# Complex Analysis: Showing analytic function is zero

How I can solve this problem:

Let $$f: D \to D$$ be an analytic function where $$D$$ is the unit open disc in $$\mathbb C$$. Suppose there is a positive number $$\delta > 0$$ such that , $$\lim_{z \to e^{iθ}} ⁡f(z)= 0; \qquad \forall \ |\theta| < \delta.$$

Show that $$f \equiv 0$$ on $$D$$.

Thanks

Note: An easier version of this Privalov's Theorem is an exercise problem from Stein and Shakarchi's Complex Analysis textbook which additionally assumes the holomorphic function converges uniformly to $$0$$ on the portion of the arc. See [1], [2], [3]. The absence of this 'uniform' non-tangential limit on the portion of the arc makes it a slightly more difficult problem.

• In order to get a good reception, please show your initial work or your thoughts/ideas about how to solve the problem, or where are you stuck. Welcome to MSE! Mar 6, 2020 at 1:39
• Use the principle of isolated zeroes. Mar 6, 2020 at 3:44
• Quick way .. consider $g(z) = \prod\limits_{k=0}^{N} f(e^{2\pi i k/N} z)$ where, $N$ is chosen such that $2\pi/N < \delta$. Then $g \equiv 0$ on $\partial D$ .. then follow Chris's hint.
– r9m
Mar 6, 2020 at 6:36

I think the Schwarz reflection principle, applied to the circular arc $$\>C\!: z=e^{i\theta}, \ |\theta|<\delta$$, does the trick.

The function $$f$$ is analytic on one side of this arc. Extending $$f$$ with the value $$0$$ to the points of $$C$$ makes it continuous on $$D\cup C$$ and real valued on $$C$$. It follows that $$f$$ can be extended analytically to the outside of $$C$$ by putting $$\tilde f(z):=\overline{f(1/\bar z)}\qquad\bigl(|z|>1\bigr)\ ,$$ and $$\tilde f(z):=f(z)$$ otherwise. As $$\tilde f$$ is now analytic in a neighborhood of $$z=1$$ and is $$\equiv0$$ on $$C$$ it follows that $$\tilde f(z)\equiv0$$.

• Can we extend f analytically to the closed unit disc D without using the Schwarz reflection principle ??
– ks1
Mar 6, 2020 at 16:02
• @ks1: Some "hard theorem" is needed to prove your claim. The Schwarz reflection principle, working locally near $z=1$, is one of them. After we have applied this principle the identity theorem guarantees that in fact $\tilde f(z)\equiv0$ on all of ${\mathbb C}$. Mar 6, 2020 at 16:19

Hint:. There's an identity theorem. If you can get that $$f$$ is zero on a subset of $$D$$ that has an accumulation point, you can apply it.

• I know the identity theorem and I tried to get a sequence {z∈D: f(z )=0} which has a limit point in D...but I stuck there and didn't get any result !!
– ks1
Mar 6, 2020 at 3:00
• I'm with you. I'm not seeing it right now. Unless you can analytically continue $f$ or something, I don't know. I might delete.
– user403337
Mar 6, 2020 at 3:24

This is a famous result of Fatou and while the usual proofs use the properties of the Poisson kernel, there is a cool proof using just Fourier series.

Let $$f(z)=\sum{a_nz^n}, |f(z)| \le 1, |z| <1$$. Since $$\sum{|a_n|^2r^{2n}}=\frac{1}{2\pi}\int_0^{2\pi}|f(re^{it}|^2dt \le 1$$, it follows immediately $$\sum{|a_n|^2} \le 1$$, hence $$f(t)=\sum{a_ne^{it}} \in L^2(dt)$$ the Hilbert (!) space of square integrable functions on the unit circle with the normalized Lebesgue measure (or if you want the usual periodic real functions etc).

The hypothesis implies that $$f(t)=0$$ on an arc, hence on a set $$E$$ of nonzero measure on the unit circle - this is actually what we need and of course, the crucial fact that $$f(t)$$ has no negative index Fourier series terms as it comes from an analytic functions inside the unit disc. If we prove that $$f=0$$ a.e. on the unit circle, we get $$a_n=0$$ for all $$n$$ hence $$f=0$$ in the unit disc too

Assume $$f$$ not identically zero (on the circle) and assume wlog $$a_0 \ne 0$$ (as otherwise, we take $$e^{-ikt}f(t)$$ where $$a_k \ne 0$$ is the first nonzero coefficient). Consider the convex set $$C_0=f(e^{it})(1+b_1e^{it}+...b_me^{imt})$$ where $$m \ge 1, b_k$$ arbitrary and take its closure $$C$$ in $$L^2$$. This has a unique element $$g$$ of minimal norm by the usual basic facts about Hilbert spaces. We claim that $$|g|$$ is constant (a.e.) and since obviously $$g$$ vanishes where $$f$$ does by construction, so in particular on a set of positive measure, we get $$|g|=0$$, hence $$g=0$$ (a.e) while $$g$$ has constant Fourier term $$a_0 \ne 0$$ which is a contradiction.

So let's prove the claim about $$g$$. By construction $$g+\alpha e^{int}g \in C, n \ge 1$$ (here it is crucial that $$f$$ hence $$C$$ consists of functions with Fourier series that start with $$a_0$$ but have only nonnegative index terms as otherwise obviously the statement above is not true as we can introduce constant terms when we multiply a $$e^{-int}$$ term with $$\alpha e^{int}$$).

By minimality $$||g+\alpha e^{int}g||^2=||g||^2(1+|\alpha|^2)+2\Re \alpha \frac{1}{2\pi}\int_0^{2\pi}{|g|^2e^{int}}$$ has a minimum at $$\alpha =0$$ and obviously this implies $$\int_0^{2\pi}{|g|^2e^{int}}dt=0, n \ge 1$$ (otherwise as usual the $$||g||^2|\alpha|^2$$ being quadratic in small $$\alpha$$, will be overwhelmed by the linear in $$\alpha$$ integral term and by choosing arguments appropriately we can make it negative of course). Conjugating we get the same equality for $$n=-1,-2,...$$, hence $$|g|^2$$ is constant as all its Fourier terms except the constant one vanish. Done!

• It seems hard proof...we didn't take any of these theorems so far in our course!! If there any simpler proof would be appreciated.
– ks1
Mar 6, 2020 at 4:09
• This is really nice proof! (+1)
– r9m
Mar 6, 2020 at 6:41
• @r9m - for some reason I thought the problem assumes only the non-tangential limit to be zero and then the answer requires the full machinery - as singular inner functions show for example - I think the $L^2$ approach which I learned from Helson is clearer than the usual Fatou approach using the Poisson kernel; if the limit is unrestricted than the problem is fairly easy as Schwarz reflection applies- the answer above with that is the one that is most pertinent Mar 6, 2020 at 13:50