Jtdylktuy

Let $$u(x,t) = frac{1}{pi^{N/2}}int_{mathbb{R}^N}
u_0(x-2sqrt{t}y)e^{-|y|^2} dy$$

Show that if there are constants $M, delta>0$ such that $$|u_0(x)|le
Me^{-delta|x|^2}, xinmathbb{R}^N$$ then $$|u(x,t)|le M(1+4delta
t)^{-N/2}e^{-delta|x|^2/(1+4delta t)}$$ Conclude that if $u_0$ has
compact support, then $lim_{tto infty} u(x,t) = 0$ uniformly

$$|u(x,t)| = frac{1}{pi^{N/2}}left|int_{mathbb{R}^N}
u_0(x-2sqrt{t}y)e^{-|y|^2} dyright|le frac{1}{pi^{N/2}}int_{mathbb{R}^N}
left|u_0(x-2sqrt{t}y)e^{-|y|^2}right| dy $$

So I need to find a bound for $left|u_0(x-2sqrt{t}y)e^{-|y|^2}right|$ using $|u_0(x)|le
Me^{-delta|x|^2}$

Ok,

$$left|u_0(x-2sqrt{t}y)e^{-|y|^2}right|le Me^{-delta|x|^2}e^{-|y|^2} = Me^{-delta|x|^2-|y|^2}$$

I tried working with $-delta|x|^2-|y|^2$ and using $|a|^2 -|b|^2 le |a-b|^2$ or something like that but it looks like it won't work.

UPDATE:

the right bound is $|u_{0}(x-2sqrt{t}y)| leq M e^{-delta |x-2sqrt{t}y|^2}$ so we get

$$|u(x,t)|le frac{1}{pi^{N/2}}int_{mathbb{R}^N}|u_{0}(x-2sqrt{t}y)e^{-|y|^2}|dy leq frac{1}{pi^{N/2}}int_{mathbb{R}^N}M e^{-delta |x-2sqrt{t}y|^2} e^{-|y|^2}dy =\ frac{1}{pi^{N/2}}int_{mathbb{R}^N}M e^{-delta |x-2sqrt{t}y|^2 -|y|^2}dy $$