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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.

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What's $\delta$ got to do with anything? It doesn't seem to serve any purpose in the question as currently written. –  George Lowther Feb 19 '11 at 0:31
    
The factor in front of the exponential should be $(4\pi t)^{-n/2}$. –  t.b. Feb 19 '11 at 0:54
    
if $x = \sqrt{t}$ the expression is clearly not bounded in $t$. –  Zarrax Feb 19 '11 at 1:38
    
I am very sorry for the typo in the exponent and that I missed to mention the domain on which to bound the kernel. I have corrected this -- the question is important when it comes to show the integrability of the convolution with the heat kernel. –  shuhalo Feb 19 '11 at 4:34

2 Answers 2

up vote 4 down vote accepted

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 $x = 0$.

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Thank you very much. It seems I misunderstood the meaning of uniformly bounded. –  shuhalo Feb 20 '11 at 2:09

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.

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Mea culpa. It has been a typo. –  shuhalo Feb 19 '11 at 4:48

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