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I'm reading Gelbart and Jacquet's paper 'forms of GL(2) from the analytic point of view', and was confused at a point in the proof of analytic continuation of Eisenstein series. On the top of page 232, after the calculation of L^2 norm for the derivative of the truncated Eisenstein series, it said it is easy to conclude the the convergence of the Taylor series of truncated series in some disc, which I don't get it. It seems to me that we can conclude the convergence in L^2 norm, but to get pointwise convergence, we need more work.

A similar argument appears also in Kubota's book 'Elementary theory of Eisenstein series', which is at the bottom of page 31.

Could someone explain the hiding part? Thanks a lot.

Gelbart and Jacquet's paper can be found via books.google.com

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If you have $L^2$ (or in general, $L^p$) convergence for say a sequence of functions $\{f_{n}\}$, you can get a.e. pointwise convergence by passing to a subsequence (this is part of the proof of the Riesz-Fischer theorem about completeness of $L^p$). As they are proving absolute convergence for the power series (they don't care about the situation at the boundary, $\| s-s_{0} \| = R$), if you proved convergence for say $\| s-s_{0} \|=T$, you have it immediately for $r\leq T$ (see in the middle of p.30 in Kubota's). –  Asaf Dec 26 '12 at 23:02
    
Thanks, and I understand your comment. But I'm still confused how to apply theorem 1.3.4 in Kubota's book? –  user1832 Dec 27 '12 at 12:42
    
There's probably a misprint in Kubota, as the relevent theorem is thm 3.1.4 about extension of the const. term of the Eisenstein series. If you believe the formula at the top of p. 31, and believe thm 3.1.4, then as long as $Re(s),Re(s')$ are larger than $1/2$ and not touching the real axis (well, unless $Re(s)>1$ that is), then the RHS of the formula will be holomorphic, because $\varphi$ is. Notice that the circle he writes about should be $|s-S_{0}|=\min (S_{0}-0.5,t_0)$, and not with capital $S$. –  Asaf Dec 27 '12 at 19:43
    
I don't think it is a misprint, if you look at the bottom of page 31. –  user1832 Dec 28 '12 at 5:26
    
Yes, I should have read thm 1.3.4 before posting. Anyway, if you cut the Dirichlet domain to two parts, one consisting of points where $y<Y$ and the other consisting of points where $y>Y$, on each part, you can apply thm 1.3.4. Therefore in each subdomain, you will get an eigenvalue of the Laplacian, and then, by thm 1.3.4, you can get that the convergence is pointwise (outside of the horocycle $y=Y$, but $Y$ was arbitrary). This holds as long $s$ is satisfying the relevant inequalities. Now by general regularity theorem, as Paul hinted, that appears in p.32, you relax the inequalities. –  Asaf Dec 28 '12 at 11:59

2 Answers 2

I don't know whether this is what any of those authors were thinking, but if a function has a derivative in an $L^2$ sense, then it certainly has a derivative in a distributional sense. A distribution that satisfies the Cauchy-Riemann equations is in fact a smooth function, by the ellipticity of the Cauchy-Riemann operator, and then we return to the Cauchy-Goursat arguments that give (complex) analyticity. This may be overkill, but it is conceptual, is certainly not a fragile argument, and can be put into words.

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I will post it as an answer because it's a bit long, although just technical.

We're looking at the Taylor series - $$ \sum_{n} \frac{1}{n!}E^{Y}_{n}(z,s_0)(s-s_0)^{n} $$ notice that the capital $Y$ won't cause any trouble, we can assume that we are in the compact part away from the cusp.

The $L^2$ estimate proved in Kubota shows that the partial sums of this sequence converges weakly to this sum (which is in $L^2$ due to the truncation). Now it is enough to show that the partial sums are (generalized-)eigenfunctions and use thm 1.3.4 in Kubota.

So we just apply the Laplacian to it. $$\Delta \sum_{n} \frac{1}{n!}E^{Y}_{n}(z,s_0)(s-s_0)^{n} = \sum_{n}\frac{1}{n!}[\frac{\partial^{n}}{\partial s^{n}}\Delta E(z,s)]_{s=s_0}(s-s_0)^{n} $$ Because the differentiation is over $s$ and the Laplacian cares only about $x,y$ and this is some instance of Schwartz' lemma of multivariable calculus.

Now $\Delta E(z,s)=s(s-1)E(z,s)$, (I'm using Kubota normalization, where $\Delta$ is positive and not negative).

Therefore we'll calculate $\frac{\partial^{n}}{\partial s^{n}} s(s-1)E(z,s)$. We'll calculate it by generalized Leibnitz rule.

$$\frac{\partial^{n}}{\partial s^{n}} s(s-1)E(z,s)=\sum_{k=0}^{n}\binom{n}{k}[s(s-1)]^{(k)}E(z,s)^{(n-k)}.$$

Notice that the first function is polynomial in $s$ of deg. $2$, hence we have that $$\sum_{k=0}^{n}\binom{n}{k}[s(s-1)]^{(k)}E(z,s)^{(n-k)}=\sum_{k=0}^{2}\binom{n}{k}[s(s-1)]^{(k)}E(z,s)^{(n-k)}.$$

By explicitly computing $k=0,1,2$ we have - $$\sum_{k=0}^{2}\binom{n}{k}[s(s-1)]^{(k)}E(z,s)^{(n-k)}=s(s-1)E(z,s)^{(n)}+n(2s-1)E(z,s)^{(n-1)}+n(n-1)E(z,s)^{(n-2)}.$$

This is of-course not applicable to the case where $n=0,1$, which separate explicit calculation is needed.

Now by grouping the proper derivatives of the Eisenstein series together you can see that the functions are indeed eigenfunctions of the Laplacian. $$\frac{s(s-1)}{n!}E(z,s)^{n}+\frac{(n+1)(2s-1)}{(n+1)!}E(z,s)^{(n+1)-1}+\frac{(n+2)(n+1)}{(n+2)!}E(z,s)^{(n+2)-2}=\frac{1}{n!}(s(s-1)+(2s-1)+1)E(z,s)^{(n)}$$

This doesn't apply to the top $2$ powers (you won't sum over the three terms as done above), and I can't seem to fix it in any reasonable way.

Nevertheless, you might still, somehow, use the machinery of thm 1.3.4. The limit, as the computation shows, is an ($L^2$-) eigenfunction of the Laplacian.

Now one needs to show that this convergence happens in a certain Sobolov space, and not just in $L^2(\Gamma \backslash H)$. If so, then the dirac measure defines a bounded functional on the Sobolov space (by say convolution against the smooth vectors, which you will always have in hand by a theorem due to Harish-Chandra, see Borel's book Ch.2) and then you will get pointwise convergence out of weak convergence.

Now, unless I'm completely wrong in the above calculations, I believe that the original proof in Kubota (and which might be also the one in Jacquet-Gelbart) was flawed, because of this iconic error that Paul is mentioning in his notes, the trucnted Eisenstein series is not an eigenfunction of the Laplacian - http://www.math.umn.edu/~garrett/m/v/iconic_error.pdf

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thanks a lot, this clarifies my question. –  user1832 Dec 31 '12 at 10:25

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