Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is the same question I asked in stackexchange: http://math.stackexchange.com/questions/519152/asymptotic-of-the-heat-kernel

I read Proposition 3.23(page 101) in Rosenberg's book "The Laplacian on a Riemannian Manifold" and not quite clear how to get the estimate $(4\pi t)^{n/2}|Q_k * H_k|\leq C \cdot t^{k+1}$ on a compact manifold. Since $|Q_k|\leq C \cdot t^{k-(n/2)}$ and $H_k = \eta\cdot e^{-r^2/4t}(4\pi t)^{-n/2}(u_0 + t u_1 + \cdots + t^k u_k)$, it amounts to show the following estimate $$ (4\pi t)^{n/2}\left|\int_0^t C \tau^{k-n/2}\int_M H_k(t-\tau,x,y)dy\right|\leq C_1 t^{k+1} $$ for some constant when $t\in [0, T]$ and $T$ is small. Since $M$ is compact the terms without $t$ are uniformly bounded and it reduces to the following one(maybe with different constants) $$ t^{n/2} \int_0^t \tau^{k - n/2}(t-\tau)^{-n/2} d\tau \leq C t^{k+1}. $$ It seems that the above inequality does not hold. I would appreciate if someone can point out where I am wrong.

share|improve this question
    
For those without the book in front of them, what are $H_k$ and $Q_k$? –  Nate Eldredge Oct 8 '13 at 18:34
    
The following is the link to his book: math.bu.edu/people/sr/articles/book.pdf For a short answer, $H_k = \eta S_k$ where $\eta$ is the cut-off function in the ball of injectivity radius(assuming positive) and $S_k$ is given by $$ S_k = (4\pi t)^{-n/2}e^{-r^2(x,y)/4t}(u_0(x,y)+\cdots +t^k u_k(x,y)) $$ where $u_i$'s satisfy certain recurrence relation and they can be thought as smooth function on $M\times M$. –  BewSMA Oct 8 '13 at 18:37
    
$Q_k(t,x,y)$ is quite complicated and it can be thought as a smooth function on $[0,\infty)\times M\times M$ with the estimate $|Q_k|\leq C t^{k-n/2}$ on $[0,T]\times M \times M$ and $C=C(T,M)$. The integer $k$ is chosen large with $k-n/2 >2$. –  BewSMA Oct 8 '13 at 18:49
add comment

1 Answer

You just need to check that $$ \int_M H_k(x, y) \mathrm{d}y = 1 + O(t^{k+1})$$ for all $x \in M$ and for all $k$. For example, this follows directly from the method of stationary phase (just take a geodesic chart around $x$ that is so large that the support fo $\eta$ is in its domain. Translate to $\mathbb{R}^n$ and use the method of stationary phase there.

Therefore, your first equation in fact reduces to $$(4 \pi t)^{n/2} \int_0^tC \tau^{k-n/2}(1 + O(\tau^{k+1}))\mathrm{d} \tau \leq C_1 t^{k+1}$$ which is obviously true.

So you just estimated to roughly when you took the sup-norm. Estimating the $L^1$-norm does the job.

share|improve this answer
add comment

Your Answer

 
discard

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.