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From John Roe: Elliptic Operators, topology and asymptotic methods, page 82-83.

Let $\mathcal{D}$ be a Dirac operator on the spin bundle $S$, then any section $s\in L^{2}(S)$ has a "Fourier expansion" as $s=\sum_{\lambda\in \sigma(\mathcal{D})}s_{\lambda}$. Now if $f$ is bounded on $\sigma(\mathcal{D})$, we can define a bounded operator on $L^{2}(S)$ by setting $$ f(\mathcal{D})s=\sum f(\lambda)s_{\lambda}, s=\sum s_{\lambda} $$ Roe claimed that the map $f\rightarrow f(\mathcal{D})$ is a unital homomorphism from the ring of bounded functions on $\sigma(\mathcal{D})$ to $B(H)$. Further, Roe claimed that $f(\mathcal{D})$ is a smoothing operator. Roe's proof is built up by the following steps:

1) If $|f(\lambda)|=O(|\lambda|)^{-k}$ for each $k$, then we know $f(\mathcal{D})$ maps $L^{2}(S)$ to $C^{\infty}(S)$. (I think Roe used Soblev embedding).

2) Notice that for $\lambda \in \sigma(\mathcal{D})$, the orthogonal projection operator $P_{\lambda}$ onto the $\lambda$-eigenspace of $\mathcal{D}$ is smoothing. (This is not entirely trivial, but it can be done).

3) Now, it is "not hard to see" that for each $k$ there is an $l(k)$ such that the Sobolev-$k$-norm on $(M\times M)$ of the smoothing kernel of $P_{\lambda}$ is bounded by $C_{k}\lambda^{l(k)}$. (this is the part I am stuck)

4) Thus, if $f$ is rapidly decreasing, the series $f(\mathcal{D}=\sum_{\lambda}f(\lambda)P_{\lambda}$ converges in the topology of smoothing kernels on $M\times M$. (Given 3, this is trivial).

So the crucial step is to prove the claim (3). John Roe give the following hint (page 85): Let $K$ be a smoothing operator on $L^{2}(S)$. Prove that the $L^{2}$ norm of the smoothing kernel of $K$ is bounded by a multiple of the operator norm of $K$ as an operator from $L^{2}(S)$ to $C^{0}(S)$. Hence prove that if $K$ is the projection operator $P_{\lambda}$, then the $W^{k}$-norm of its kernel is bounded by $C_{k}\lambda^{l}$, for some $l>k+\frac{n}{2}$.

If I am not mistaken, the $L^{2}$-norm of the smoothing kernel of $K$ is given by $$|K|^{2}_{2}=\int_{M\times M}K(x,y)^{2}dxdy$$ while the operator norm of $K:L^{2}(S)\rightarrow C^{0}(S)$ is given by $$ |K|_{0}=|\int_{M\times M} K(x,y)f(y)dy|_{max, |f|_{2}=1} $$ But I do not really know how to achieve this bound other than fixing one coordinate and treat $k(y)=K(x,y)$, then normalizing all $k(y)$, taking the maximum over all possible $x$. This procedure is obviously not what Roe wanted, for it would give us a bound depending on $K$. So I want to ask a hint for Part (3).

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Here's an idea for the case when the bundle is trivial, so $K(x,y)$ is just a function on $M \times M$:

I will use $K$ to denote both the operator and its kernel. Think of the square of the $L^2$ norm of $K(x,y)$ on $M \times M$ as follows: $$ \int \int |K(x,y)|^2 \, dy \, dx = \int K(K(x, \cdot)) \,dx,$$ where by $K(K(x, \cdot))$ I mean the operator $K$ applied to the function $K(x, \cdot)$, where we view $x$ as fixed. (Maybe we really want the complex conjugate of $K(x, \cdot)$.)

Now bound the integrand pointwise using the operator norm of $K$ that Roe describes times the $L^2$ norm in just the $y$-variable of $K(x,y)$. Then use Holder's inequality to deal with the square root in the definition of the $L^2$ norm. I think the constant that pops out is the volume of $M$.

I would have to think about the general case; I would guess one could use local trivializations.

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  • $\begingroup$ Thanks! I thought my idea was nonsense, I did not know you would think the same. $\endgroup$ – Bombyx mori Oct 15 '14 at 2:31
  • $\begingroup$ Sorry, I didn't realize that was also your idea. What did you mean by a "bound depending on $K$"? Am I missing something, or does the bound you get not depend on $K$? $\endgroup$ – Phillip Andreae Oct 15 '14 at 15:00
  • $\begingroup$ I thought since I fixed one coordinate in $K$, the result would be depending on $K$ as well. But obviously that was a mistake. $\endgroup$ – Bombyx mori Oct 15 '14 at 17:17

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