Cauchy estimates and Liouville

Since \(f\) has no zeros on \(\overline{\mathbb D}\), the function \(1/f\) is holomorphic on \(\mathbb D\) and continuous on \(\overline{\mathbb D}\). On the unit circle, \[ |f|=1, \qquad |1/f|=1. \] The maximum modulus principle applied to \(f\) gives \(|f(z)|\le1\) in \(\mathbb D\), while applied to \(1/f\) it gives \(|f(z)|\ge1\). Hence \[ |f(z)|=1 \qquad(z\in\mathbb D). \] A nonconstant holomorphic function is an open map, but the unit circle has empty interior. Therefore \(f\) must be constant.

Write \(f(z)=\sum_{k\geq0}c_kz^k\), and choose \(C\) and \(R_0\) with \(\abs{f(\xi)}\le C\abs\xi^p\) for \(\abs\xi\ge R_0\). For \(R\ge R_0\), Cauchy’s integral formula for \(c_k=f^{(k)}(0)/k!\) gives \[ \abs{c_k} &\leq {1\over 2\pi}\int_{\abs{\xi} = R} {\abs{f(\xi)} \over \abs{\xi}^{k+1}}\abs{\dxi}\\ &\leq {1 \over 2\pi}\int_{\abs \xi = R}{ C\abs{\xi}^{p} \over \abs{\xi}^{k+1} }\abs{\dxi} \\ &= {C \over 2\pi} {1\over R^{k+1-p}} \cdot 2\pi R \\ &= C R^{p-k} ,\] which converges to \(0\) as \(R\to \infty\) provided that \(k>p\). So every coefficient \(c_k\) with \(k> p\), that is, \(k\geq \floor{p}+1\), vanishes, and \(f\) is a polynomial of degree at most \(\floor p\).

Let \(\omega_1,\omega_2\) be two real-linearly independent periods of the entire function \(f\). Consider the closed fundamental parallelogram \[ P=\{t_1\omega_1+t_2\omega_2:0\le t_1,t_2\le1\}. \] It is compact, so continuity of \(f\) gives \[ M=\max_{z\in P}|f(z)|<\infty. \]

Every \(z\in\mathbb C\) can be written \[ z=z_0+m\omega_1+n\omega_2 \] with \(m,n\in\mathbb Z\) and \(z_0\in P\). By periodicity, \[ f(z)=f(z_0), \] so \(|f(z)|\le M\) for all \(z\in\mathbb C\). Thus \(f\) is bounded and entire; Liouville’s theorem implies that \(f\) is constant.

Let \(f\) be entire with \(\abs{f(z)}\le M\abs z^n\) for \(\abs z\ge R\). Taylor expand at \(z=0\) to get \(f(z) = \sum_{j\geq 0}c_j z^j\) everywhere. Claim: \(c_{n+k} = 0\) for all \(k\geq 1\). By the formula for Taylor coefficients, it suffices to show \(f^{(n+k)}(0) = 0\) for all \(k\geq 1\). Apply the Cauchy estimate on the circle of radius \(\rho\geq R\): \[ \abs{ f^{(n+k)} (0)} &\leq {(n+k)! \over 2\pi} \int_{\abs{\xi} = \rho} \abs{f(\xi) \over \xi^{n+k+1}}\abs{\dxi}\\ &\leq {(n+k)! \over 2\pi} \int_{\abs{\xi} = \rho} {M\abs{\xi}^n \over \abs{\xi}^{n+k+1}}\abs{\dxi}\\ &= {(n+k)! \over 2\pi} {M\over \rho^{k+1}} \cdot 2\pi \rho \\ &= (n+k)!\,M\rho^{-k} \convergesto{\rho\to\infty} 0 .\] So \(f\) is a polynomial of degree at most \(n\).

If the bound holds on all of \(\CC\setminus\theset0\), then \(h(z) \da f(z)/z^n\) is bounded near \(0\), so its singularity at \(0\) is removable, and \(h\) is a bounded entire function. By Liouville \(h\) is constant, and \(f(z) = cz^n\).

The inequality implies \(f\) has no zeros, so \(g(z) \da 1/f(z)\) is entire. Moreover it is bounded on \(\CC\), since \[ \abs{g(z)} \leq {1\over \abs{z} + 1} \leq 1 ,\] so \(g\equiv c\) is constant by Liouville. Since \(\abs{g(z)}\le 1/(\abs z+1)\to0\) as \(\abs z\to\infty\), \(c=0\), which contradicts \(g=1/f\) having no zeros. So there are no such entire functions.

Apply a Cauchy estimate over a contour of radius \(R> M\) to obtain \[ \abs{f^{(n)}(0)} \leq n!{A+B R^k \over R^n} \asymptotic 1/R^{n-k} ,\] and if \(n>k\) then this goes to zero in \(R\) and \(c_n = 0\) for all \(n>k\).

If \(g\equiv0\), then the inequality forces \(f\equiv0\), so take \(c=0\).

Assume \(g\not\equiv0\) and set \(h=f/g\) away from the zeros of \(g\). There \(\abs{h}\leq1\). If \(a\) is a zero of \(g\), then \(h\) is bounded on a punctured neighborhood of \(a\), so \(a\) is a removable singularity of \(h\). Thus \(h\) extends to an entire function on \(\CC\) with \(\abs h\leq1\) everywhere. By Liouville’s theorem, \(h\equiv c\) for some \(c\in\CC\) with \(\abs c\leq1\). Hence \[ f=cg. \]

Write \(g(z) \da f(z) / \sin(z)\), which is meromorphic with singularities at the zeros of \(\sin(z)\) and bounded by 1. By boundedness, these singularities are removable, so \(g\) extends to a bounded entire and thus constant function. So \(f(z) = c\sin(z)\) where \(\abs{c} \leq 1\).

Write \(g(z) \da f(z) - z_0\), so \(\abs{g(z)} \geq r\). Now \(\abs{1/g(z)} \leq 1/r\) for all \(z\), so \(1/g\) is bounded. Moreover it is entire since \(f(z) \neq z_0\) for any \(z\), and so \(1/g = c\) is constant. Now unwind to get \(f(z) = z_0 + {1\over c}\), which is also constant.

The inequality forces \(f\neq 0\) anywhere, so \(1/f\) is entire and bounded by 1. By Liouville, \(1/f\) is constant, and thus so is \(f\).

Induct on \(k\ge0\). For \(k=0\), \(\abs f\le A+B\) outside a disk and \(f\) is bounded on the disk by continuity, so \(f\) is constant by Liouville.

For \(k\geq1\), assume the statement for \(k-1\) and let \(\abs{f(z)}\le A+B\abs z^k\) for \(\abs z\ge R\). Consider \[ g(z) \da \begin{cases} {f(z) - f(0) \over z} & z\neq 0, \\ f'(0) & z=0, \end{cases} \] which is entire. For \(\abs z\geq\max(R,1)\), \[ \abs{g(z)} \leq {A+\abs{f(0)}+B\abs{z}^k \over \abs{z}} \leq \qty{A+\abs{f(0)}}+B\abs{z}^{k-1} .\] By the induction hypothesis \(g\) is a polynomial of degree at most \(k-1\), so \(f(z)=f(0)+zg(z)\) is a polynomial of degree at most \(k\).

Suppose that \(f(\mathbb C)\) were not dense in \(\mathbb C\). Then there would be a point \(a\in\mathbb C\) and an \(r>0\) such that \[ B(a,r)\cap f(\mathbb C)=\varnothing. \] Thus \[ |f(z)-a|\ge r\qquad\text{for every }z\in\mathbb C. \] The function \[ g(z)=\frac1{f(z)-a} \] is therefore entire and bounded by \(1/r\). By Liouville’s theorem, \(g\) is constant. Hence \(f\) is constant, contradicting the hypothesis.

Therefore every nonempty open disk meets \(f(\mathbb C)\), so \(f(\mathbb C)\) is dense in the plane.