To limit an integral by a function Being $$u(x,t)=\frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4t}} u(y,0) dy$$
suppose $0\leq u(x,0)$, $||u(-,0)||_1 < \infty$, and $u(x,0)$ nonzero.
Prove that there is a value $x_0$ and a constant $C>0$ such that $u(x_0,t)\geq \frac{C}{\sqrt{t}}$ for all $t>0$
I tried:
$$u(x,t)=\frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^{\infty} e^{\frac{-(x-y)^2}{4t}} u(y,0) dy$$ $$>$$ $$\frac{1}{\sqrt{4 \pi t}} \int_{0}^{\epsilon} e^{\frac{-(x-y)^2}{4t}} min(u(y,0))dy$$ $$=$$ $$\frac{min_{y\in[0, \epsilon]}(u(y,0))}{\sqrt{4 \pi t}} \int_{0}^{\epsilon} e^{\frac{-(x-y)^2}{4t}} dy$$
but don't know how to proceed next.
Thank you very much

I think I solved the problem I stated some time ago, and I share it for if it can be helpful for someone else. Anything wrong please tell me. Couldn't prove for all t, but for $t\ge1/4$
for $t\ge1/4$, we have
$$e^\frac{-(x-y)^2}{4t}\ge e^{-(x-y)^2}$$
So:
$$\frac{min_{y\in[0, \epsilon]}(u(y,0))}{\sqrt{4 \pi t}} \int_{0}^{\epsilon} e^{\frac{-(x-y)^2}{4t}} dy$$ $$\ge$$
$$\frac{min_{y\in[0, \epsilon]}(u(y,0))}{\sqrt{4 \pi t}} \int_{0}^{\epsilon} e^{-(x-y)^2} dy$$ for $t\ge1/4$.
Since $$\int_{0}^{\epsilon} e^{-(x-y)^2} dy=erf(\epsilon)$$
and it is limited by $[-1,1]$, 
$$\frac{min_{y\in[0, \epsilon]}(u(y,0))}{\sqrt{4 \pi t}} \int_{0}^{\epsilon} e^{-(x-y)^2} dy        =          \frac{min_{y\in[0, \epsilon]}(u(y,0))}{\sqrt{4 \pi t}}erf(\epsilon)$$
Since this is true for all $\epsilon$, choosing a concrete value of it holds the inequality.
