analytic function on unit disk

This seems to be the correct place for complex analysis. Here's the problem I'm working on:

Let $\displaystyle \Delta (a,r)$ be the disk $\displaystyle \{z \in \mathbb{C}:|z-a|<r \}$, and let $\displaystyle \Delta$ be the unit disk $\displaystyle \Delta (0,1)$. Let $\displaystyle f$ be an analytic function on $\displaystyle \Delta$.

(a) Let $\displaystyle a \in \Delta$. Show that for any $\displaystyle r$, $\displaystyle 0<r<1-|a|$, we have:

$\displaystyle f(a)=\frac{1}{\pi{r}^2}\int_{\Delta (a,r)}f(z)\ dxdy$

(b) Show that if $\displaystyle M=\int_{\Delta}|f(z)|\ dxdy$, then:

$\displaystyle |f(a)| \le \frac{M}{\pi(1-|a|)^2}$, $\displaystyle a \in \Delta$

(c) Let $\displaystyle \mathcal{F}$ be a family of analytic functions on $\displaystyle \Delta$ for which there exists $\displaystyle M>0$ such that:

$\displaystyle \int_{\Delta}|f(z)|\ dxdy \le M\text{ for all }f \in \mathcal{F}$.

Show that $\displaystyle \mathcal{F}$ is a normal family.

For (c), a normal family is one in which every sequence in $\displaystyle \mathcal{F}$ has a subsequence that converges uniformly on compact subsets.

Unless I'm missing something, (b) follows from (a) by just taking the absolute value:

$\displaystyle |f(a)| \le \frac{1}{\pi{r}^2}\int_{\Delta (a,r)}|f(z)|\ dxdy$

$\displaystyle \le \frac{1}{\pi{r}^2}\int_{\Delta}|f(z)|\ dxdy$

$\displaystyle = \frac{1}{\pi{r}^2}M$

and since this must be true for all $\displaystyle r<1-|a|$, taking the limit as $\displaystyle r$ goes to $\displaystyle 1-|a|$ gives

$\displaystyle |f(a)| \le \frac{M}{\pi{(1-|a|)}^2}$

I'm still lost on (a) and (c), though.

Thanks,

Hollywood

Re: analytic function on unit disk

Just skimming it, it looks like (a) is just the statement of the mean value property for the harmonic functions Re(f(z)) and Im(f(z)). I think that's proven in the "integral over a bounding circle" form by citing Cauchy's Theorem that the counter clockwise integral of an holomorphic function (throughout the interior too) around a closed curve of f(z) / ( f(z)-f(a) ) = 2 pi i f(a). Then put that "mean value property:boundary circle" into "mean value property:disk" form by applying Stokes theorem. Something like that - I think that's how it can be done. Maybe I'll try to work it out later. Good luck.

Re: analytic function on unit disk

Thank you. The "integral over bounding circle" version follows directly from the Cauchy integral formula, as you said.

It turns out to be just a special case of the Poisson integral formula: If a real-valued function $\displaystyle u$ is harmonic in a neighborhood of the closed disk $\displaystyle \overline{B}(a,\rho)$, then for any point $\displaystyle a+re^{it}$ in the open disk $\displaystyle B(a,\rho)$, $\displaystyle u(a+re^{it})=\frac{1}{2\pi}\int_0^{2\pi}u(a+\rho{e }^{i \theta})$$\displaystyle \frac{\rho^2-r^2}{\rho^2-2\rho{r}\cos(\theta-t)+r^2}\ d\theta$.

Setting $\displaystyle r$ and $\displaystyle t$ to zero gives $\displaystyle u(a)=\frac{1}{2\pi}\int_0^{2\pi}u(a+\rho{e}^{i \theta})\ d\theta$.

I think now all I need to do is integrate with respect to $\displaystyle \rho$:

$\displaystyle 2 \pi \rho\ u(a)=\int_0^{2\pi} u(a+\rho{e}^{i \theta})\ \rho\ d\theta$

$\displaystyle \pi r^2 u(a)=\int_{B(a,r)} u(a+\rho{e}^{i \theta})\ dA$

since the area element in polar coordinates is $\displaystyle \rho\ d\rho\ d\theta$. This gives the real part, and the imaginary part is the same, so we get the required result.

Any ideas for part (c)?

Re: analytic function on unit disk

Going from the usual mean value (average on the bounding circle) to the area form by integrating - that's elegant. It's been a while, but surely that's how it's usually derived - too pretty not to be.

I've an idea for how to attack (c), though I haven't sat down and worked it out, so maybe it doesn't work. And it's just to get the limit function; I haven't thought about the uniformity of the convergence.

Start with a compact subset of B(0,1). It's a subset of a closed ball D = closure of B(0,r) for some r<1. From part (b), all f in the family share the same bound for |f| on D. My thought is to exploit that they all have power series centered at 0 converging on all of B(0,1), in order to find a limit function by examining the coefficients of the power series, one at a time.

From Cauchy's Integral Formula, that bound on all the |f| means that all the functions in the family also share a bound for |f(0)|, call it M1. Then all the f(0), for f in that family, are in the closed disk D(0,M1). Thus they have a limit point, and so some subfamily can be choosen so that its associated set of f(0) values has a limit point point.

Now with that subfamily, do the same thing for each f'(0) = K integral(Circle,r) f(z)/z^2, so that can get a bound for the |f'(0)| on that subfamily, so that all the f'(0)'s are in some closed ball D(0,M2). Thus choose a sub-sub-family that has their f'(0) values converging.

Now do f''(0), and so forth, forever. Voila - a limit function.

Since this is all happening in D(0,r), the error terms should go to 0. If doing it "one coefficient at a time" is somehow a problem when worked out, could perhaps examine the countably infinite product of compact D(0,Mi)'s, which is compact (Tychonoff), and the points (f(0), f'(0), f''(0), ... ) in it for each function in the family in order to get the limit function.

For the issue of the original compact K growing out toward the boundary of the unit disk: this limit function would remain the same, right? Previously found functions on "smaller" compact K's would still work, since they're simply a power series based at 0 - so a bunch of coefficients that won't be changing. Besides, at 0 is far away from where the K is changing. Also, since these functions are analytic in B(0,1), they're entirely determined by their values in any B(0,epsilon). At each step, the value of the bounds that guaranteed a limit point would change as these compact K's grew, but the bounds themselves didn't matter (though perhaps they would in some argument for uniformity?) Only that they were bound was what really mattered, as that guaranteed that there was some limit point providing the needed convergent subsequence.

Does all this actually work? Even if it does, does it give uniform convergence on all compact sets? I dunno. It's just what's come to mind.

Re: analytic function on unit disk

I think I've gotten (c). I kept trying to force it into a "Weierstrass M Test", but, if there's a way to do it like that, I didn't find it.

My previous post outlined how to get some convergent subsequence, and it works out. I won't do it here, as this is already really long, and it's less interesting than the uniform convergence. For the uniform convergence on compact subsets, the details of the function converged to turn out not to matter, so my thoughts about trying to control the f were wrong.

You only need to say enough about f to make it continuous in order to get the uniform convergence - though to get the "normal family", you need to also show it's holomorphic in the open unit disk. The later actually isn't hard, since f is defined as a power series.

(If you try to do it, use Cauchy integral formula for the derivatives at 0, on a radius of r = sqrt(n/(n+2)) for the nth derivative, which gives the smallest values. Ratio test for the radius of convergence will then rely on the (1+1/m)^m limit for e, and will give you convergence in the open unit ball about 0, making the f so defined holomorphic there.)

**Prop:** Suppose $\displaystyle \{f_n\} \subset \mathcal{F} (\mathcal{F}$ as given in the problem), and $\displaystyle f_n \rightarrow f$ pointwise on $\displaystyle \Delta(0,1)$, to some $\displaystyle f \in C(\Delta(0,1))$.

Then for every compact subset $\displaystyle K$ of $\displaystyle \Delta(0,1)$, then the convergence is uniform on $\displaystyle K$.

Outline of proof:

$\displaystyle \lVert f_n(z) - f(z) \rVert \le \lVert f_n(z) - f_n(w) \rVert + \lVert f_n(w) - f(w) \rVert + \lVert f(w) - f(z) \rVert$.

Locally: will control the 1st term by a lemma below (the real meat of this problem), the 2nd term by pointwise convergence, and the 3rd term by "on compact spaces, continous implies uniformly continuous."

Then will use that K is compact to show that the same N works for all of K.

**Proof:**

Fix $\displaystyle \epsilon > 0$.

Since $\displaystyle f \in C(\Delta(0,1))$, have that $\displaystyle f \in C(K)$. Thus f is uniformly continuous on K, so $\displaystyle \exists \delta_0 >0$ such that:

if $\displaystyle z, z' \in K, \lVert z - z' \rVert < \delta_0$ then $\displaystyle \lVert f(z) - f(z') \rVert < \epsilon/3$.

Now, for each $\displaystyle w \in K$, do the following:

By the lemma, $\displaystyle \exists B > 0$, and $\displaystyle \exists r > 0$ ($\displaystyle B$ and $\displaystyle r$ depending on $\displaystyle w$) such that

$\displaystyle \forall n, \lVert z - w \rVert < r \Rightarrow \lVert f_n(z) - f_n(w) \rVert \le B \lVert z - w \rVert$.

Let $\displaystyle \delta_1 = \min \{ \frac{\epsilon}{3B}, r \}$. Clearly, $\displaystyle \delta_1 > 0$. Then for every n:

$\displaystyle \lVert z - w \rVert < \delta_1 \Rightarrow \lVert z - w \rVert < r$,

and so $\displaystyle \lVert f_n(z) - f_n(w) \rVert \le B \lVert z - w \rVert < B \delta_1 \le B (\frac{\epsilon}{3B}) = \frac{\epsilon}{3}$.

Have shown that for every n, $\displaystyle \lVert z - w \rVert < \delta_1 \Rightarrow \lVert f_n(z) - f_n(w) \rVert < \frac{\epsilon}{3}$.

The last part is to observe that $\displaystyle f_n \rightarrow f$ pointwise on $\displaystyle \Delta(0,1)$, so $\displaystyle \exists N = N(w, \epsilon) \ni n \ge N \Rightarrow \lVert f_n(w) - f(w) \rVert < \frac{\epsilon}{3}$.

Let $\displaystyle \delta = \min \{ \delta_1, \delta_0\}$. Obviously $\displaystyle \delta > 0$. Let $\displaystyle U = K \cap \Delta(w, \delta)$, obviously $\displaystyle U$ open in $\displaystyle K$.

Then $\displaystyle z \in U$ and $\displaystyle n > N$ implies that:

1) $\displaystyle z \in \Delta(w, \delta)$, so $\displaystyle \lVert z - w \rVert < \delta \le \delta_1$, so $\displaystyle \lVert f_n(z) - f_n(w) \rVert < \frac{\epsilon}{3}$.

2) $\displaystyle n > N$, so $\displaystyle \lVert f_n(w) - f(w) \rVert < \frac{\epsilon}{3}$.

3) $\displaystyle z \in \Delta(w, \delta)$, so $\displaystyle \lVert z - w \rVert < \delta \le \delta_0$. Thus since $\displaystyle z, w \in K$, have $\displaystyle \lVert f(z) - f(w) \rVert < \frac{\epsilon}{3}$.

Putting 1, 2, 3 together:

$\displaystyle \lVert f_n(z) - f(z) \rVert \le \lVert f_n(z) - f_n(w) \rVert + \lVert f_n(w) - f(w) \rVert + \lVert f(w) - f(z) \rVert$

$\displaystyle < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$

(*) Therefore, for each $\displaystyle w \in K$, there's a pair $\displaystyle U, N$ such that $\displaystyle z \in U$ and $\displaystyle n > N$ implies $\displaystyle \lVert f_n(z) - f(z) \rVert < \epsilon$.

Now label these, for each $\displaystyle w \in K$, as $\displaystyle U_{w}$ and $\displaystyle N_{w}$. Since $\displaystyle U_{w}$ is an open neighborhood of $\displaystyle w$ in $\displaystyle K$,

have that $\displaystyle \{U_w\}_{w \in K}$ is an open cover of the compact K, hence it admits a finite subcover $\displaystyle \{U_{w_i} \}_{i = 1}^{n_0}$.

Let $\displaystyle N^* = \max \{ N_{w_i} \}_{i = 1}^{n_0}$.

Let $\displaystyle n > N^*$.

Then for any $\displaystyle z \in K$, have $\displaystyle z \in U_{w_i}$ for some $\displaystyle i \in \{1, 2, ..., n_0 \}$.

But also $\displaystyle n > N^*$, so $\displaystyle n > N_{w_i}$. By (*), this proves that $\displaystyle \lVert f_n(z) - f(z) \rVert < \epsilon$.

Thus given $\displaystyle \epsilon > 0$, have produced an $\displaystyle N^*$ such that for all $\displaystyle n > N^*$ and for all $\displaystyle z \in K$, $\displaystyle \lVert f_n(z) - f(z) \rVert < \epsilon$.

Therefore the convergence is uniform on K.

**Lemma:** $\displaystyle \forall z_0 \in \Delta(0,1), \exists B >0, r >0 \ni \forall g \in \mathcal{F},$

if $\displaystyle \lVert z_1 - z_0 \rVert < r$, then $\displaystyle \lVert g(z_1) - g(z_0) \rVert \le B \lVert z_1 - z_0 \rVert$.

Proof:

Let $\displaystyle 0 < r_0 < 1 - \lVert z_0 \rVert$ (to keep everything inside $\displaystyle \Delta(0,1)$). Let $\displaystyle z_1\in \Delta(z_0, r_0)$.

Then $\displaystyle g(z_1) = \frac{1}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{z-z_1}dz$, and $\displaystyle g(z_0) = \frac{1}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{z-z_0}dz$.

Thus $\displaystyle g(z_1) - g(z_0) = \frac{1}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } \left(\frac{g(z)}{z-z_1} - \frac{g(z)}{z-z_0} \right) dz = \frac{1}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } g(z)\left(\frac{1}{z-z_1} - \frac{1}{z-z_0} \right) dz$

$\displaystyle = \frac{1}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } g(z)\left(\frac{z_1 - z_0}{(z-z_1)(z-z_0)} \right) dz = \frac{z_1 - z_0}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{(z-z_1)(z-z_0)} dz$.

Will prove the lemma by showing that that integral is bounded by a value that depends only on M and $\displaystyle z_0$, (actually, only on $\displaystyle \lVert z_0 \rVert$).

Now, it's geometrically clear that if $\displaystyle z \in \partial \Delta(z_0, r_0) }$, then $\displaystyle \lVert z - z_1 \rVert \ge (r_0 - \lVert z_1 - z_0 \rVert)$. (Triangle inequality: $\displaystyle r_0 = d(z_0,z) \le d(z_0, z_1) + d(z_1, z)$).

Thus $\displaystyle \frac{1}{\lVert z - z_1 \rVert} \le \frac{1}{r_0 - \lVert z_1 - z_0 \rVert} \ \forall z \in \partial \Delta(z_0, r_0)$.

Integrating over $\displaystyle \partial \Delta(z_0, r_0)$ uses $\displaystyle z(\theta) = z_0 + r_0e^{i \theta}$ for $\displaystyle 0 \le \theta \le \2 \pi$, so will have:

$\displaystyle \int_{ \partial \Delta(z_0, r_0) } \left(\frac{g(z)}{(z-z_1)(z-z_0)} \right) dz = \int_0^{2 \pi}\frac{g(z(\theta))}{(z(\theta)-z_1)(r_0e^{i \theta})} \right) (ir_0e^{i \theta} d \theta = i \int_0^{2 \pi}\frac{g(z(\theta))}{(z(\theta)-z_1)} \right) d \theta$.

Now, by part (b) and $\displaystyle g \in \mathcal{F}$, have that $\displaystyle \lVert g(z) \rVert \le \frac{M}{\pi (1 - \lVert z \rVert ^2)}$ on $\displaystyle \Delta(0,1)$, so on $\displaystyle \boundary \Delta(z_0, r_0)$.

But $\displaystyle z \in \partial \Delta(z_0, r_0) \Rightarrow \lVert z \rVert \le \lVert z_0 \rVert + r_0$ (again, geometrically obvious, proven via triangle inequality).

Thus for $\displaystyle z \in \partial \Delta(z_0, r_0)$ have $\displaystyle \lVert g(z) \rVert \le \frac{M}{\pi (1 - \lVert z \rVert ^2)}\le \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)}$.

So $\displaystyle \left\lVert\int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{(z-z_1)(z-z_0)} dz \right\rVert \le \int_0^{2 \pi} \left\lVert i \frac{g(z(\theta))}{(z(\theta)-z_1)} \right\rVert d \theta$

$\displaystyle \le \int_0^{2 \pi} \frac{\lVert g(z(\theta)) \rVert }{\lVert z(\theta)-z_1 \lVert } d \theta = \int_0^{2 \pi} \lVert g(z(\theta)) \rVert \left(\frac{1}{\lVert z(\theta)-z_1 \lVert } \right) d \theta$

$\displaystyle \le (2 \pi) \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)} \right) \left(\frac{1}{r_0 - \lVert z_1 - z_0 \rVert} \right)$.

Therefore $\displaystyle \lVert g(z_1) - g(z_0) \rVert = \left\lVert \frac{z_1 - z_0}{2 \pi i} \int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{(z-z_1)(z-z_0)} dz \right\rVert$

$\displaystyle = \frac{\lVert z_1 - z_0\rVert}{2 \pi} \ \left\lVert \int_{ \partial \Delta(z_0, r_0) } \frac{g(z)}{(z-z_1)(z-z_0)} dz \right\rVert$

$\displaystyle \le \frac{\lVert z_1 - z_0\rVert}{2 \pi} (2 \pi) \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)} \right) \left(\frac{1}{r_0 - \lVert z_1 - z_0 \rVert} \right)$

$\displaystyle = \lVert z_1 - z_0\rVert \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)(r_0 - \lVert z_1 - z_0 \rVert)} \right)$.

So $\displaystyle \lVert g(z_1) - g(z_0) \rVert \le \lVert z_1 - z_0\rVert \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)(r_0 - \lVert z_1 - z_0 \rVert)} \right)$.

Now let $\displaystyle r = \frac{r_0}{2} > 0$. And let $\displaystyle B = \frac{M}{\pi r (1 - (\lVert z_0 \rVert + 2r)^2)}$.

If $\displaystyle w \in \Delta(z_0, r)$, then $\displaystyle w \in \Delta(z_0, r_0)$, and so $\displaystyle \lVert g(w) - g(z_0) \rVert \le \lVert w - z_0\rVert \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)(r_0 - \lVert w - z_0 \rVert)} \right)$.

But so $\displaystyle \lVert w - z_0 \rVert < r$, so $\displaystyle r_0 - \lVert w - z_0 \rVert > r_0 - r = r$, so $\displaystyle \frac{1}{r_0 - \lVert w - z_0 \rVert} < \frac{1}{r}$.

Thus $\displaystyle w \in \Delta(z_0, r) \Rightarrow\lVert g(w) - g(z_0) \rVert \le \lVert w - z_0\rVert \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)} \frac{1}{(r_0 - \lVert w - z_0 \rVert)} \right)$.

$\displaystyle \le \lVert w - z_0\rVert \left( \frac{M}{\pi (1 - (\lVert z_0 \rVert + r_0)^2)} (\frac{1}{r}) \right) = B \lVert w - z_0\rVert$.

Have shown that, for an $\displaystyle z_0 \in \Delta(0, 1), \exists B >0, r >0 \ni \forall g \in \mathcal{F},$

if $\displaystyle \lVert z_1 - z_0 \rVert < r$, then $\displaystyle \lVert g(z_1) - g(z_0) \rVert \le B \lVert z_1 - z_0 \rVert$.

That proves the lemma.

Re: analytic function on unit disk

Wow! That's a lot of work.

I think I might have found a shorter proof based on Montel's Theorem:

Montel's Theorem (Theorem 6.5.5 in Greene & Krantz):

Let $\displaystyle U \subseteq \mathbb{C}$ be an open set and let $\displaystyle \mathcal{F}$ be a family of holomorphic functions on $\displaystyle U$ that is bounded on compact sets. Then for every sequence $\displaystyle \{f_j\} \subseteq \mathcal{F}$ there is a subseqence $\displaystyle \{f_{jk}\}$ that converges normally on $\displaystyle U$ to a limit (necessarily holomorphic) function $\displaystyle f_0$.

We are given that $\displaystyle \mathcal{F}$ is a family of holomorphic functions on the open set $\displaystyle \Delta$. To apply Montel's Theorem, then, we only need to show that $\displaystyle \mathcal{F}$ is bounded on compact sets. If $\displaystyle K \subseteq \Delta$ is a compact set, then the continuous function $\displaystyle |z|$ achieves a maximum on $\displaystyle K$. Let $\displaystyle a_0$ be that maximum. Then since $\displaystyle K \subseteq \Delta$, $\displaystyle a_0$ is strictly less than 1. So there is $\displaystyle a_1$ strictly less than 1 for which $\displaystyle |z| < a_1$ for all $\displaystyle z \in K$.

Now the trick is to invoke part (b): We are given that $\displaystyle \int_\Delta |f(z)|\ dxdy \le M$ for all $\displaystyle f \in \mathcal{F}$. So for all $\displaystyle z \in K$,

$\displaystyle |f(z)| \le \frac{I}{\pi (1-|z|)^2}$ where $\displaystyle I = \int_\Delta |f(z)|\ dxdy$, so

$\displaystyle |f(z)| \le \frac{M}{\pi (1-|z|)^2}$ since we are given that $\displaystyle I \le M$, so

$\displaystyle |f(z)| \le \frac{M}{\pi (1-a_1)^2}$ since $\displaystyle |z| < a_1$.

Since $\displaystyle M$ and $\displaystyle a_1$ are fixed, $\displaystyle f(z)$ is bounded on $\displaystyle K$, and since $\displaystyle K$ was arbitrary, $\displaystyle f(z)$ is bounded on compact sets, so we can apply Montel's Theorem.

The conclusion of Montel's Theorem, that for every sequence $\displaystyle \{f_j\} \subseteq \mathcal{F}$ there is a subseqence $\displaystyle \{f_{jk}\}$ that converges uniformly on compact subsets of $\displaystyle \Delta$ to a holomorphic limit function $\displaystyle f_0$, matches the definition of a normal family. So we have our result.

Re: analytic function on unit disk

Yep - Montel's theorem does it quickly using your argument.

Actually, what I did is a proof of half of Montel's Theorem - the uniform convergence half (not the existence of a convergent subsequence half) - since it all takes place on a compact subset of the domain. It's basically word for word a proof, with only the 3 lines in my lemma that would need to change. The line in the lemma starting with "Now, by part (b) and $\displaystyle g \in \mathcal{F}$, have ..." is the first of 3 successive lines who's only point is to establish that there's bound for that family on a certain subset of that compact set. Those three lines would be replaced by one sentence saying that there's a bound on that family based directly on the premise of Montel's Thm. (Well, now that I think about it, I'd have to worry about points in compact sets having empty interiors - though I could "grow" the compact set to a "nice" one to avoid that - but, I suppose there would be some additional arguments I'd need to add.)

It actually isn't as long as it seems - it's just that I filled in virtually all the details. It could be written up much more briefly. I think the Lemma is the interesting part; the main proof is just a standard "epsilon over three argument".

Re: analytic function on unit disk

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Re: analytic function on unit disk

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