# Which of the following statements are true

Which of the following statements are true?

1. There exists an entire function $f : \mathbb C\rightarrow \mathbb C$ which takes only real values and is such that $f(0) = 0$ and $f(1) = 1$.

2. There exists an entire function $f : \mathbb C\rightarrow \mathbb C$ such that $f(n + {1\over n}) = 0$ for all positive integers $n$.

3. There exists an entire function $f : \mathbb C\rightarrow \mathbb C$ which is onto and which is such that $f({1\over n}) = 0$ for all positive integers $n$.

• What do you mean by $f(n+1n)$? – Gerry Myerson Aug 2 '12 at 4:02
• What have you tried? What relevant facts do you know? (For example, are you familiar with Picard's theorems?) – bradhd Aug 2 '12 at 5:17

1. No such function exists. By Little Picard's Theorem (see wikipedia), if $f$ is entire and non-constant, then $f$ can miss at most one value.

2. Here I am assuming you want an entire function with zeros at $n + \frac{1}{n}$. Since $n + \frac{1}{n}$ does not have a limit point (goes to $\infty$), you can use the Hadamard product formula to produce an entire function with exactly those values as it zeros (and you can even control its order of growth).

3. No such function exists. Since $(\frac{1}{n})_{n \in \omega}$ has a limit point in $\mathbb{C}$, by analytic continuation, this function must be the constant zero function.

• Picard's theorem is overkill for this one. It's easy to prove that a real-valued analytic function is constant. – Robert Israel Aug 2 '12 at 6:55
• To follow up on Robert's comment: just look at the Cauchy-Riemann equations. – t.b. Aug 2 '12 at 7:23
• @RobertIsrael sir in the third part, function values are zero at countably many points only. why can't other points approach every point in the complex plane( may or may not have an exception) and form an entire function? Please explain. Thanks in advance. – Unknown x Oct 15 '17 at 14:23
• @N.Maneesh An entire function is analytic everywhere. The fact that zeros of $f$ have a limit at $0$ is enough to show that $f$ is identically $0$ in a neighbourhood of $0$, and therefore everywhere. – Robert Israel Oct 15 '17 at 15:31
• @RobertIsrael Thank you for your help.May I know "The fact that zeros of $f$ have a limit at 0 is enough to show that $f$ is identically 0 in a neighbourhood of 0, and therefore everywhere. ." How does this statement come true?. In the case of a linear function, it is easy to prove. May I know the proof in this case. Thanks you in advance. – Unknown x Oct 15 '17 at 16:12

$1$ is false: Non-constant entire function takes every value in $\mathbb C$ with one possible exception.

$2$ is true: $f=0$.

$3$ is false: Here $f$ is a non-constant entire function & so $f$ takes every value in $\mathbb C$ with one possible exception. {$1\over n$} is a sequence of distinct points converging to $0\implies f=0$ $!$

• I like the proof for 2) :D – CBenni Dec 16 '12 at 12:00
• Beginners always try to go for trivial examples :) – Sugata Adhya Dec 16 '12 at 12:03