Fatou's theorem should not be confused with Fatou's lemma.
In mathematics, specifically in complex analysis, Fatou's theorem, named after Pierre Fatou, is a statement concerning holomorphic functions on the unit disk and their pointwise extension to the boundary of the disk.
If we have a holomorphic function
f
D=\{z:|z|<1\}
r
\begin{cases}
1 | |
f | |
r:S |
\to\Complex\ fr(ei\theta)=f(rei\theta)\end{cases}
where
S1:=\{ei\theta:\theta\in[0,2\pi]\}=\{z\in\Complex:|z|=1\},
is the unit circle. Then it would be expected that the values of the extension of
f
fr
r\to1
Lp
fr
Theorem. Let
f:D\to\Complex
\sup0<r<1\|fr\|
Lp(S1) |
<infty,
where
fr
fr
f1\inLp(S1)
Lp
\begin{align} \left
i\theta | |
|f | |
r(e |
)-f1(ei\theta)\right|&\to0&&foralmostevery\theta\in[0,2\pi]\\ \|fr-f1\|
Lp(S1) |
&\to0 \end{align}
Now, notice that this pointwise limit is a radial limit. That is, the limit being taken is along a straight line from the center of the disk to the boundary of the circle, and the statement above hence says that
f(rei\theta)\to
i\theta | |
f | |
1(e |
) foralmostevery\theta.
The natural question is, with this boundary function defined, will we converge pointwise to this function by taking a limit in any other way? That is, suppose instead of following a straight line to the boundary, we follow an arbitrary curve
\gamma:[0,1)\toD
ei\theta
f
f1(ei\theta)
\gamma(t)=tei\theta
\gamma
\gamma
Definition. Let
\gamma:[0,1)\toD
\lim\nolimitst\to\gamma(t)=ei\theta\inS1
\begin{align} \Gamma\alpha&=\{z:\argz\in[\pi-\alpha,\pi+\alpha]\}\\ \Gamma\alpha(\theta)&=D\capei\theta(\Gamma\alpha+1) \end{align}
\Gamma\alpha(\theta)
2\alpha
ei\theta
\gamma
ei\theta
0<\alpha<\tfrac{\pi}{2}
\gamma
\Gamma\alpha(\theta)
\lim\nolimitst\to\gamma(t)=ei\theta
Fatou's Theorem. Let
f\inHp(D).
\theta\in[0,2\pi],
\limt\to
i\theta | |
f(\gamma(t))=f | |
1(e |
)
for every non-tangential limit
\gamma
ei\theta,
f1