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Say we have a power series of two variables, with an associated function $f$ defined as $$ \begin{split} f(x, y) =\, & \sum_{n,m} a_{n,m}x^ny^m,\\ & a_{n,m} \geq 0 \quad \forall n, m \in\mathbb{N}, \end{split} $$ which is known to converge in each point of a compact set $\mathcal{C} \subset \mathbb{R_+}^2$. We can also assume $f$ to be bounded on $\mathcal{C}$.

Can $f$ have a point of discontinuity (or more) on $\mathcal{C}$?

Edit : As shown by Tom Goodwillie with this nice counter-example, it turns out to be possible for $f$ to be discontinuous. Thanks!

Edit : clarified question as per the inputs from Tom Goodwillie, David E Speyer and Daniele Tampieri.

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  • $\begingroup$ Did you really mean to write $a_{n,m}\ge 0$? $\endgroup$ Jul 20, 2022 at 0:44
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    $\begingroup$ If not, what you mean by convergence? There is the question of absolute versus relative convergence. $\endgroup$ Jul 20, 2022 at 0:45
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    $\begingroup$ I don't understand, $(1-2xy) \sum (x^2+y^2)^n$ is not absolutely convergent at $(1/\sqrt{2}, 1/\sqrt{2})$. Writing it as $\sum (x^2+y^2)^n - 2 xy \sum (x^2+y^2)^n$, the first sum only contains terms $x^{2a} y^{2b}$ and the second only contains terms $x^{2a+1} y^{2b+1}$, so there is no cancellation and the original sum is absolutely convergent if and only if each of $\sum (x^2+y^2)^n$ and $2 xy \sum (x^2+y^2)^n$ are absolutely convergent. These latter sums clearly diverge at $(1/\sqrt{2}, 1/\sqrt{2})$. $\endgroup$ Jul 20, 2022 at 4:58
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    $\begingroup$ I admit I phrased it in a confusing way Daniele, because I'm indeed interested in positive values of $(x,y)$. It is my understanding that absolute convergence in $\mathcal{C}$ means convergence on a domain $\Omega = \{z = (z_1,z_2) \in \mathbb{C}^2 | (|z_1|,|z_2|) \in \mathcal{C}\}$ in $\mathbb{C}^2$. I wanted to say that we can assume this set to be a complete log-convex Reinhardt domain, so that $\mathcal{C}$ is not necessarily part of the interior of the domain of convergence (which would give continuity). But this is actually not useful information, so I'll just remove it! $\endgroup$
    – Raphael B
    Jul 20, 2022 at 10:09
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    $\begingroup$ What do we know for one complex variable? If a power series with radius of convergence $1$ actually converges everywhere on the unit circle, must it be continuous there? $\endgroup$ Jul 20, 2022 at 10:32

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The series $f(x,y)=y+xy+x^2y+x^3y+\dots$ converges to $0$ when $y=0$, and converges to $y/(1-x)$ when $|x|<1$. This function is not continuous at $(x,y)=(1,0)$.

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    $\begingroup$ Thank you Tom, this is an interesting series! It definitely shows that continuity is not necessarily happening if there is absolute convergence on every point, so I'll edit the original post with a link to your counter-example for this more general condition. However, it still seems continuous to me on every compact set containing only points of absolute convergence, including $(1,0)$ (it looks like we need to stay sufficiently close to the line $x=1$ to get a divergent sequence, and this is what the compacity prevents from happening here). $\endgroup$
    – Raphael B
    Jul 20, 2022 at 13:31
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    $\begingroup$ The sum is absolutely convergent on the compact set $x \geq 0$, $y \geq 0$, $x+y \leq 1$, and is still discontinuous when restricted to this triangle. $\endgroup$ Jul 20, 2022 at 15:56
  • $\begingroup$ Thanks for pointing this out! I didn't realise it but indeed, the series' value is constantly 1 along the line $y =1-x$, so its limit in $(1,0)$ is $1 \neq 0 = f(1,0)$. My bad! I'll edit and mark this as solved then. $\endgroup$
    – Raphael B
    Jul 20, 2022 at 16:07

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