Asymptotic of the heat kernel

Question

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.

Answer 1

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.