# The identity theorem at the boundary (complex analysis)

Let $$\mathbb{D}^2$$ be the closed unit disk, and let $$f:\mathbb{D}^2 \to \mathbb{C}$$ be a smooth map, which is holomorphic on the open unit disk $$\text{int}(\mathbb{D}^2)$$.

Suppose that there exists a sequence $$z_ n \in \text{int}(\mathbb{D}^2)$$, $$z_n \to z_0 \in \partial \mathbb{D}^2$$ such that $$f(z_n)=0$$. Is $$f$$ identically zero on $$\mathbb{D}^2$$?

The usual formulation of the identity theorem is for open connected domains; it states that a holomorphic function whose zero set has an accumulation point (inside the open domain) is identically zero.

Note that I assumed that $$f$$ is smooth on the closed disk. (In a sense it is "holomorphic" at the boundary too, as the condition of being conformal is a closed one).

Edit:

If $$f$$ could be extend $$f$$ holomorphically to an open neighbourhood of $$\mathbb D^2$$, then the answer would be positive, by the usual identity theorem (as the accumulation point would now be in the interior of the new extended domain).

I am not sure if such an extension is always possible. There are certainly continuous examples that cannot be extended: e.g. $$f(z) = \sum_{n=1}^\infty \frac{z^{n!}}{n!}$$. (See here for details). However, I don't know any smooth example which cannot be extended.

• What exactly is your definition of "smooth" for a function on the closed disk that is different to the open disk? Smoothness is generally defined in terms of the derivatives, and derivatives require open regions to define. – or1426 May 1 '19 at 12:58
• There is such a thing called "smooth maps between manifolds with boundary". In our present context, this means that around every boundary point $p \in \partial \mathbb D^2$ we can locally extend $f$ to a smooth function on an open neighbourhood. (Equivalently, all the partial derivatives of $f$ , of all orders, have continuous extensions up to the boundary). – Asaf Shachar May 1 '19 at 13:10
• @or1426: In general, there is no holomorphic extension. For instance, take a bounded simply connected domain $U$ with $C^\infty$ boundary in the complex plane but the boundary is nowehere real analytic. Then use the Riemann mapping $f: D\to U$. This mapping has a $C^\infty$ extension to the boundary. This extended function has no holomorphic extension at any boundary point of the unit disk. mathoverflow.net/questions/82613/… – Moishe Kohan May 1 '19 at 15:17
• I am not sure about the question itself but Luzin-Privalov theorem might suffice for your purposes of study of harmonic functions: If $f$ is zero at a subset of positive linear measure on the boundary of $D$ then $f$ is identically zero. I suggest, you post the question at MO and ping Alex Eremenko, I am sure he knows either a counterexample or a reference. – Moishe Kohan May 1 '19 at 15:43
• On the second thought, no need to ask at MO, such functions doe exist and are not hard to find. See math.stackexchange.com/questions/74295/… for an example which is continuous along the boundary. One can use the same idea to find smooth examples. – Moishe Kohan May 1 '19 at 21:30

This is not a complete answer, but it was too long for a comment. As such, I am making it community wiki, if anyone wants to fill the missing steps.

An easy example (provided you know the Ostrowski-Hadamard gap theorem) of a $$C^{\infty}$$ map on the closed disk, holomorphic in the interior which cannot be extended (i.e. the circle is its natural boundary):

$$f(z)=\sum_{n=0}^{\infty} \frac{z^{2^n}}{n!}$$

Back to the original problem:

Consider the orizontal strip $$\Omega=\{-\frac\pi2<\Im(z)<\frac\pi2\}$$, and the conformal mapping $$\varphi:\Omega\to \text{int}(\mathbb{D}^2)$$, which is easily seen to be $$\varphi(z)=\frac{e^z-1}{e^z+1}=\text{tanh}\left(\frac z2\right)$$.

On the horizontal strip, we can consider the function $$g(z):=\sin(z)h(z)$$, where $$h$$ is a map that goes to $$0$$ suitably fast as $$z\in \Omega\to \infty$$. This function has infinite zeros, and thus precomposing with $$\varphi^{-1}$$ gives a holomorphic map on the unit disc with infinite zeros. It remains only to prove that it is smooth on the boundary: if $$z\neq \pm 1$$ this is trivial, and if we choose an $$h$$ that goes to $$0$$ fast enough, this should suffice to ensure smoothness. It surely is enough to ensure $$n$$-th differentiability

• – Moishe Kohan Dec 27 '19 at 1:53