Let $u \in C^\infty (\mathbb R^n \times (0,\infty)) \cap C^0(\mathbb R^n \times [0,\infty))$ be the only bounded solution for the problem
$$ \partial_t u = \partial_x ^2 u, \quad \forall (x,t )\in \mathbb R^n \times (0,\infty) \quad \text{and} \quad u (x,0) = u_0(x), \forall x \in \mathbb R^n, $$
where $u_0 \in C^\infty(\mathbb R^n)$ is bounded.
Suppose that exists $M, \delta > 0$ such that
$$ |u(x,0)| \leq M \, \exp {(-\delta \|x\|^2)}, \quad \forall x \in \mathbb R^n $$
I'm trying to proof that
$$|u(x,t)| \leq \frac{M}{(1 + 4\delta t)^n/2} \, \exp{\left (\frac{\delta \| x\| ^2}{1+4\delta t} \right )}, \quad \forall (x,t) \in \mathbb R^n \times [0,\infty) \quad \quad (I).$$
My attempt:
Since the only solution for this problem is $$ u(x,t) = \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}} \, u_0(y)\, dy$$ we have
$$ |u(x,t)| \leq \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}} |u_0(y)|\, dy \leq \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{\frac{-\| x-y \|^2}{4t}}Me^{-\delta\|y\|^2} \, dy $$ $$ \therefore \quad |u(x,t)| \leq \frac{M}{(4\pi t)^{n/2}}\int_{\mathbb R^n} \exp{\left (\frac{-\| x-y \|^2}{4t} - \delta\|y\|^2\right )} \, dy $$
How can I conclude the argument, i.e showing $(I)$?
Help?