Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 5 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 $x = 0$.

share|cite|improve this answer
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.

share|cite|improve this answer
Mea culpa. It has been a typo. – shuhalo Feb 19 '11 at 4:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.