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  1. $f: \mathbb{C} \rightarrow \mathbb{C}$ be a continuous function and assume that $f(z) = f(2z)$ for all $z \in \mathbb{C}$. Prove that f is constant...

Then we are supposed to use this result to solve the second question which is ...

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be analytic throughout $\mathbb{C}$ and satisfy $f(2z) = 2f(z)$ for all z. Prove that there exists $c \in \mathbb{C}$ such that $f(z) = cz$ for all z.

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  • $\begingroup$ do you mean $f$ is constant when you say $Z$ is constant? what have you tried? $\endgroup$ – user87543 Feb 7 '14 at 2:04
  • $\begingroup$ ya sorry i changed that typo, I think the second part has something to do with cauchy-reimann equations, since thats what we've been doing in class. But in part 1 the function is just continuous, not analytic. So I'm not really sure. $\endgroup$ – user2253455 Feb 7 '14 at 2:10
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Note that $f(z) = f({1 \over 2^n} z)$. Letting $n \to \infty$ gives $f(z) = f(0)$.

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  • $\begingroup$ Could you please explain to me how is it possible to take $f(z)=f(\frac{z}{2^n})$? $\endgroup$ – user114873 Apr 28 '17 at 9:31
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    $\begingroup$ We are given that $f(z) = f ({1 \over 2} z )$, repeating gives $f(z) = f ({1 \over 2} z ) = f ({1 \over 2^2} z ) = f ({1 \over 2^3} z ) = \cdots$. $\endgroup$ – copper.hat Apr 28 '17 at 12:45
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compute $$ f'(z)=\frac{1}{2}f'(2z)2=f'(2z) $$ apply the answer to the first question to conclude $f'(z)=C$ for all $z$, so $f(z)=Cz+D$ and $$ D=f(0)=2f(0)=2D \Rightarrow D=0 $$

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Working backward... $$ f(z) = f(z/2) = f(z/4) = f(z/8) \cdots$$

Let $\epsilon >0$ be given. Then there is a $delta >0$ such that $|z|<\delta$ implies $|f(z)-f(0)|<\epsilon$. So for any we can chose an $n$ such that $z/2^n$ is as small as we please, so $|f(z)-f(0)|$ is less than any $\epsilon>0$. Hence

$$ f(z) \equiv f(0) $$

For part (2), it is easy to see that $f(0)=2f(0)$ and hence $f(0)=0$. Let $g(z)=f(z)/z$. This has a removable singularity at $z=0$, so we can assume that $g$ is continuous.

$$g(2z)= f(2z)/(2z) = (2 f(z))/(2z) = f(z)/z= g(z)$$

Hence $g(z) = c$ a constant, and $f(z) = cz$

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    $\begingroup$ I do not really understand why would most of the users post full solutions without missing any details when the OP has not showed any of his work.... you could have given some hint... It is your personal opinion anyways.... $\endgroup$ – user87543 Feb 7 '14 at 2:14
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    $\begingroup$ Point taken. Here is the condition under which I try to give full answer: The OP is new, and OP has not accepted or voted on an answer and the question is more than 10 minutes old. Under these conditions, I feel that the OP should be encouraged and not feel unwelcome. Many first time users of the site do feel intimidated. $\endgroup$ – user44197 Feb 7 '14 at 2:22
  • $\begingroup$ don't worry, the full solution doesn't make any sense to me either... i would have preferred some form of hint. $\endgroup$ – user2253455 Feb 7 '14 at 2:23
  • $\begingroup$ Could you please explain to me how is it possible to take $f(z)=f(\frac{z}{2^n})?$ $\endgroup$ – user114873 Apr 28 '17 at 9:32

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