1
$\begingroup$

I am having trouble constructing a function with the following property: Define the function $f_a: B(0,1)\to\mathbb{C}$ by $f_a(z) = z+a(1-z)^3$, where $a>0$. Is it possible to choose $a$ such that $|f_a(z)| < 1$ for all $z$?

What I have tried so far is:

1) Let $f(z) = z$ and define $f_n(z) = z+\frac{1}{n}(1-z)^3$ and note $f_n \to f$ uniformly. This seemed to be a dead end, since this only gives $|f_n(z)-z| < \varepsilon$ for sufficiently large $n$.

2) For $a>0$ apply the fundamental theorem of algebra to $f_a$ to obtain $$ f_a(z) = -a(z-r_1)(z-r_2)(z-r_3) $$ but I take the roots of $f_a$ depend on $a$, so I cannot begin to pick $a$ small without changing the roots.

I would very much appreciate a hint or a solution. Thanks!

  • 0
    $a$ must be fixed, no? – 2017-01-18
  • 0
    mh, I see my explanation might not be quite clear. I am looking at the class of functions $f_a(z) = z+a(1-z)^3$ with $a\in\mathbb{R}_+$ and I want to show there is an $a$, such that $|f_a(z)|<1$ for all $z\in B(0,1)$. – 2017-01-18
  • 0
    The point is I don't know what $a$ is - I want to find an $a$ (or show that one exists) with the property that $|f_a(z)| < 1$. – 2017-01-18
  • 0
    As far as I know, for $|a|<\dfrac{1}{4}$ we have $f(\mathbb{D})\subset\mathbb{D}$. – 2017-01-18
  • 0
    Can you point me in the direction of a proof? – 2017-01-18
  • 0
    For $a = -\frac{1}{8}$, you have $f(-1) = -1 - \frac{1}{8}(2^3) = -1 -1 = -2.$ So your conjecture about $|a| < 0.25$ must be wrong. In fact, by looking at $f(-1)$, you can see that $a$ must be positive. (I'm assuming that $a$ is real here). – 2017-01-18
  • 0
    BTW, I suspect that the Schwarz Lemma (or its proof) might be relevant here, but this is a completely wild guess – 2017-01-18
  • 0
    @JohnHughes Thanks for the nice suggestion! I have tried applying Schwarz Lemma to $g_a(z) = f_a(z)-a$, since this satisfies $g_a(0) = 0$, but this only gives $|f_a(z)-a| < |z|$, which is only close to being the desired. I don't really see how I can modify the proof to my needs. – 2017-01-18
  • 0
    But $f$ not necessarily univalent to apply Koebe theorem! – 2017-01-18
  • 0
    Any $a$ such that $0 < a < 1/6$ works. – 2017-01-18
  • 0
    @quasi Can you point me in the direction of a proof? – 2017-01-18
  • 0
    @Mads Friis -- I'm working on writing it up -- may take a while. If it survives the write-up, I'll post it. – 2017-01-18
  • 0
    @Mads Friis -- Unless I made a mistake, a futher restriction on $a$ is needed, namely $a_1 < a < 1/6$ where $a_1$ is irrational but is approximately 0.0173705485. As far as posting my solution, I may hold off, since the way I did it is very messy, and most likely, someone else, using a more sophisticated method, will probably be able to do it with less effort. – 2017-01-18
  • 0
    @Mads Friis -- Yes, I made a mistake about $a_1$. My original claim $0 < a < 1/6$ appears correct, and is further supported by Kelenner's solution. – 2017-01-18

1 Answers 1

2

A second try:

By the maximum modulus theorem, for any $z$ such that $|z|<1$, we have $|f(z)|\leq {\rm Max}_{\theta}|f(\exp(i\theta))|$ Using $1-\exp(i\theta)=-2i\sin(\theta/2)\exp(i\theta/2)$, and factorizing by $\exp(3i\theta/2)$, we get

$$|f(\exp(i\theta))|^2=|\exp(-i\theta/2)+i8a\sin(\theta)/2)^3|^2$$ and this is $1-16a(\sin(\theta/2))^4+64a^2(\sin(\theta/2))^6$.

Now put $t=\sin(\theta/2)^2\in [0,1]$, and $g(t)=1-16at^2+64a^2t^3$; On $[0,1]$, and for $0

  • 0
    Thanks, I just went through your calculations and I very much agree! – 2017-01-19