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Let $f(z)$ be an entire function such that for some constant $K$ , $|f(z)| ≤ K$ $|z|^3<$ for $|z|\ge1$ and $f(z) = f(iz)$ $∀z∈C$ , then which of the following are correct ?

(A)$ f(z) = Kz^3$ $∀ z∈\Bbb{C}$
(B) $f(z)$ is a constant function
(C) $f(z)$ is quadratic function
(D) No such $f$ exists

My attempt : I can deduce that since $f(z) = f(iz)$ and since $i^4=1 \implies f$ must involve only fourth powers of $z$ . Hence $f(z)=a_1z^4+b_1z^8+c_1z^{16}+ \cdots $-------(1)

Also for $|z| \ge 1 $: let C.

Then $|g(z)| = |f(z)|/|z^3| \le k$
=> By the Cauchy estimate theorem we can prove that : $f(z)/z^3$ is a constant function $= c $

$\implies f(z)=cz^3$ for $|z| > 1$ ...........(2)

From (1) and (2) ; we get there does not exist a function like this ?

My textbook answer says it's a constant function ?

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MathMan
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1 Answers1

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We cannot immediately use Cauchy's estimates because the theorem requires a bound on $D(0, R)$ and our bound is on $|z| \ge 1$. But we can fix that. Since $f$ is entire (in particular, continuous) and the closed unit disk is compact, there exists $M > 0$ so that $f(z) \le M$ for all $|z| \le 1$. Putting both bounds together, we have: $$ \left|f(z)\right| \le K|z|^3 + M \quad \forall z \in \Bbb C $$

Now we can apply Cauchy's estimates as in this answer and conclude that $f$ is a polynomial of degree $3$ at most.

We also have the fact that $f(z) = f(i z)$ for all $z \in \Bbb C$. This forces $f$ to be of the form: $$ f(z) = \sum_{n=0}^\infty a_n z^{4n} $$

Notice that the series may contain a constant term.

Putting both results together, we conclude that $f$ must be constant.

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Ayman Hourieh
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  • got it . I got one more query. I reckon one of my profs just giving us a hint on this one as , since f(z) = f(iz) => every z has the same functional value as that of z rotated by 90 degrees. From there he concluded that this could lead to a deduction of f being a constant function ? Can anything on similar lines be thought of without writing all the above equations ? Wondering if it can be that easy .

    Thanks.

     – MathMan May 26 '13 at 20:13
  • @VishakHPillai The condition $f(z) = f(iz)$ alone isn't sufficient to conclude that $f$ is constant, as demonstrated by $f(z) = z^4$. Are you sure you aren't missing something? – Ayman Hourieh May 26 '13 at 20:15
  • uhmm,yeah, true. I might be missing something, I just reckon a very quick hint given by him which did not mention of any calculation as such. He said it can be proved just by theory . I am thinking about it . – MathMan May 26 '13 at 20:21