I'm working on this one: (Fritz John "Partial Differential Equations" page 226, $\#1$ and generalization: Jurgen Jost "Partial Differential Equations" page 125 $\# \, 5.3$):
Define for $x,y, t \in \mathbb{R}, \, t \neq 0$ $$K(x,y,t)= \dfrac{1}{(4\pi |t|)^{\frac{1}{2}}}\, e^{-\frac{(x-y)^2}{4t}} $$ Show that $$K(x,0,s+t) = \int_{\mathbb{R}} K(x,y,t) \cdot K(y,0,s) \, dy \quad (\star) $$ holds
(a) when $s> 0, \, t>0$;
(b) when $0<t<-s$.
My attempt
I'm using this version of Fourier Transform: $$\mathcal{F}(f)(\omega) = \dfrac{1}{2\pi} \int_{\mathbb{R}} f(x) e^{i\omega x} \, dx $$
Part (a):
Note that $$\int_{\mathbb{R}} K(x,y,t) \cdot K(y,0,s) \, dy = \int_{\mathbb{R}} \dfrac{1}{4\pi\sqrt{|st|}}e^{-\frac{(x-y)^2}{4t}} \cdot e^{-\frac{y^2}{4s}} \, dy =\dfrac{1}{4\pi\sqrt{|st|}} e^{-\frac{x^2}{4s}}* e^{-\frac{x^2}{4t}} $$
Applying Fourier transform we have that, for $s,t \,>0 $, $$\mathcal{F}\left[e^{-\frac{x^2}{4s}}* e^{-\frac{x^2}{4t}}\right] =2\pi \cdot \mathcal{F}\left[e^{-\frac{x^2}{4s}}\right] \cdot \mathcal{F}\left[ e^{-\frac{x^2}{4t}}\right] $$ $$= 2\pi \cdot \dfrac{1}{\sqrt{4 \pi (\frac{1}{4s})}} \cdot \dfrac{1}{\sqrt{4 \pi (\frac{1}{4t})}} \cdot e^{-\omega^2/(4(\frac{1}{4s}))} \cdot e^{-\omega^2/(4(\frac{1}{4t}))} $$ $$= 2\sqrt{st}\cdot e^{-\omega^2(s+t)}$$
Applying inverse Fourier transform we have that $$\mathcal{F}^{-1}\left[2\sqrt{st}\cdot e^{-\omega^2(s+t)} \right] = 2\sqrt{st} \cdot \dfrac{\sqrt{\pi}}{\sqrt{s+t}}\cdot e^{\frac{-x^2}{4(s+t)}}$$
Hence, $$\int_{\mathbb{R}} K(x,y,t) \cdot K(y,0,s) \, dy= \dfrac{1}{4\pi\sqrt{st}}\cdot 2\sqrt{st} \cdot \dfrac{\sqrt{\pi}}{\sqrt{s+t}}\cdot e^{\frac{-x^2}{4(s+t)}} $$ $$= \dfrac{1}{2\sqrt{\pi(s+t)}}\cdot e^{\frac{-x^2}{4(s+t)}} = \dfrac{1}{\sqrt{4\pi(s+t)}}\cdot e^{\frac{-x^2}{4(s+t)}}= K(x,0,s+t)$$
My question is:
Can I prove with Fourier transform the condition $0<t<-s$ holds $(\star)$?
Thanks for any suggestions (and edits because i'm not english speaker).