbb{R}.$$
First, I want to show $u$ is well defined. My reasoning is that $\Phi_t$ is integrable and $f$ is a rapidly decreasing function and so is bounded. Hence the integral is well defined.
Next, I want to show $u(t,\cdot)$ is in $\mathcal{S}(\mathbb{R})$. I am not sure what is the proper way to do this...is the convolution of a integral function and a schwartz function a schwartz function? How to verify an arbitrary function is indeed a Schwartz function? I saw this is somewhat mentioned in page 4 of the following enter link description here
But they do not have an answer for my question.
For the first you're right: Since $\Phi_t$ is a positive $L^1$ function with $\int_{\mathbb{R}} \Phi_t(x)dx=1$ for all $t> 0$, we get that the convolution integral is well defined and by Holder's inequality we get $$ |u(t,x)|\leq \sup_{y\in\mathbb{R}}|f(y)|. $$
Using the Dominated Convergence Theorem you can also see that if $f\in \mathcal{S}(\mathbb{R})$ then you can differentiate under the integral sign in the definition of $u$ and put all derivatives on $f$ to get that $u\in C^\infty(\mathbb{R})$ and moreover $$ |\partial^\alpha_x u(t,x)|\leq \sup_{y\in \mathbb{R}} |\partial^\alpha f(y)|. $$ The easiest way to get the decay is through the Fourier transform: We get that if we take the FT on the spatial variables, then $\hat{u}(t,\xi)=\hat{\Phi}_t(\xi)\hat{f}(\xi)$ so that decay of $u(t,\cdot)$ is equivalent to differentiability of $\hat{u}$, which is "clear" since $\hat{\Phi}_t$ and $\hat{f}$ are smooth.