# Asymptotic of the heat kernel

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.

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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

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.