- Describe all holomorphic functions on \(\mathbb { C } \backslash \{ 0 \}\) with the property that
\[ | f ( z ) | \leq | z | ^ { 2 } + { \frac { 1 } { | z | ^ { { \frac { 1 } { 2 } } } } } , \quad \forall z \in \mathbb { C } \backslash \{ 0 \} . \]
Solution
Every solution is the restriction of a polynomial \(p(z)=az^2+bz+c\) whose coefficients satisfy the criterion below. The constant polynomial \(3\), for example, violates the bound at \(|z|=1\).
For given \(a,b,c\in\mathbb C\), define real polynomials \[ \begin{aligned} A(x,y)&=\operatorname{Re}(a\overline b(x+iy)),\\ B(x,y)&=\operatorname{Re}(b\overline c(x+iy)),\\ C(x,y)&=\operatorname{Re}(a\overline c(x+iy)^2),\\ Q(t,x,y)&=(1-|a|^2)t^{10}-2A(x,y)t^8 -(|b|^2+2C(x,y))t^6\\ &\hspace{12mm}+2t^5-2B(x,y)t^4-|c|^2t^2+1. \end{aligned} \] Here real and imaginary parts of \(a,b,c\) are parameters; \(t,x,y\) are real variables. The admissible triples \((a,b,c)\) are exactly those with:
- \(|a|=1\) and \(b=c=0\); or
- \(|a|<1\) and the following polynomial system has no real solution: \[ \begin{gathered} t>0,\qquad x^2+y^2=1,\qquad Q(t,x,y)<0,\\ Q_t(t,x,y)=0,\qquad xQ_y(t,x,y)-yQ_x(t,x,y)=0. \end{gathered} \]
Step 4 gives the criterion in closed form when \(b=0\) or \(c=0\).
1. The singularity at zero is removable and the extension has degree at most two.
The bound gives \[ |zf(z)|\leq |z|^3+|z|^{1/2}\longrightarrow0 \quad(z\to0). \] The holomorphic function \(zf(z)\) therefore extends across zero with value zero by the removable singularity theorem [@SS03]. Its Taylor series is divisible by \(z\), so division by \(z\) extends \(f\) itself holomorphically. Denote the resulting entire function by \(p\).
For its Taylor coefficient \(d_n\) at zero, Cauchy’s estimate on a circle of radius \(R\) gives \[ |d_n|\leq R^{2-n}+R^{-n-1/2} \] [@SS03]. For every integer \(n\geq3\), let \(R\to\infty\) to obtain \(d_n=0\). Hence \(p(z)=az^2+bz+c\).
2. The leading coefficient satisfies \(|a|\leq1\), with equality only for \(p(z)=az^2\).
Divide the original bound on the positive real ray by \(R^2\) and let \(R\to\infty\). This gives \(|a|\leq1\). Suppose \(|a|=1\) and \(b\ne0\). Choose a unit complex number \(w\) so that \(aw^2\) and \(bw\) have the same argument. Then for all \(R>0\), \[ |p(Rw)|\geq |aR^2w^2+bRw|-|c| =R^2+|b|R-|c|. \] The required bound would imply \(|b|R-|c|\leq R^{-1/2}\), impossible for large \(R\). Thus \(b=0\). If \(c\ne0\), choose \(w\) so that \(aw^2\) and \(c\) have the same argument. Then \(|p(Rw)|=R^2+|c|\), forcing \(|c|\leq R^{-1/2}\) for every \(R\), again impossible. Therefore \(c=0\). Conversely, \(|a|=1\) and \(b=c=0\) plainly satisfy the bound.
3. For \(|a|<1\), the stated root-exclusion criterion is necessary and sufficient.
Every nonzero complex number has the form \(z=t^2w\), where \(t>0\) and \(w=x+iy\) has modulus one. Both sides of the desired inequality are nonnegative, so squaring and multiplying by \(t^2\) is reversible. It gives \[ |at^5w^2+bt^3w+ct|^2\leq(t^5+1)^2. \] Expanding the two sides shows that this is precisely \(Q(t,x,y)\geq0\) on \(t>0\), \(x^2+y^2=1\).
If the polynomial system has a solution, its first three conditions give a point where the bound fails. For the converse, suppose \(Q\) is negative at some point of this cylinder. As \(t\downarrow0\), \(Q\to1\) uniformly on the circle. As \(t\to\infty\), \(Q\to+\infty\) uniformly on the circle: its leading coefficient is \(1-|a|^2>0\) and all other powers have smaller degree with coefficients bounded on that compact circle. Therefore \(Q\) has a negative global minimum on the cylinder, attained at some \(t>0\) and some point of the circle.
At this minimum the radial derivative is \(Q_t=0\). Parametrizing the circle by \((x,y)=(\cos\theta,\sin\theta)\) gives angular derivative \(-yQ_x+xQ_y=0\). Thus the minimizing point satisfies every condition in the displayed polynomial system. This proves the converse and the criterion.
4. When one lower coefficient vanishes, the criterion has a simple explicit form.
Write \(\delta=1-|a|>0\). If \(b=0\), the maximum of \(|az^2+c|\) on \(|z|=R\) is \(|a|R^2+|c|\): the phases can be aligned by choosing the argument of \(z\). Hence the condition is exactly \[ |c|\leq\inf_{R>0}(\delta R^2+R^{-1/2}) =\frac5{4^{4/5}}\delta^{1/5}. \] The derivative \(2\delta R-\tfrac12R^{-3/2}\) changes sign from negative to positive at \(R=(4\delta)^{-2/5}\), which gives the displayed minimum.
If \(c=0\), the circular maximum is \(|a|R^2+|b|R\), so the condition is exactly \[ |b|\leq\inf_{R>0}(\delta R+R^{-3/2}) =\frac5{2^{2/5}3^{3/5}}\delta^{3/5}. \] Here the unique minimizing radius is \(R=(3/(2\delta))^{2/5}\), obtained by setting \(\delta-\tfrac32R^{-5/2}=0\). For \(|a|=1\), step 2 supplies the boundary case. In particular, with \(a=b=0\), a constant polynomial \(c\) is a solution exactly when \(|c|\leq5/4^{4/5}\).