Determine all entire functions \(f\) that satisfy
\[ |f(z)|\leq |\operatorname{Re}z|^2+|z|^{3/2}\qquad\text{whenever }|z|>1. \]
Solution
The solutions are precisely the affine functions \(f(z)=az+b\) whose coefficients satisfy \[ |aw+b|\leq1+(\operatorname{Re}w)^2\qquad(|w|=1). \tag{1} \] This compact-circle condition has the following finite algebraic form. For \(a,b\in\mathbb C\), define the real polynomial in the real variable \(t\) \[ Q_{a,b}(t)=4(1+t^4)^2-(1+t^2)^2 \left|(a+b)+(b-a)t^2+2iat\right|^2. \tag{2} \] Here the squared modulus means multiplication by the coefficientwise conjugate polynomial, since \(t\) is real. The exact coefficient test is \[ |a|^2+|b|^2\leq1, \qquad Q_{a,b}(t)\geq0\text{ at every real root of }Q_{a,b}'(t). \tag{3} \] Under the first inequality, \(Q_{a,b}'\) has degree seven, so this requires checking at most seven real numbers. Thus (3) is a necessary and sufficient condition on the coefficients \(a,b\).
1. Every solution is affine.
On \(|z|=R>1\) the hypothesis gives \(|f(z)|\leq R^2+R^{3/2}\). If \(c_k\) is the \(k\)th Taylor coefficient of \(f\) at zero, Cauchy’s estimate gives \[ |c_k|\leq R^{2-k}+R^{3/2-k} \] [@SS03]. For \(k\geq3\), let \(R\to\infty\) to obtain \(c_k=0\). Hence \(f(z)=cz^2+az+b\). On the imaginary ray the given bound is \(|f(iR)|\leq R^{3/2}\). Divide by \(R^2\) and let \(R\to\infty\). The expression \(f(iR)/R^2=-c+ia/R+b/R^2\) tends to \(-c\), whereas its modulus is at most \(R^{-1/2}\). Therefore \(c=0\).
2. For affine functions, (1) is equivalent to the original inequality.
Necessity follows by setting \(z=rw\), \(|w|=1\), and letting \(r\downarrow1\) in the original inequality. Conversely assume (1). At \(w=i\) and \(w=-i\) it gives \(|ai+b|\leq1\) and \(|-ai+b|\leq1\). Adding the squares yields \[ 2(|a|^2+|b|^2)=|ai+b|^2+|-ai+b|^2\leq2, \] so in particular \(|a|\leq1\). For any \(r>1\) and \(|w|=1\), put \(x=\operatorname{Re}w\). Using a convex combination and then (1), \[ \begin{aligned} |arw+b|/r &=\left|(1-r^{-1})aw+r^{-1}(aw+b)\right|\\ &\leq (1-r^{-1})|a|+r^{-1}|aw+b|\\ &\leq1+r^{-1}x^2. \end{aligned} \] Thus \(|arw+b|\leq r+x^2\leq r^{3/2}+r^2x^2\) since \(r>1\). This last expression is precisely \(|rw|^{3/2}+|\operatorname{Re}(rw)|^2\), proving sufficiency.
3. The finite coefficient test (3) is equivalent to (1).
Parametrize the unit circle except \(-1\) by \[ w(t)=\frac{1-t^2+2it}{1+t^2},\qquad t\in\mathbb R. \] Then \[ aw(t)+b=\frac{(a+b)+(b-a)t^2+2iat}{1+t^2}, \qquad 1+(\operatorname{Re}w(t))^2=\frac{2(1+t^4)}{(1+t^2)^2}. \] Squaring (1) and multiplying by the strictly positive \((1+t^2)^4\) gives exactly \(Q_{a,b}(t)\geq0\). These operations are reversible because both sides of (1) are nonnegative. Continuity supplies (1) also at \(w=-1\) by letting \(|t|\to\infty\).
Step 2 already shows that (1) implies \(|a|^2+|b|^2\leq1\). Under this inequality the leading coefficient of \(Q_{a,b}\) is \[ 4-|b-a|^2\geq4-2(|a|^2+|b|^2)\geq2. \] Hence \(Q_{a,b}\) has degree eight and tends to positive infinity as \(t\to\pm\infty\). It attains a global minimum on the real line; differentiability makes each such minimizer a root of \(Q_{a,b}'\). Therefore \(Q_{a,b}\) is nonnegative everywhere exactly when it is nonnegative at every real critical point. Its degree-seven derivative has at most seven distinct real roots. This proves (3) and completes the classification.