Let \(D=\{z\in\mathbb C:|z|<1\}\) and let \(f:D\to\mathbb C\) be \(C^1\) as a function of two real variables. Require \(f(1/n)=0\) and \(Df(1/n)=0\) for each integer \(n\geq2\), where \(Df\) denotes the real differential.
Determine all holomorphic functions \(f\) with this property. [The terms “holomorphic” and “complex analytic” have the same meaning.]
Give an example of a non-holomorphic \(C^1\) function with this property. (You must explain why your example has this property.)
Only \(n\geq2\) gives points in the open unit disk. The vanishing first-jet condition specifies the meaning of “double zero” for a \(C^1\) function. If exactly order two is required, rather than vanishing to at least second order, there are no holomorphic examples in (a): the identity theorem forces the zero function, which has no zero of exact order two. The function \(F\) of step <1>2 has exact quadratic order at every prescribed point and therefore answers (b) under either interpretation.
Solution
1. Under the first-jet interpretation, the only holomorphic function is zero.
The distinct zeros \(1/n\), \(n\geq2\), accumulate at the interior point zero of the connected disk. The identity theorem gives \(f\equiv0\) [@SS03]. This function has zero value and differential at all points, so it does satisfy the stated first-jet condition. The exact-order alternative follows as explained in the remark.
2. A nonholomorphic example is \(C^1\) even at the accumulation point.
Define \[ F(z)=\begin{cases} e^{-1/|z|^2}\sin^2(\pi/z),&z\ne0,\\ 0,&z=0. \end{cases} \] It is smooth away from zero. For \(r=|z|>0\), the exponential formulas for sine and cosine give \(|\sin(\pi/z)|,|\cos(\pi/z)|\leq e^{\pi/r}\). Hence \[ |F(z)|\leq e^{-1/r^2+2\pi/r},\qquad |F_x(z)|,|F_y(z)| \leq(2r^{-3}+2\pi r^{-2})e^{-1/r^2+2\pi/r}. \] For the derivative estimate, the Gaussian factor has each first partial derivative bounded by \(2r^{-3}e^{-1/r^2}\); the holomorphic factor \(\sin^2(\pi/z)\) has complex derivative of modulus at most \(2\pi r^{-2}e^{2\pi/r}\). The product rule gives the displayed bounds.
When \(r\leq1/(4\pi)\), the exponent is at most \(-1/(2r^2)\). For every fixed \(k\geq0\), \(r^{-k}e^{-1/(2r^2)}\to0\); this follows, for example, by bounding the exponential below by an arbitrarily high power in its power series. It follows that \(F(z)=o(|z|)\) and both first partial derivatives tend to zero. Thus \(DF(0)=0\), the partial derivatives extend continuously by zero, and \(F\in C^1(D)\).
3. The prescribed zeros have exact quadratic order, and \(F\) is not holomorphic.
At \(a=1/n\), \(n\geq2\), the holomorphic function \(s(z)=\sin(\pi/z)\) has value zero and derivative \[ s'(a)=-\pi n^2\cos(\pi n)=-\pi n^2(-1)^n\ne0. \] Locally write \(s(z)=(z-a)h_a(z)\) with \(h_a\) holomorphic and \(h_a(a)=s'(a)\ne0\). Then \[ F(z)=(z-a)^2 u_a(z),\qquad u_a(z)=e^{-1/|z|^2}h_a(z)^2, \] where \(u_a\) is smooth near \(a\) and \(u_a(a)\ne0\). Consequently \(F(a)=0\), \(DF(a)=0\), and \(F(z)/(z-a)^2\to u_a(a)\ne0\).
Finally \(F(i/2)\ne0\), since \(\sin(-2\pi i)=-i\sinh(2\pi)\ne0\). If \(F\) were holomorphic on \(D\), the zeros accumulating at zero would force \(F\equiv0\) by step 1, contradicting this value. Thus it is the required nonholomorphic example.