Complex Analysis: Showing analytic function is zero

Question

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.

Answer 1

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 $\tilde f(z)\equiv0$.

Answer 2

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.

Answer 3

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!