Closed subspace of square-integrable functions on D

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Let ##\mathbb{D}## be the open unit disk in ##\mathbb{C}##. Show that the space ##\mathcal{O}(\mathbb{D}) \cap L^2(\mathbb{D})## of square-integrable, holomorphic functions on ##\mathbb{D}## is a closed subspace of ##L^2(\mathbb{D})##.
 
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  • #2
Euge said:
Let ##\mathbb{D}## be the open unit disk in ##\mathbb{C}##. Show that the space ##\mathcal{O}(\mathbb{D}) \cap L^2(\mathbb{D})## of square-integrable, holomorphic functions on ##\mathbb{D}## is a closed subspace of ##L^2(\mathbb{D})##.
What is ##\mathcal{O}(\mathbb{D})## in this context? The sheaf of holomorphic functions?
 
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  • #3
jbergman said:
What is ##\mathcal{O}(\mathbb{D})## in this context? The sheaf of holomorphic functions?
Nevermind. You clearly state that.
 
  • #4
Euge said:
Let ##\mathbb{D}## be the open unit disk in ##\mathbb{C}##. Show that the space ##\mathcal{O}(\mathbb{D}) \cap L^2(\mathbb{D})## of square-integrable, holomorphic functions on ##\mathbb{D}## is a closed subspace of ##L^2(\mathbb{D})##.
This seems clearly true for finite linear combinations of functions in this set. Do we have to also show this is true for infinite combinations?

My functional analysis knowledge is basically zero.
 
  • #5
I've been thinking about this problem and tried to make some progress. If anyone has some hints to get me unstuck, I'd appreciate it.

Imagine we have a Cauchy sequence of square integrable, holomorphic functions ##f_n##. Then we know that they converge to ##f##, but we have to show that ##f## is holomorphic.

Since ##f_n## is holomorphic, we know that we can write it as
$$f_n = \sum_{k=0}^{\infty} a_n^k z^k$$
Now we can look at the holomorphic functions of the following terms which I will call ##f_n^k = a_n^k z^k##. For each ##k##,
$$\lim_{n\rightarrow \infty} a_n^k = b^k$$ This can easily be shown by the fact that $$||f_{n+1}^k - f_{n}^k||^2 \leq ||a_{n-1}^k-a_n^k||^2 ||z^k||^2$$
So we know that ##f_n \rightarrow f## with ##f(z)=\sum_{k=0}^{\infty} b^k z^k##.
Unfortunately, we still have to somehow show that ##f## is bounded... At this point I am stuck. The two ideas that I am trying to explore are somehow to use a Fourier inversion or try and find a way to use the dominated convergence theorem.
 
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  • #6
I may have proven that on the open disc ##\mathbb{D}_{R-\delta} = \{ z : 0 \leq |z| < {R-\delta} \}## where ##R < 1## for any ##\delta > 0## the functions ##f_n (z)## uniformly converge to a ##L^2## holomorphic function ##f(z)##. Not sure it can be used to prove the full result.

Let ##\{ f_n \}## be a Cauchy sequence in ##\mathcal{S} = \mathcal{O} (\mathbb{D}) \cap L^2 (\mathbb{D})##. That is, for every ##\epsilon > 0##, there exists an integer ##N## such that for all ##m,n > N##, ##d(f_m,f_n) = \| f_m (z) - f_n (z) \|_2 < \epsilon##. To prove that ##\mathcal{S}## is closed in ##L^2 (\mathbb{D})## we must show that any accumulation point of ##\mathcal{S}## is in ##\mathcal{S}##.

Norm explicitly is for open disc ##\mathbb{D}_R = \{ z : 0 \leq |z| < R \leq 1\}##:

\begin{align*}
\| f (z) \|_2^R &= \left( \int_{\mathbb{D}_R} |f (z)|^2 dA \right)^{1/2}
\nonumber \\
& = \left( \int_0^R \int_0^{2 \pi} |f (re^{i \theta})|^2 r d \theta dr \right)^{1/2}
\end{align*}

By the Cauchy integral formula

\begin{align*}
f_n(z) & = \frac{1}{\pi (1-R^2)} \int_R^1 \int_0^{2 \pi} \dfrac{ f_n(re^{i \theta}) }{1 - z r^{-1} e^{-i \theta}} dA
\nonumber \\
& = \frac{1}{\pi (1-R^2)} \int_R^1 \int_0^{2 \pi} \sum_{l=0}^\infty z^l (f_n(re^{i \theta})r^{-l} e^{-i l \theta}) dA
\end{align*}

By the Cauchy-Schartz inequality:

\begin{align*}
\left( \int |f_n \cdot 1 | d A \right)^2 \leq \int |f_n|^2 d A \int |1|^2 d A = Area \int |f_n|^2 d A < \infty
\end{align*}

we have

\begin{align*}
\frac{1}{\pi (1-R^2)} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} \dfrac{|f_n(re^{i \theta})|}{r^l} dA |z|^l & < \frac{1}{\pi (1-R^2)} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} |f_n(re^{i \theta})| dA \frac{|z|^l}{R^l}
\nonumber \\
& \leq \frac{1}{\sqrt{\pi (1-R^2)}} \sum_{l=0}^\infty \left( \int_R^1 \int_0^{2 \pi} |f_n(re^{i \theta})|^2 dA \right)^{1/2} \frac{|z|^l}{R^l}
\nonumber \\
& < \infty
\end{align*}

where the first inequality arises from replacing the ##r^{-l}## by ##R^{-l}##. By Fubini it is legitimate to swap summation and integration:

\begin{align*}
f_n(z) & = \frac{1}{\pi (1-R^2)} \sum_{l=0}^\infty \left[ \int_R^1 \int_0^{2 \pi} (f_n(re^{i \theta}) (R/r)^l e^{-i l \theta}) dA \right] \frac{z^l}{R^l} < \infty
\end{align*}

for ##|z| < R##.

But, using the same estimation on ##f_m(z)-f_n(z)##,

\begin{align*}
|f_m (z) - f_n(z)| & \leq \frac{1}{\pi (1-R^2)} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} |f_m(re^{i \theta}) - f_n(re^{i \theta})| \left( \frac{R}{r} \right)^l dA \frac{(R-\delta)^l}{R^l}
\nonumber \\
& \leq \frac{1}{\sqrt{\pi(1-R^2)}} \sum_{l=0}^\infty \left( \int_R^1 \int_0^{2 \pi} |f_m(re^{i \theta}) - f_n(re^{i \theta})|^2 dA \right)^{1/2} \frac{(R-\delta)^l}{R^l}
\nonumber \\
& = \frac{1}{\sqrt{\pi(1-R^2)}} \dfrac{1}{1 - \frac{R-\delta}{R}} \| f_m - f_n \|_2^{Annul_R^1}
\nonumber \\
& < \frac{R}{\sqrt{\pi(1-R^2)} \delta} \| f_m - f_n \|_2^R
\end{align*}

for any ##\delta > 0##, implying that we have uniform convergence, ##f_n \overset{u}{\rightarrow} f##, on ##\overline{\mathbb{D}}_{R-\delta}##.

By uniform convergence, ##f## is continuous in ##\mathbb{D}_{R-\delta}##. Consider ##\gamma## in ##\mathbb{D}_{R-\delta}##, we also have by uniform convergence

\begin{align*}
\left| \oint_\gamma f_n dz - \oint_\gamma f dz \right| & \leq |\gamma| \| f_n - f \|_\infty
\end{align*}

where ##|\gamma|## is the length of ##\gamma##. This implies

\begin{align*}
\lim_{n \rightarrow \infty} \oint_\gamma f_n dz = \oint_\gamma f dz \qquad (*)
\end{align*}

By Cauchy's theorem:

\begin{align*}
\oint_\gamma f_n (z) dz & = 0
\end{align*}

for all triangles, ##\gamma##, in ##\mathbb{D}_{R-\delta}##.

Morera's theorem. Suppose that ##f(z)## is continuous on an open set ##G## and that ##\int f_\gamma ( z ) dz = 0## for all triangles ##\gamma## in ##G##. Then ##f(z) \in \mathcal{O} (G)##.

We can apply Morera's theorem to ##(*)## to conclude that ##f(z)## is holomorphic on ##\mathbb{D}_{R-\delta}##.

We now show that ##\| f \|_2^{R-\delta} < \infty##. As ##f_n (z)## converges to ##f(z)##, ##|f_n(z)|^2## converges pointwise to ##|f(z)|^2##, as:

\begin{align*}
|f_n (z) f_n^* (z) - f (z) f^* (z)| & = |f_n (z) f_n^* (z) - f(z) f_n^* (z) + f(z) f_n^* (z) - f (z) f^* (z)|
\nonumber \\
& \leq (|f_n (z)| + |f(z)|) |f_n (z) - f(z)|
\end{align*}

We have ##|f_n (z)|^2 = u_n^2 (x,y) + v_n^2 (x,y) \geq 0##. By Fatou's lemma ##\| f \|_2^{R-\delta} < \infty##.

So we have proven that on the open disc ##\mathbb{D}_{R-\delta} = \{ z : 0 \leq |z| < R - \delta \}## where ##R < 1## for any ##\delta > 0## the functions ##f_n (z)## uniformly converge to a ##L^2## holomorphic function ##f(z)##.
 
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  • #7
julian said:
I may have proven that on the open disc ##\mathbb{D}_{R-\delta} = \{ z : 0 \leq |z| < {R-\delta} \}## where ##R < 1## for any ##\delta > 0## the functions ##f_n (z)## uniformly converge to a ##L^2## holomorphic function ##f(z)##. Not sure it can be used to prove the full result.

Let ##\{ f_n \}## be a Cauchy sequence in ##\mathcal{S} = \mathcal{O} (\mathbb{D}) \cap L^2 (\mathbb{D})##. That is, for every ##\epsilon > 0##, there exists an integer ##N## such that for all ##m,n > N##, ##d(f_m,f_n) = \| f_m (z) - f_n (z) \|_2 < \epsilon##. To prove that ##\mathcal{S}## is closed in ##L^2 (\mathbb{D})## we must show that any accumulation point of ##\mathcal{S}## is in ##\mathcal{S}##.

Norm explicitly is for open disc ##\mathbb{D}_R = \{ z : 0 \leq |z| < R \leq 1\}##:

\begin{align*}
\| f (z) - g (z) \|_2^R &= \left( \int_{\mathbb{D}_R} |f (z) - g (z)|^2 dA \right)^{1/2}
\nonumber \\
& = \left( \int_0^R \int_0^{2 \pi} |f (re^{i \theta})|^2 r d \theta dr \right)^{1/2}
\end{align*}

By the Cauchy integral formula

\begin{align*}
f_n(z) & = \frac{1}{\pi (1-R^2)} \int_R^1 \int_0^{2 \pi} \dfrac{f_n(re^{i \theta}) r^{-1}}{1 - z r^{-1} e^{-i \theta}} dA
\nonumber \\
& = \frac{1}{\pi (1-R^2)} \int_R^1 \int_0^{2 \pi} \sum_{l=0}^\infty z^l (f_n(re^{i \theta})r^{-l-1} e^{-i l \theta}) dA
\end{align*}

By the Cauchy-Schartz inequality:

\begin{align*}
\left( \int |f_n \cdot 1 | d A \right)^2 \leq \int |f_n|^2 d A \int |1|^2 d A = Area \int |f_n|^2 d A < \infty
\end{align*}

we have

\begin{align*}
\frac{1}{\pi (1-R^2)} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} \dfrac{|f_n(re^{i \theta})|}{r^{l+1}} dA |z|^l & < \frac{1}{\pi (1-R^2) R} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} |f_n(re^{i \theta})| dA \frac{|z|^l}{R^l}
\nonumber \\
& \leq \frac{1}{\sqrt{\pi (1-R^2)} R} \sum_{l=0}^\infty \left( \int_R^1 \int_0^{2 \pi} |f_n(re^{i \theta})|^2 dA \right)^{1/2} \frac{|z|^l}{R^l}
\nonumber \\
& < \infty
\end{align*}

By Fubini it is legitimate to swap summation and integration:

\begin{align*}
f_n(z) & = \frac{1}{\pi (1-R^2)R} \sum_{l=0}^\infty \left[ \int_R^1 \int_0^{2 \pi} (f_n(re^{i \theta}) (R/r)^{l+1} e^{-i l \theta}) dA \right] \frac{z^l}{R^l} < \infty
\end{align*}

for ##|z| < R##.

But, using the same estimation on ##f_m(z)-f_n(z)##,

\begin{align*}
|f_m (z) - f_n(z)| & \leq \frac{1}{\pi (1-R^2)R} \sum_{l=0}^\infty \int_R^1 \int_0^{2 \pi} |f_m(re^{i \theta}) - f_n(re^{i \theta})| \left( \frac{R}{r} \right)^{l+1} dA \frac{(R-\delta)^l}{R^l}
\nonumber \\
& \leq \frac{1}{\sqrt{\pi(1-R^2)} R} \sum_{l=0}^\infty \left( \int_R^1 \int_0^{2 \pi} |f_m(re^{i \theta}) - f_n(re^{i \theta})|^2 dA \right)^{1/2} \frac{(R-\delta)^l}{R^l}
\nonumber \\
& = \frac{1}{\sqrt{\pi(1-R^2)} R} \dfrac{1}{1 - \frac{R-\delta}{R}} \| f_m - f_n \|_2^{Annul_R^1}
\nonumber \\
& < \frac{1}{\sqrt{\pi(1-R^2)} \delta} \| f_m - f_n \|_2^R
\end{align*}

for any ##\delta > 0##, implying that we have uniform convergence, ##f_n \overset{u}{\rightarrow} f##, on ##\overline{\mathbb{D}}_{R-\delta}##.

By uniform convergence, ##f## is continuous. Also,

\begin{align*}
\left| \oint_\gamma f_n dz - \oint_\gamma f dz \right| & \leq |\gamma| \| f_n - f \|_\infty
\end{align*}

where ##|\gamma|## is the length of ##\gamma##. This implies

\begin{align*}
\lim_{n \rightarrow \infty} \oint_\gamma f_n dz = \oint_\gamma f dz \qquad (*)
\end{align*}

By Cauchy's theorem:

\begin{align*}
\oint_\gamma f_n (z) dz & = 0
\end{align*}

for all triangles, ##\gamma##, in ##\mathbb{D}_{R-\delta}##.

Morera's theorem. Suppose that ##f(z)## is continuous on an open set ##G## and that ##\int f_\gamma ( z ) dz = 0## for all triangles ##\gamma## in ##G##. Then ##f(z) \in \mathcal{O} (G)##.

We can apply Morera's theorem to ##(*)## to conclude that ##f(z)## is holomorphic on ##\mathbb{D}_{R-\delta}##.

We now show that ##\| f \|_2^{R-\delta} < \infty##. As ##f_n (z)## converges to ##f(z)##, ##|f_n(z)|^2## converges pointwise to ##|f(z)|^2##, as:

\begin{align*}
|f_n (z) f_n^* (z) - f (z) f^* (z)| & = |f_n (z) f_n^* (z) - f(z) f_n^* (z) + f(z) f_n^* (z) - f (z) f^* (z)|
\nonumber \\
& \leq (|f_n (z)| + |f(z)|) |f_n (z) - f(z)|
\end{align*}

We have ##|f_n (z)|^2 = u_n^2 (x,y) + v_n^2 (x,y) \geq 0##. By Fatou's lemma ##\| f \|_2^{R-\delta} < \infty##.

So we have proven that on the open disc ##\mathbb{D}_{R-\delta} = \{ z : 0 \leq |z| < R - \delta \}## where ##R < 1## for any ##\delta > 0## the functions ##f_n (z)## uniformly converge to a ##L^2## holomorphic function ##f(z)##.
Why is ##g(z)## in your norm formulas? Specifically, I don't see it in the second one. I was also thinking to use the Cauchy Integral formula but was afraid of that integral. Still digesting what you have written.

Also, do you have a reference to the version of the Cauchy integral formula you used? I have only seen the single integral around a loop version.
 
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  • #8
For a suggestion, try to prove first that if ##u\in \mathcal{O}(\mathbb{D}) \cap L^2(\mathbb{D})##, then for every compact ##K \subset \mathbb{D}##, there is a constant ##M = M(K) > 0## such that ##\sup_{z\in K} |u(z)| \le M\|u\|_{L^2(\mathbb{D})}##.
 
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  • #9
jbergman said:
Why is ##g(z)## in your norm formulas? Specifically, I don't see it in the second one. I was also thinking to use the Cauchy Integral formula but was afraid of that integral. Still digesting what you have written.

Also, do you have a reference to the version of the Cauchy integral formula you used? I have only seen the single integral around a loop version.
The ##g(z)## is a typo, removed it now.

As you know the Cauchy integral formula can be written in terms of ##\theta## by putting ##w=re^{i \theta}##:

\begin{align*}
f_n(z) & = \frac{1}{2 \pi i} \oint_{|w|=r} \dfrac{f_n (w)}{w-z} dw
\nonumber \\
& = \frac{1}{2 \pi i} \int_0^{2 \pi} \dfrac{f_n (r e^{i \theta})}{re^{i \theta}-z} r i e^{i \theta} d \theta
\nonumber \\
& = \frac{1}{2 \pi} \int_0^{2 \pi} \dfrac{f_n (r e^{i \theta})}{1-zr^{-1}e^{-i \theta}} d \theta
\end{align*}

I then multiply both sides by ##r## and integrate from ##R## to ##1##:

\begin{align*}
f_n(z) \int_R^1 r dr & = \frac{1}{2 \pi} \int_R^1 \int_0^{2 \pi} \dfrac{f_n(re^{i \theta})}{1 - z r^{-1} e^{-i \theta}} r d \theta dr .
\end{align*}

Then divide both sides by ##\int_R^1 r dr##. I've made a very slight mistake when writing the end formula down in my previous post, I've corrected that now.
 
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  • #10
Euge said:
For a suggestion, try to prove first that if ##u\in \mathcal{O}(\mathbb{D}) \cap L^2(\mathbb{D})##, then for every compact ##K \subset \mathbb{D}##, there is a constant ##M = M(K) > 0## such that ##\sup_{z\in K} |u(z)| \le M\|u\|_{L^2(\mathbb{D})}##.
Isn't this obviously true because, we can take ##K## to be closed disc, and then ##\sup_{z\in K} |u(z)| ## is some positive constant ##C## (which follows from a theorem from Hardy? about bounded holomorphic functions on the open disc being continuous at the boundary). So we have, for any compact set ##K## in the unit disc.
$$sup_{z\in K} |u(z)| \le C = C \frac{\|u\|_{L^2(\mathbb{D})}}{\|u\|_{L^2(\mathbb{D})}}$$

Therefore ##M = \frac{C}{\|u\|_{L^2(\mathbb{D})}}##
 
  • #11
Still thinking about what else I want to say.

Say ##K \subset \mathbb{D}## is compact. So it is bounded and closed. Say ##R_{min}## is the minimum distance from ##K## to ##\partial \mathbb{D} = \{ z : |z| = 1 \}##. Say ##2R = R_{min}##. There is a cover of ##K## by a collection of open balls: ##K \subseteq \bigcup_{a \in K} B(a,R)##.

We can use that holomorphic functions are analytic and as such can be expanded about any point ##a \in K##:

\begin{align*}
u(z) = \sum_{n=0}^\infty a_n (z-a)^n
\end{align*}

Note ##u(a) = a_0##. The norm squared of ##u(z)##, ##\| u \|_{L^2 (B(a,R))}^2## is:

\begin{align*}
\| u \|_{L^2 (B(a,R))}^2 & = \| \sum_{m=0}^\infty a_m (z-a)^m \|_{L^2 (B(a,R))}^2
\nonumber \\
& = \int_0^R \int_0^{2 \pi} |\sum_{m=0}^\infty a_m r^m e^{i m \theta}|^2 r d\theta dr < \infty
\nonumber \\
& = \int_0^R \int_0^{2 \pi} \sum_{m,n=0}^\infty a_m^* a_n r^{m+n+1} e^{-i (m-n) \theta} d\theta dr
\nonumber \\
& = \sum_{m,n=0}^\infty a_m^* a_n \int_0^Rr^{m+n+1} \int_0^{2 \pi} e^{-i (m-n) \theta} d\theta dr
\nonumber \\
& = \pi \sum_{m=0}^\infty \dfrac{|a_m|^2}{m+1} R^{2m+2}
\end{align*}

We have

\begin{align*}
\pi R^2 |u(a)|^2 & = \pi R^2 |a_0|^2
\nonumber \\
& \leq \pi R^2 |a_0|^2 + \pi \sum_{m=1}^\infty \dfrac{|a_m|^2}{m+1} R^{2m+2}
\nonumber \\
& = \| u \|_{L^2 (B(a,R))}^2
\nonumber \\
& \leq \| u \|_{L^2 (\mathbb{D})}^2
\end{align*}

So we have

\begin{align*}
\sup_{z \in K} |u(z)| & \leq \frac{1}{\sqrt{\pi} R} \| u \|_{L^2 (\mathbb{D})}
\end{align*}

So ##M(K) = \frac{1}{\sqrt{\pi} R}##.

We can say

\begin{align*}
\sup_{z \in K} |f_m (z) - f_n (z)| & \leq \frac{1}{\sqrt{\pi} R} \| f_m - f_n \|_{L^2 (\mathbb{D})}
\end{align*}

So we can say we have uniform convergence, ##f_n \overset{u}{\rightarrow} f##, on every compact ##K##. Implying that ##f## is continuous on ##K##.

Denote the interior of ##K## by ##\text{int } K##. Uniform convergence implies continuity of ##f(z)## on int ##K##. It also implies:

\begin{align*}
\lim_{n \rightarrow \infty} \oint_\gamma f_n dz = \oint_\gamma f dz \qquad (*)
\end{align*}

By Cauchy's theorem:

\begin{align*}
\oint_\gamma f_n (z) dz & = 0
\end{align*}

for all triangles, ##\gamma##, in ##\text{int } K##.

Morera's theorem. Suppose that ##f(z)## is continuous on an open set ##G## and that ##\int f_\gamma ( z ) dz = 0## for all triangles ##\gamma## in ##G##. Then ##f(z) \in \mathcal{O} (G)##.

We can apply Morera's theorem to ##(*)## to conclude that ##f(z)## is holomorphic on ##\text{int } K##.


Converges to same function on all compact sets:

For two compact sets ##K_A## and ##K_B## write ##f_n \overset{u}{\rightarrow} f_A## and ##f_n \overset{u}{\rightarrow} f_B##. Do we have on the overlap, ##K_A \cap K_B##, that ##f_A (z) = f_B (z)##? We have:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| & = \sup_{z \in K_A \cap K_B} |f_A (z) - f_n (z) - f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A \cap K_B} |f_n (z) - f_A (z)| + \sup_{z \in K_A \cap K_B} |f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A} |f_n (z) - f_A (z)| + \sup_{z \in K_B} |f_n (z) - f_B (z)|
\end{align*}

So that:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| \leq \lim_{n \rightarrow \infty} \sup_{z \in K_A} |f_n (z) - f_A (z)| + \lim_{n \rightarrow \infty} \sup_{z \in K_B} |f_n (z) - f_B (z)| = 0
\end{align*}

Implying ##f_A(z) = f_B(z) = f(z)## on ##K_A \cap K_B##.

If we have two compact sets ##K_A## and ##K_B## that do not overlap, we can always find a third compact set, ##K_C##, that intersects both ##K_A## and ##K_B##. By a similar argument, this implies that ##f(z)## is the same function on the compact sets ##K_A## and ##K_B##. This implies that the function ##f(z)##, to which ##f_n (z)## converges uniformly on each compact set, is the same function on all compact sets. So if, for example, ##f(z)= z + z^2## on one compact set, you would have ##f(z)= z + z^2## on all compact sets.
 
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  • #12
julian said:
Still thinking about what else I want to say.

Say ##K \subset \mathbb{D}## is compact. So it is bounded and closed. Say ##R_{min}## is the minimum distance from ##K## to ##\partial \mathbb{D} = \{ z : |z| = 1 \}##. Say ##2R = R_{min}##. There is a cover of ##K## by a collection of open balls: ##K \subseteq \bigcup_{a \in K} B(a,R)##.

We can use that holomorphic functions are analytic and as such can be expanded about any point ##a \in K##:

\begin{align*}
u(z) = \sum_{n=0} a_n (z-a)^n
\end{align*}

Note ##u(a) = a_0##. The norm squared of ##u(z)##, ##\| u \|_{L^2 (B(a,R))}^2## is:

\begin{align*}
\| u \|_{L^2 (B(a,R))}^2 & = \| \sum_{m=0}^\infty a_m (z-a)^m \|_{L^2 (B(a,R))}^2
\nonumber \\
& = \int_0^R \int_0^{2 \pi} |\sum_{m=0}^\infty a_m r^m e^{i m \theta}|^2 d\theta dr < \infty
\nonumber \\
& = \int_0^R \int_0^{2 \pi} \sum_{m,n=0}^\infty a_m^* a_n r^{m+n} e^{-i (m-n) \theta} d\theta dr
\nonumber \\
& = \sum_{m,n=0}^\infty a_m^* a_n \int_0^Rr^{m+n} \int_0^{2 \pi} e^{-i (m-n) \theta} d\theta dr
\nonumber \\
& = \pi \sum_{m=0}^\infty \dfrac{|a_m|^2}{m+1} R^{2m+2}
\end{align*}

We have

\begin{align*}
\pi R^2 |u(a)|^2 & = \pi R^2 |a_0|^2
\nonumber \\
& \leq \pi R^2 |a_0|^2 + \pi \sum_{m=1} \dfrac{|a_m|^2}{m+1} R^{2m+2}
\nonumber \\
& = \| u \|_{L^2 (B(a,R))}^2
\nonumber \\
& \leq \| u \|_{L^2 (\mathbb{D})}^2
\end{align*}

So we have

\begin{align*}
\sup_{z \in K} |u(z)| & \leq \frac{1}{\sqrt{\pi} R} \| u \|_{L^2 (\mathbb{D})}
\end{align*}

So ##M(K) = \frac{1}{\sqrt{\pi} R}##.

We can say

\begin{align*}
\sup_{z \in K} |f_m (z) - f_n (z)| & \leq \frac{1}{\sqrt{\pi} R} \| f_m - f_n \|_{L^2 (\mathbb{D})}
\end{align*}

So we can say we have uniform convergence, ##f_n \overset{u}{\rightarrow} f##, on every compact ##K##. Implying that ##f## is continuous on ##K##.

Denote the interior of ##K## by ##\text{int } K##. Uniform convergence implies continuity of ##f(z)## on int ##K##. It also implies:

\begin{align*}
\lim_{n \rightarrow \infty} \oint_\gamma f_n dz = \oint_\gamma f dz \qquad (*)
\end{align*}

By Cauchy's theorem:

\begin{align*}
\oint_\gamma f_n (z) dz & = 0
\end{align*}

for all triangles, ##\gamma##, in ##\text{int } K##.

Morera's theorem. Suppose that ##f(z)## is continuous on an open set ##G## and that ##\int f_\gamma ( z ) dz = 0## for all triangles ##\gamma## in ##G##. Then ##f(z) \in \mathcal{O} (G)##.

We can apply Morera's theorem to ##(*)## to conclude that ##f(z)## is holomorphic on ##\text{int } K##.


Converges to same function on all compact sets:

For two compact sets ##K_A## and ##K_B## write ##f_n \overset{u}{\rightarrow} f_A## and ##f_n \overset{u}{\rightarrow} f_B##. Do we have on the overlap, ##K_A \cap K_B##, that ##f_A (z) = f_B (z)##? We have:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| & = \sup_{z \in K_A \cap K_B} |f_A (z) - f_n (z) - f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A \cap K_B} |f_n (z) - f_A (z)| + \sup_{z \in K_A \cap K_B} |f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A} |f_n (z) - f_A (z)| + \sup_{z \in K_B} |f_n (z) - f_B (z)|
\end{align*}

So that:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| \leq \lim_{n \rightarrow \infty} \sup_{z \in K_A} |f_n (z) - f_A (z)| + \lim_{n \rightarrow \infty} \sup_{z \in K_B} |f_n (z) - f_B (z)| = 0
\end{align*}

Implying ##f_A(z) = f_B(z) = f(z)## on ##K_A \cap K_B##.

If we have two compact sets ##K_A## and ##K_B## that do not overlap, we can always find a third compact set, ##K_C##, that intersects both ##K_A## and ##K_B##. By a similar argument, this implies that ##f(z)## is the same function on the compact sets ##K_A## and ##K_B##. This implies that the function ##f(z)##, to which ##f_n (z)## converges uniformly on each compact set, is the same function on all compact sets. So if, for example, ##f(z)= z + z^2## on one compact set, you would have ##f(z)= z + z^2## on all compact sets.
Wow. Nicely done!
 
  • #13
jbergman said:
Isn't this obviously true because, we can take ##K## to be closed disc, and then ##\sup_{z\in K} |u(z)| ## is some positive constant ##C## (which follows from a theorem from Hardy? about bounded holomorphic functions on the open disc being continuous at the boundary). So we have, for any compact set ##K## in the unit disc.
$$sup_{z\in K} |u(z)| \le C = C \frac{\|u\|_{L^2(\mathbb{D})}}{\|u\|_{L^2(\mathbb{D})}}$$

Therefore ##M = \frac{C}{\|u\|_{L^2(\mathbb{D})}}##
Note that ##M## is to depend only on ##K##, but your ##M## depends on ##u##.
 
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  • #14
I have proven that the Cauchy sequence ##\{ f_n \}## converges uniformly on every compact set contained in ##\mathbb{D}##. Here's an attempt at sketching the rest of the proof.

Converges to same function on all compact sets:

For two overlapping compact sets ##K_A , K_B## contained in ##\mathbb{D}## write ##f_n \overset{u}{\rightarrow} f_A## and ##f_n \overset{u}{\rightarrow} f_B##. Do we have on the overlap, ##K_A \cap K_B##, that ##f_A (z) = f_B (z)##? We have:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| & = \sup_{z \in K_A \cap K_B} |f_A (z) - f_n (z) - f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A \cap K_B} |f_n (z) - f_A (z)| + \sup_{z \in K_A \cap K_B} |f_n (z) - f_B (z)|
\nonumber \\
& \leq \sup_{z \in K_A} |f_n (z) - f_A (z)| + \sup_{z \in K_B} |f_n (z) - f_B (z)|
\end{align*}

So that:

\begin{align*}
\sup_{z \in K_A \cap K_B} |f_A (z) - f_B (z)| \leq \lim_{n \rightarrow \infty} \sup_{z \in K_A} |f_n (z) - f_A (z)| + \lim_{n \rightarrow \infty} \sup_{z \in K_B} |f_n (z) - f_B (z)| = 0
\end{align*}

Implying ##f_A(z) = f_B(z) = f(z)## on ##K_A \cap K_B##.

If we have two compact sets ##K_A## and ##K_B## that do not overlap, we can always find a third compact set, ##K_C##, that intersects both ##K_A## and ##K_B##. By a similar argument, this implies that ##f(z)## is the same function on the compact sets ##K_A## and ##K_B##. This implies that the function ##f(z)##, to which ##f_n (z)## converges uniformly on each compact set, is the same function on all compact sets. So if, for example, ##f(z)= z + z^2## on one compact set, you would have ##f(z)= z + z^2## on all compact sets.


Convergence to function ##f(z)## in limit ##\mathbb{D} = \lim_{i \rightarrow \infty} B (0 , 1-\frac{1}{i})##:

In particular, as closed bounded sets are compact, we can say that ##f_n (z)## converges uniformly on every closed ball ##\overline{B (0 , 1-\frac{1}{i})}## for ##i \in \mathbb{N}##, and to the same function ##f(z)##. So we have that ##f_n (z)## converges uniformly on every open ball ##B (0 , 1-\frac{1}{i})## for ##i \in \mathbb{N}##, and to the same function ##f(z)##.

Is the following formal argument needed? For ##B_i = B (0 , 1-\frac{1}{i})## we have ##\lim_{i \rightarrow \infty} B_i = \mathbb{D}##. Write ##f_n (z) |_{B_i}## for the restriction of ##f_n (z)## to ##B_i##. Then ##\lim_{i \rightarrow \infty} f_n (z) |_{B_i} = f_n (z)##. Do we have ##\lim_{n \rightarrow \infty} \lim_{i \rightarrow \infty} f_n (z) |_{B_i} = \lim_{i\rightarrow \infty} \lim_{n \rightarrow \infty} f_n (z) |_{B_i}##?

Choose ##z \in \mathbb{D}##. There is an ##I## such that ##z \not\in B_I## but ##z \in B_{I+1}##. There exists ##N_1 \in \mathbb{N}## independent of ##n## such that for all ##n \in \mathbb{N}##, ##i,j > N_1## implies

\begin{align*}
|f_n (z)|_{B_i} - f_n (z)|_{B_j}| = 0 .
\end{align*}

(obviously, ##N_1##, is any integer such that ##N_1 \geq I##). As ##n \rightarrow \infty##, ##|f (z)|_{B_i} - f (z)|_{B_j}| = 0## which means ##f (z) |_{B_i}## is a Cauchy sequence which converges to a unique limit ##L##. In addition, as ##j \rightarrow \infty##, we have ##|f (z) |_{B_i} - L| = 0##.

On the other hand, if we take ##j \rightarrow \infty## first, we have ##|f_n (z) |_{B_i} - f_n (z)| = 0##.

For ##i > N_1##, given ##\epsilon > 0## there exists ##N_2## such that for all ##n > N_2## we have ##|f_n (z)|_{B_i}| - f (z)|_{B_i}| < \epsilon##. So

\begin{align*}
|f_n (z) - L| & \leq |f_n (z) - f_n (z)|_{B_i}| + |f_n (z)|_{B_i} - f (z)|_{B_i}| + |f(z)|_{B_i} - L|
\nonumber \\
& < 0 + \epsilon + 0
\end{align*}

This proves that

\begin{align*}
\lim_{n \rightarrow \infty} \lim_{i \rightarrow \infty} f_n (z)|_{B_i} = L = \lim_{i \rightarrow \infty} f(z)|_{B_i}
\end{align*}


##f(z)## is Holomorphic:

If we show that in every open ball ##B(a,r)## whose closure is contained in ##\mathbb{D}##, that ##f(z)## is holomorphic, we find that ##f(z)## is holomorphic in all of ##\mathbb{D}##.

Let ##\overline{B (a,r)}## be a closed ball contained in ##\mathbb{D}##. Let ##\gamma## be an arbitrary triangle contained in ##B(a,r)##. The closed ball ##\overline{B (a,r)}## is closed and bounded and so compact. So ##f_n (z)## uniformly converges to ##f(z)## in ##\overline{B (a,r)}##. Implying

\begin{align*}
\lim_{n \rightarrow \infty} \oint_\gamma f_n (z) dz = \oint_\gamma f (z) dz
\end{align*}

and ##f(z)## is continuous on ##\overline{B (a,r)}##. For all triangles, ##\gamma \subset B(a,r)##, by Cauchy's theorem:

\begin{align*}
\oint_\gamma f_n (z) dz & = 0
\end{align*}

So we have for our arbitrary triangle:

\begin{align*}
\oint_\gamma f (z) dz & = 0 .
\end{align*}

Morera's theorem. Suppose that ##f(z)## is continuous on an open set ##G## and that ##\int f_\gamma ( z ) dz = 0## for all triangles ##\gamma## in ##G##. Then ##f(z) \in \mathcal{O} (G)##.

We have that for all triangles ##\gamma \subset B(a,r)## that ##\oint_\gamma f (z) dz = 0## and that ##f(z)## is continuous on ##B(a,r)##. We can apply Morera's theorem to conclude that ##f(z)## is holomorphic on ##B(a,r)##.


Square-integrable: ##f(z) \in L^2 (\mathbb{D})##

We now show that ##\| f \|_2 < \infty##. As ##f_n (z)## converges to ##f(z)## in ##\mathbb{D}##, ##|f_n(z)|^2## converges pointwise to ##|f(z)|^2##, because:

\begin{align*}
|f_n (z) f_n^* (z) - f (z) f^* (z)| & = |f_n (z) f_n^* (z) - f(z) f_n^* (z) + f(z) f_n^* (z) - f (z) f^* (z)|
\nonumber \\
& \leq (|f_n (z)| + |f(z)|) |f_n (z) - f(z)|
\end{align*}

We have ##|f_n (z)|^2 = u_n^2 (x,y) + v_n^2 (x,y) \geq 0##. By Fatou's lemma ##\| f \|_2 < \infty##.
 
Last edited:
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