Heat kernel has uniformly bounded derivatives Let the heat kernel on $(0,\infty)\times \mathbb R^n$ be given by 
$\Psi(x,t) = (4\pi t )^{-\frac{n}{2}} e^{ -\dfrac{|x|^2}{4t} }$
for $t>0$, otherwise $0$ except at the origin of space-time.
It is clear that the heat kernel is smooth everywhere except at the origin. Let $\delta > 0$. How can you show that ( the absolute values of ) the heat kernel and all of its derivatives are uniformly bounded on $[\delta, \infty ) \times \mathbb R^n$, say, by some constant $C_\delta > 0$?
It seems clear as each derivative does not change the relation between the exponents in $x$ and $t$. However, I am stuck and don't know how to proceed. Search on the web provided me only with statements of this fact without a proof.
Thank you.
 A: The Fourier transform in the $x$-variables of ${\displaystyle h(x,t) = {1 \over (4\pi t)^{n \over 2}}e^{-{\vert x\vert ^2 \over t}}}$ is given by ${\displaystyle 2^{-n}e^{-\pi|\xi|^2t}}$ (The $2^{-n}$ factor may not be there depending on your normalization).
Taking an $x_i$-derivative of $h(x,t)$ corresponds to multiplying by $i\xi_i$ on the Fourier transform side, while taking a $t$ derivative corresponds on the Fourier transform side to once again taking a $t$-derivative. Thus the $x$-Fourier transform of the partial derivative $\partial_x^{\alpha}\partial_t^{\beta}h(x,t)$ is of the form ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$ where $p(\xi)$ is a monomial. One can evaluate $\partial_x^{\alpha}\partial_t^{\beta}h(x,t)$ at any point by using the Fourier inversion formula on ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$. This implies that $|\partial_x^{\alpha}\partial_t^{\beta}h(x,t)|$ is at most the $L^1$ norm of ${\displaystyle p(\xi)e^{-\pi|\xi|^2t}}$ (in the $\xi$-variables). Since this function is decreasing in $t$, $|\partial_x^{\alpha}\partial_t^{\beta}h(x,t)|$ is going to be bounded by 
$${\displaystyle \int_{R^n} |p(\xi)|e^{-\pi|\xi|^2\delta}}\,d\xi$$
Thus one has a uniform bound for a given derivative. But the bounds will not be uniform over all derivatives as equality above is achieved for $x = 0$.
A: Even with the exponent on $t$ being negaitve I don't think it is true.  At the origin, you just need to show that the $\exp(\frac{-1}{t})$ term dominates every polynomial.  But for $t$ a little bit positive, $\frac {d}{dt}t^{-n}\exp(\frac{-1}{t})\approx t^{-n-2}\exp(\frac{-1}{t})$ which grows unboundedly with $n$.  Similarly taking more derivatives puts more factors of $t$ in the denominator, which increases the derivative.  So it is not uniformly bounded.
