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Originally asked on stackexchange since I figured it was a very elementary and silly error, but since I haven't had any response or interest in a couple days, I thought I'd post it here.

I have a "proof" of the false fact that the Hecke regularized Eisenstein series $$ E_2(\tau) = \lim_{s\to 0} \sum_{(m,n)\ne (0,0)} \frac{1}{(m+\tau n)^2 |m+\tau n|^s}$$ is holomorphic; could someone help me figure out where I'm being too naive?

If we write $$ E_2(\tau, s) = \sum_{(m,n)\ne (0,0)} \frac{1}{(m+\tau n)^2 |m+\tau n|^s},$$ it converges absolutely and uniformly on compact subsets of the upper half-plane for $\text{Re }s>0$. Thus, for $\text{Re }s>0$, $$ \frac{\partial}{\partial \overline{\tau}}E_2(\tau, s) = \sum_{(m,n)\ne (0,0)} \frac{\partial}{\partial \overline{\tau}}\frac{1}{(m+\tau n)^2 |m+\tau n|^s} $$ and the RHS is still an absolutely convergent sum which approaches $0$ as $s\to 0$ since we pick up a factor of $s$ from the antiholomorphic derivative. Thus, by continuity, $ \frac{\partial}{\partial \overline{\tau}}E_2(\tau, 0)=0$, i.e. it's holomorphic.

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From your uniform convergence on compact subsets of $\Re s > 0$ you only know that $$ \frac{\partial}{\partial \overline{\tau}} E_2(\tau, s)$$ is continuous on the domain $\Re s > 0$ but it can still be discontinuous at $s = 0$.

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  • $\begingroup$ isn't the definition of $E_2(\tau,s)$ at $s=0$ via analytic continuation in $s$? and therefore wouldn't the dolbeault operator applied to this function of $s$, analytic at $s=0$, be analytic (and thus continuous) at $s=0$? $\endgroup$
    – xir
    Commented Aug 30, 2022 at 2:41
  • $\begingroup$ Yes but being analytic in $s$ does not imply anything about differentiation in $\tau$ direction. $\endgroup$ Commented Aug 31, 2022 at 9:23
  • $\begingroup$ i guess the thing i'm implicitly assuming is that applying a differential operator in $\tau$ to an analytic function of $s$ remains analytic in $s$. i'd like to understand better why this fails; is there a simple example of a function $f(x,y)$ analytic in $y$ such that its $x$-derivative acquires a jump discontinuity in $y$ (for some fixed $x$)? $\endgroup$
    – xir
    Commented Aug 31, 2022 at 20:41
  • $\begingroup$ i guess a better way to phrase this is: i'm assuming del bar of the analytic continuation equal to to the analytic continuation of del bar, which is not necessarily accurate. $\endgroup$
    – xir
    Commented Aug 31, 2022 at 23:02

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