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

Question

anonymous · Answer

If you're looking for a proof here, I *think* you can try this by contradiction. Assume $f(z)$ is non-constant, perhaps by saying it's monotonically increasing or decreasing for some interval. Is it possible then for $f(z)=f(2z)$ to be true? Just an idea, I don't know if it'd take you anywhere.

anonymous · Answer

I should take that back. I'm not even sure monotonicity applies to complex-valued functions...

kirbykirby · Answer

Yeah I'm not too sure about that either

kinggeorge · Answer

If you had the fact that $f$ was holomorphic, I might be able to do something, but I don't see anything (right now) if it's just continuous.

kinggeorge · Answer

I think I have an idea. Since it's continuous everywhere, it's continuous at 0. So let $\{z_i\}_{i=1}^\infty$ be a sequence of complex numbers approaching 0 along a straight line so that $z_i=2z_{i+1}$. So since $f$ is continuous, $$\lim_{n\to\infty}f(z_n)=f(0).$$But from the way we defined our sequence, $\lim_{n\to\infty}f(z_n)=f(z_1)$. So $f(z_1)=f(0)$. Now, notice that we can choose $z_1$ arbitrarily in $\mathbb{C}$ and define the same sequence, and it will always approach $0$. But that means that $f(0)=f(z)$ for all $z\in\mathbb{C}$, so the function must be constant.

kirbykirby · Answer

how do you say that $\lim_{n\rightarrow \infty} f(z_n)=f(z_1)$

kinggeorge · Answer

That's from the condition that $f(z)=f(2z)$. The way we defined the sequence, we know that $f(z_i)=f(z_j)$ for all $i,j\in\mathbb{Z}^+$. So if we say $f(z_1)=L$, then we could rewrite the limit as $\lim_{n\to\infty}L$. But $L$ is a constant, so the limit is clearly equal to $L=f(z_1)=f(0)$.

kirbykirby · Answer

Gosh I feel really dumb but I'm not sure I quite understand :( 

Maybe because I have trouble understand how complex sequences work? Honestly our prof spent like 30 mins on that... can we imagine that they work similarly as in real-valued functions?

Also, if you constructed this particular sequences, how does it imply it's true in all cases?

kinggeorge · Answer

Convergent sequences in the complex numbers are slightly different from sequence in the real numbers, but the particular sequence I constructed is almost exactly like a sequence in the real numbers since every point in the sequence lies on the same line.

As for continuous functions, I'm using the definition: A function $f$ is continuous at a point $c$ if for any sequence $\{x_1\}_{n=1}^\infty$ that converges to $c$, then the sequence $\{f(x_n)\}_{n=1}^\infty$ converges to $f(c)$ (so the limit as $n\to\infty$ of $f(x_n)$ will be $f(c)$.

It's true in all cases, since if I choose ANY $z_1\in\mathbb{C}$, then define the sequence $z_{i+1}=z_i/2$, the sequence $\{z_i\}_{i=1}^\infty$ will approach the origin. Since the function is continuous, we use the above definition to see that $f(z_i)$ approaches $f(0)$ as $i\to\infty$. Finally, from the condition that $f(z)=f(2z)$, or equivalently $f(z)=f(z/2)$, we notice that $f(z_i)$ is constant for all $i\in\mathbb{Z}^+$. So the sequence must approach that same constant. But we already know it approaches $f(0)$, so $f(0)$ must be that constant. So $f(0)=f(z_1)$ for all $z_1\in\mathbb{C}$, so $f$ must be constant.

kirbykirby · Answer

ohhh oh my I had to read this MANY times but it has finally sunk in! 
I get it now. Thank you sooooo much!!!

kinggeorge · Answer

You're welcome. Out of curiosity, what is this for? Complex analysis? Real analysis? Or something different?

kirbykirby · Answer

It is for complex analysis

kinggeorge · Answer

Nice. You'll be surprised at how much nicer things work in the complex plane rather than the real line.

kirbykirby · Answer

Hehe good to know