Show that if $f$ is a non-constant entire function,it cannot satisfy the condition: $$f(z)=f(z+1)=f(z+i)$$  My line of argument so far is based on Liouville's theorem that states that every bounded entire function must be a constant.  So I try-to no avail-to show that if $f$ satisfies the given condition, it must be bounded. I haven't made much progress with this, so any hints or solutions are welcome.
826
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
97116
I came across this old exam problem.     Suppose $f(z)$ is entire and $|f(z)|&lt; e^{|z|}$, and also $f(z)=f(z+1)$. Show $f(z)$ is a constant.   I am able to show the singularity at infinity is not a pole.   But I can't rule out it being an essential singularity.
21472
Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.  Any hint/idea for proving that $f$ has to be constant?
6711
Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic
77964
Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b&gt;1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.
101791
Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$  I have no idea how to solve this one!  $g(z)={f(z)\over \log|z|}$ Then Can I say $g$ is constant by Liouville Theorem?
85733
Let $f$ be an entire function such that $f\circ f$ has no fixed points. Prove that $f$ is a translation   $$z\mapsto f(z)=z+b \qquad (b\neq 0)$$   Firstly, we prove that there exists a constant $c\in \mathbb{C}\backslash \{0,1\}$ such that  $$f(f(z))-z=c(f(z)-z) $$ applying Picard's little theorem. If $c=0$, then $f(f(z))=z$, so $f\circ f$ has a fixed point (absurd). If $c=1$, then $f(f(z))=f(z)$, so $f$ is the identity $f(z)=z$ and of course it has fixed point (absurd). Then,  $$F(z)=\frac{f(f(z))-z}{(f(z)-z)}$$ is an entire function which does not take the values 0 and 1 so, by Picard's little theorem, it must be constant.  Also, I've proved that $f'\circ f$ is a constant function. Let's see this. Differentiating  $$f(f(z))-z=c(f(z)-z) $$ we have $$f'(z)f'(f(z))-1=cf'(z)-c$$ $$f'(z)[f'(f(z))-c]=1-c$$ Again, the entire function  $$G(z)=f'(z)[f'(f(z))-c]$$ does not take the values $0$ and $1$ so, by Picard's Little Theorem, then is constant.   However, I don't know how to prove this problem. Any help would be appreciate.
109651
Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated.
10437
So the question asks me to prove that an entire function with positive real parts is constant, and I was thinking that this might somehow be related to showing an entire bounded function is constant (Liouville's theorem), but are there any other theorems that might help me prove this fact?
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
118113
Let $f$ a entire function such that there exists $3$ real numbers not   all zero $a$, $b$ and $c$ which $a \Re(f(z))+ b\Im(f(z)) \leq c$, $z \in \mathbb{C}$. Show that $f$ is constant.   I think we could consider the entire function $e^{\alpha f(z)}$ with $\alpha=a+bi$. I know that $|e^{\alpha f(z)}|= |e^{\Re({\alpha f(z)})+i \Im({\alpha f(z)})}|=e^{\Re({\alpha f(z)})}$, but I don't know how to finish it.  Is anyone could help at this point? Maybe we have to use Liouville theorem.
80576
$f(z)$ is a function such that $|f(z)| &gt; |z|$ for all $z\in\mathbb{C}$  How would you prove the function $\frac{1}{f(z)}$ is bounded. I think you have to use the $|f(z)|\leq M$ but I'm not sure.
86441
The question is  Let $f $ be analytical and not constant in $\{z\in\mathbb C:|z|\leq 1\}$. If there is some $z_0\in \{|z|\leq 1\}$ such that $$|f(z_0)|=\max_{|z|\leq 1} |f(z)|\,,$$ then there must be $|z_0|=1 $ and $f’(z_0)\neq 0$.  The first part is just the maximum principle, but for the derivative, I have no idea how solve it. Looking forward to some hints or answers, thank you.
102668
Suppose that $f$ is an entire function and that there is a nonempty open ball $B(w_0,r)$ such that $w\notin \mathrm{Range}(f)$ for all $w\in B(w_0,r)$. Show that $f$ is constant.    I'm not sure how to start this. This looks like maybe something to do with Liouville's Theorem, but this is under a topic where we try not to apply other theorems, so I'm not sure if I can use it here if it's even possible.
60820
Is there a characterization of the nonconstant entire functions $f$ that satisfy $|f(z)|=1$ for all $|z|=1$?   Clearly, $f(z)=z^n$ works for all $n$. Also, it's not difficult to show that if $f$ is such an entire function, then $f$ must vanish somewhere inside the unit disk. What else can be said about those functions?  Thank you.
115504
Problem: Suppose that $f$ is a bounded, real-valued function on $[a,b]$ such that $f^2\in R$ (i.e. it is Riemann-Integrable). Must it be the case that $f\in R$ ?      Thoughts: I think that this is not necessarily true, but I am having trouble refuting or even proving the above. Of course, the simplest way to prove that it is not necessarily true would be to give an example, but I am unable to think of one! I also have tried using $\phi(y)=\sqrt y$ and composing this with $f^2$ (to try show $f$ is continuous); however, the interval $[a,b]$ may contain negative numbers so I can't utilise $\phi$ in that case.  Question: Does there exist a function $f$ such that $f^2\in R$ but $f$ $\not\in R$ ? Or conversely, if $f^2\in R$ does this always imply $f$ $\in R$ ? (If so, could you provide a way of proving this).
86893
Let $f: \mathbb C \to \mathbb C$ be an analytic function, such that for all $z\in\mathbb C$, $$| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.$$   I have to show that $f$ is constant.   I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated.