About convergence of complex power series on the circle ##|z - z_0|=r##

[cianfa72]

TL;DR Summary

Does convergence of power series on the circle $|z - z_0|=r$ imply absolute convergence there ?

Whether a complex power series $\sum_{k=0}^\infty c_k (z - z_0)^k$ converges at a point $\tilde z \in \mathbb C$ then it converges absolutely in the open disk $|z−z_0|<|\tilde z−z_0|=r$.

Assume now a power series convergent on the circle $|z−z_0|=r$, does it imply absolute convergence on the closed disk $|z−z_0|≤r$ ?

Searching on net, I found a negative answer as in this link from MSE: Sierpinski (1916) produced an example where the power series converges everywhere on the unit circle, yet the function $f(z)$ it converges to isn't bounded on the closed disk $|z|≤1$, hence isn't continuous on it and certainly does not converge absolutely on the circle $|z|=1$.

In the interior of the unit disk the function $f(z)$ is continuous, for absolute convergence there imply uniform convergence that in turn imply continuity. It follows that $f(z)$ is bounded inside the unit disk, hence it must be unbounded on some point/points on the circle $|z|=1$.

Now my question is: how is actually possible that this power series converges also on this point/points on the circle $|z|=1$ as assumed by hypothesis ?

 

[pasmith]

TL;DR Summary: Does convergence of power series on the circle $|z - z_0|=r$ imply absolute convergence there ?

Since $|z-z_0| = r$, you can set $z = z_0 + re^{i\theta}$. The power series is then $$\sum_{n=0}^\infty c_k r^ke^{ik\theta}.$$ You are therefore considering the absolute convergence of a convergent fourier series, with the restriction that $c_k = 0$ for $k < 0$. Note that a necessary condition for convergence is that $|c_k|r^k \to 0$ as $k \to \infty$.

 

[cianfa72]

Thinking again, I believe I figured it out.

The point is that the function $f(z)$ the power series converges to on the closed unit disk is not bounded at least in a neighborhood of a point on the boundary (i.e. on $|z|=1$ circle). However on that/those points the series actually converges, hence $f(z)$ is discontinuous there.

 

[FactChecker]

Thinking again, I believe I figured it out.

The point is that the function $f(z)$ the power series converges to on the closed unit disk is not bounded at least in a neighborhood of a point on the boundary (i.e. on $|z|=1$ circle). However on that/those points the series actually converges, hence $f(z)$ is discontinuous there.

I am not a collector of counterexamples, but the behavior of an analytic function (a function with a complex derivative) around a singularity can be very bizarre. I don't think you can make any universal statement about it that will work for all functions.

 

[cianfa72]

I don't think you can make any universal statement about it that will work for all functions.

Yes of course. However for the specific case (Sierpinski counterexample), I think my understanding is correct.

 

[FactChecker]

Yes of course. However for the specific case (Sierpinski counterexample), I think my understanding is correct.

I'm not familiar with that counterexample. In general, suppose a function is analytic in the neighborhood of an essential singularity. That function comes arbitrarily close to every number in the complex plane in every neighborhood of the essential singularity. So you can expect such a function to have bizarre behavior around the essential singularity. It is easy to make an example. The function $e^{\frac{1}{1-z}}$ is analytic in the complex plane except for an essential singularity at $z=1$.
(It's not immediately clear to me how to translate that bizarre behavior to the behavior of its Taylor series expansion at $z=0$. I know it will converge inside $|z|=1$ and diverge outside, but I don't know what it does for $|z|=1, z \ne 1$.)

 

[cianfa72]

In general, suppose a function is analytic in the neighborhood of an essential singularity. That function comes arbitrarily close to every number in the complex plane in every neighborhood of the essential singularity. So you can expect such a function to have bizarre behavior around the essential singularity.

You mean pick a complex number $w$ with arbitrary "high" module $|w|$ then, in every neighborhood of an essential singularity, there is a point where the analytic function takes values arbitrary close to $w$.

Ps. Merry Xmas !

 

[FactChecker]

You mean pick a complex number $w$ with arbitrary "high" module $|w|$ then, in every neighborhood of an essential singularity, there is a point where the analytic function takes values arbitrary close to $w$.

That sounds a little confusing but correct when you say "arbitrarily high modulus". Suppose $f(z)$ has an essential singularity at $z=z_0$. Pick any complex number, $w$, at all and any $\epsilon \gt 0, \delta \gt 0$ no matter how small. Then there exists some complex number, $z'$, where $|z_0-z'|\lt\delta$ and $|f(z')-w|\lt\epsilon$.

Ps. Merry Xmas !

Thanks! Merry Xmas to you!

 

[cianfa72]

Suppose $f(z)$ has an essential singularity at $z=z_0$. Pick any complex number, $w$, at all and any $\epsilon \gt 0, \delta \gt 0$ no matter how small. Then there exists some complex number, $z'$, where $|z_0-z'|\lt\delta$ and $|f(z')-w|\lt\epsilon$.

Ak ok, you mean an essential singularity as explained here: The latter [Picard's great theorem] says that in every neighborhood of an essential singularity $a$, the function $f$ takes on every complex value, except possibly one, infinitely many times. (The exception is necessary; for example, the function $e^{1/z}$ never takes on the value 0).

Btw according to Radius of convergence, the (Sierpinski) power series in example 4 converges uniformly on the boundary $|z|=1$ but does not converge absolutely on it. It looks weird to me since uniform convergence imply continuity and since it converges absolutely (hence uniformly) inside the unit disk $|z| \lt 1$ then it should converge to a continuous function $f(z)$ on the entire closed disk $|z| \leq 1$ (that is not).

 

[FactChecker]

Ak ok, you mean an essential singularity as explained here: The latter [Picard's great theorem] says that in every neighborhood of an essential singularity $a$, the function $f$ takes on every complex value, except possibly one, infinitely many times. (The exception is necessary; for example, the function $e^{1/z}$ never takes on the value 0).

Yes. I forgot that the theorem was that strong. Sorry.

 

[cianfa72]

Yes. I forgot that the theorem was that strong. Sorry.

No problem, thank you.

What about the last part of my post #9 ? I'm not sure whether the claim in the wikipedia link is too strong in the sense that the Sierpinski power series actually can't converge uniformly on the boundary $|z|=1$, otherwise it would converge to a continuous function $f(z)$ on the closed disk $|z| \leq 1$.

 

[FactChecker]

What about the last part of my post #9 ? I'm not sure whether the claim in the wikipedia link is too strong in the sense that the Sierpinski power series actually can't converge uniformly on the boundary $|z|=1$, otherwise it would converge to a continuous function $f(z)$ on the closed disk $|z| \leq 1$.

And you think that would prevent there being a singularity on the unit circle? I don't know if that is true. There might still be bad behavior outside the unit circle from a singularity on the circle.

 

[cianfa72]

And you think that would prevent there being a singularity on the unit circle?

No, I was just saying that since the Sierpinski power series is unbounded in a neighborhood of $z=1$ inside the unit disk, it can't converge uniformly on the boundary $|z|=1$.

Sorry to repeat my argument: that power series has 1 as radius of convergence. Therefore it converges absolutely in the open disk $|z| \lt 1$ and diverges for $|z| \gt 1$. We also know it doesn't converge to a continuous function $f(z)$ in the closed disk $|z| \leq 1$. Hence it can't converge uniformly on the boundary $|z|=1$ (since uniform convergence implies continuity).

 

[cianfa72]

I think I managed to get the point. As per the Wikipedia link, the fact that the Sierpinski power series uniformly converges on the boundary circle $|z|=1$ doesn't imply it must converge to a continuous function $f(z)$ on the closed disk $|z| \leq 1$, yet it converges uniformly in any closed disk $|z| \leq c, c \lt 1$.

On the other hand, it can't converge absolutely on the boundary circle $|z| =1$, for it would converge uniformly to a continuous function $f(z)$ on the closed disk $|z| \leq 1$, that is not.

 

[FactChecker]

On the other hand, it can't converge absolutely on the boundary circle $|z| =1$, for it would converge uniformly to a continuous function $f(z)$ on the closed disk $|z| \leq 1$, that is not.

I am not an expert in this, but I am not aware of any general principle that says this. Are you saying this based on knowledge of that particular series or are you applying a Theorem?

 

[cianfa72]

Are you saying this based on knowledge of that particular series or are you applying a Theorem?

I'm saying that based on properties of that particular series, as described in Wikipedia link Ex. 4 (in particular it doesn't converge to a bounded/continuous function on the closed disk $|z| \leq 1$).

 

[FactChecker]

On the other hand, it can't converge absolutely on the boundary circle $|z| =1$, for it would converge uniformly to a continuous function $f(z)$ on the closed disk $|z| \leq 1$, that is not.

I'm saying that based on properties of that particular series, as described in Wikipedia link Ex. 4 (in particular it doesn't converge to a bounded/continuous function on the closed disk $|z| \leq 1$).

In Post #14, you seem to be implying that absolute convergence on the boundary implies uniform convergence in the closure. I'm not arguing, but I don't immediately see how that is true in general.

 

[cianfa72]

In Post #14, you seem to be implying that absolute convergence on the boundary implies uniform convergence in the closure. I'm not arguing, but I don't immediately see how that is true in general.

See for instance the proof here at minute 05:00. You can easily extend the proof for $c=r=1$ assuming, by hypothesis, the power series $$\sum_{k=0}^\infty |a_k| c^k$$ converges absolutely on the boundary $|z| = 1$.

 

[FactChecker]

See for instance the proof here at minute 05:00. You can easily extend the proof for $c=r=1$ assuming, by hypothesis, the power series $$\sum_{k=0}^\infty |a_k| c^k$$ converges absolutely on the boundary $|z| = 1$.

Is it so easy to extend? If so, why do they bother with the smaller closed disk and never prove the stronger, simpler-stated theorem?

 

[cianfa72]

Is it so easy to extend? If so, why do they bother with the smaller closed disk and never prove the stronger, simpler-stated theorem?

Because, in general, there is no guarantee that a power series converges absolutely on the boundary (circle) of radius of convergence $r$. If it does than you can prove it converges uniformly in the closed disk of radius $r$.

In that proof, assuming $\sum_{k=0}^\infty |a_k| c^k$ converges by hypothesis, then the series $$\sum_{k=n+1}^\infty |a_k| c^k$$ converges to zero for $n \to \infty$.

 

[Gavran]

Sierpinski provided an example of a power series, which converges uniformly in an open disk and which converges on a boundary of the disk. This means that the power series can be replaced with a function $f(z)$ which is continuous in the open disk and which is defined on the boundary.
$$ f(z)=
\begin{cases}
f_1(z),&|z-z_0|\lt r\\
f_2(z),&|z-z_0|=r
\end{cases} $$
The next two equations hold
$$ \begin{align}
&\lim_{z\to z_b}\sum_{k=0}^{\infty}c_k(z-z_0)^k=\lim_{z\to z_b}f(z)\nonumber\\
&\sum_{k=0}^{\infty}c_k(z_b-z_0)^k =f(z_b)\nonumber
\end{align} $$
where $|z-z_0|<r$ and $|z_b-z_0|=r$.

The power series on the boundary converges but it does not converge uniformly, so there will be $$ \lim_{z\to z_b}\sum_{k=0}^{\infty}c_k(z-z_0)^k\neq\sum_{k=0}^{\infty}c_k(z_b-z_0)^k $$
And finally it can be said that a condition for one-sided continuity of the function $f(z)$ on the boundary is not satisfied because $$ \lim_{z\to z_b}f(z)\neq f(z_b) $$ or $$ \lim_{z\to z_b}f_1(z)\neq f_2(z_b) $$

 

[cianfa72]

The next two equations hold
$$ \begin{align}
&\lim_{z\to z_b}\sum_{k=0}^{\infty}c_k(z-z_0)^k=\lim_{z\to z_b}f(z)\nonumber\\
&\sum_{k=0}^{\infty}c_k(z_b-z_0)^k =f(z_b)\nonumber
\end{align} $$
where $|z-z_0|<r$ and $|z_b-z_0|=r$.

The power series on the boundary converges but it does not converge uniformly, so there will be $$ \lim_{z\to z_b}\sum_{k=0}^{\infty}c_k(z-z_0)^k\neq\sum_{k=0}^{\infty}c_k(z_b-z_0)^k $$
And finally it can be said that a condition for one-sided continuity of the function $f(z)$ on the boundary is not satisfied because $$ \lim_{z\to z_b}f(z)\neq f(z_b) $$

From the MSE link, the Sierpinski power series is unbounded in a neighborhood of $z=1$ inside the unit disk. Therefore the one-sided limit from inside the open unit disk $$\left | \lim_{z\to 1}\sum_{k=0}^{\infty}c_kz^k \right | =\left | \lim_{z\to 1}f(z) \right | = \infty$$

 

[WWGD]

IIRC, the series $\Sigma z^{n!}$ on the unit disk blows up on the boundary of the unit disk, but it's Analytic inside.