entire function

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November 11th 2008, 07:05 AM
PvtBillPilgrim

entire function

Let f(z) be an entire function and f(z) = 0 on a subinterval a < x < b of the real axis. Show that f(z) = 0 for all z in C (set of complex numbers). Conclude that if f and g are entire functions and that they agree on a segment of the real axis, then they must be the same.

How do I answer this question? Thanks for help in advance.
November 11th 2008, 08:08 AM
shawsend

Quote:

Originally Posted by PvtBillPilgrim

Let f(z) be an entire function and f(z) = 0 on a subinterval a < x < b of the real axis. Show that f(z) = 0 for all z in C (set of complex numbers). Conclude that if f and g are entire functions and that they agree on a segment of the real axis, then they must be the same.

How do I answer this question? Thanks for help in advance.

First go over showing the following theorem:

A non-constant analytic function cannot have a non-isolated set of $k$ -points: Given the set $S=\{z_i\}$ with $f(z_i)=k$ with $\displaystyle \lim_{n\to\infty} z_i=z_p$ , then $f(z)$ must be the constant function $f(z)=k$ .

(I'd need to work on it too a while to show this but it's a very important property of entire functions and I think a very important theorem in Complex Analysis)
November 11th 2008, 09:07 AM
shawsend

Ok, I'd like to try:

Prove that if $f(z)=0$ on a subset of the real axis $S=\{z: a<z<b\}$ , then $f(z)=0$ throughout $\mathbb{C}$ .

Since $f(z)$ is analytic, then we can write for some $c\in S$ :

$\begin{aligned}f(z)&=\sum_{n=0}^{\infty} a_n(z-c)^n \\ &=a_m(z-c)^m+a_{m+1}(z-c)^{m+1}+\cdots \\<br /> &=(z-c)^m\left[a_m+(z-c)Q(z)\right]<br /> \end{aligned}$

where $Q(z)$ is analytic and bounded in some neighborhood of $c$ , that is $|Q|<M$ for $|z-c|<R$ , and $m$ is the first non-zero term. Now:

$|f(z)|=|z-c|^m\left|a_m+(z-c)Q(z)\right|$

so that:

$|f(z)\geq |z-c|^m\left||a_m|-|z-c||Q|\right|$

Now recall that we have a dense set of points over which $f(z)=0$ . Therefore, we can find a $z_i\in S$ such that $|z_i-c|<\frac{|a_m|}{M}$ . This means:

$|z_i-c||Q|<\frac{|a_m|}{M} M<|a_m|$

This would imply $|f(z_i)||\geq |z_i-c||a_m|>0$ since $z_i\ne c$ and $a_m\ne 0$ but by assumption $f(z_i)=0$ . This must mean we've reached a contradiction and all $a_m=0$ so that $f(z)=0$ .

I'll leave it to you to show the function $f-g=0$ on the same domain implies $f=g$ since $h(z)=f(z)-g(z)$ can still be written as $\sum_{n=0}^{\infty} a_n(z-c)^n$ for some (new) set of $a_n$ .

This may need a little more work . . .
November 11th 2008, 11:10 AM
ThePerfectHacker

Quote:

Originally Posted by PvtBillPilgrim

How do I answer this question? Thanks for help in advance.

Here is a stronger statement as shawsend was referring to. Let $f,g$ be entire functions and $\{ z_n\}$ be a convergent sequence of distinct points. If $f(z_n)=g(z_n)$ then it means $f(z) = g(z)$ for all $z\in \mathbb{C}$ . By considering $f-g$ it is sufficient to prove that if $h$ is entire function and $h(z_n) = 0$ then $h(z) = 0$ for all $z\in \mathbb{C}$ . To prove this let $z_0$ be the limit of the sequence $\{ z_n\}$ .

Since the function $h$ is entire it means $h(z) = \sum_{n\geq 0}a_n (z-z_0)^n$ . It thereby suffices to show $a_n = 0 \text{ for }n\geq 0$ . We see that $0=h(z_1) = a_0$ . If we were to define $\xi (z) = \tfrac{f(z)}{(z-z_0)} = \sum_{n\geq 1} a_n (z-z_0)^{n-1}$ with $\xi (z_0) = a_1$ then we see that $\xi (z)$ is continous and therefore $a_1=\xi (z_0)=\lim ~ \frac{ f(z_n)}{z_n} = 0$ . Therefore, $a_1=0$ . From here an induction argument can be used. If $a_j = 0$ for $j < k$ then we can define $\mu (z) = \tfrac{f(z)}{(z-z_0)^k} = \sum_{n\geq k} a_n (z-z_0)^{n-k}$ with $\mu (z_0) = a_k$ is a continous function. It follows that $a_k = \mu (z_0) = \lim \frac{f(z_n)}{z_n^k} = 0$ . And this shows that $a_k = 0$ . Thus, in general, $a_k = 0$ for all $k\geq 0$ . This shows that $h$ must be the zero function.
November 11th 2008, 11:20 AM
Laurent

Quote:

Originally Posted by shawsend

Ok, I'd like to try:

Prove that if $f(z)=0$ on a subset of the real axis $S=\{z: a<z<b\}$ , then $f(z)=0$ throughout $\mathbb{C}$ .

Since $f(z)$ is analytic, then we can write for some $c\in S$ :

$\begin{aligned}f(z)&=\sum_{n=0}^{\infty} a_n(z-c)^n \\ &=a_m(z-c)^m+a_{m+1}(z-c)^{m+1}+\cdots \\<br /> &=(z-c)^m\left[a_m+(z-c)Q(z)\right]<br /> \end{aligned}$

where $Q(z)$ is analytic and bounded in some neighborhood of $c$ , that is $|Q|<M$ for $|z-c|<R$ , and $m$ is the first non-zero term. Now:

$|f(z)|=|z-c|^m\left|a_m+(z-c)Q(z)\right|$

so that:

$|f(z)\geq |z-c|^m\left||a_m|-|z-c||Q|\right|$

Now recall that we have a dense set of points over which $f(z)=0$ . Therefore, we can find a $z_i\in S$ such that $|z_i-c|<\frac{|a_m|}{M}$ . This means:

$|z_i-c||Q|<\frac{|a_m|}{M} M<|a_m|$

This would imply $|f(z_i)||\geq |z_i-c||a_m|>0$ since $z_i\ne c$ and $a_m\ne 0$ but by assumption $f(z_i)=0$ . This must mean we've reached a contradiction and all $a_m=0$ so that $f(z)=0$ .

I'll leave it to you to show the function $f-g=0$ on the same domain implies $f=g$ since $h(z)=f(z)-g(z)$ can still be written as $\sum_{n=0}^{\infty} a_n(z-c)^n$ for some (new) set of $a_n$ .

This may need a little more work . . .

That's a good proof. I knew another way to conclude, which you may like to read:
So we have chosen $c\in S$ and we have written $f(z)=\sum_{n=0}^\infty a_n (z-c)^n$ for any $z\in\mathbb{C}$ . We want to show that $a_n=0$ for all $n.$ It is well-known that $a_n=\frac{f^{(n)}(c)}{n!}$ (derivation term by term, and then $z=c$ ). This derivative is a complex derivative, and since $c$ is real, it coincides with the derivative of $f$ restricted to the real line (same expression as a limit). But $f_{|\mathbb{R}}$ is 0 near $c$ , hence $f^{(n)}(c)=0$ , and subsequently $a_n=0$ .

You used the word "dense", it is a bit improper, and it could be " $c$ is an accumulation point of $S$ ", or " $c$ is not an isolated point in $S$ ", to underline the tight relation with what you said about non-isolated subsets where $f$ is zero. The proof of the fact you quoted is then the very same.
November 12th 2008, 05:32 AM
shawsend

I wish to attempt a summary. Please someone correct if necessary. I struggle with this.

First note the Taylor series for a function is unique. That is, anyway we can determine the Taylor series for the function centered at a point will be the Taylor series. Now, $f(z)$ is real on the real axis in the set $S=\{x: a<x<b\}$ . This mean we can calculate the Taylor series of $f(z)$ by calculating the Taylor series for the real function $f(x)$ over $S$ . but $f(x)$ is zero in that interval which means all the derivatives are zero as well (it's flat). This means all the constants in the real Taylor expansion are zero. But since Taylor series are unique, this Taylor series must be the Taylor series for $f(z)$ as well, thus $f(z)=0$ throughout $\mathbb{C}$ .
November 12th 2008, 05:44 AM
ThePerfectHacker

Quote:

Originally Posted by shawsend

I wish to attempt a summary. Please someone correct if necessary. I struggle with this.

First note the Taylor series for a function is unique. That is, anyway we can determine the Taylor series for the function centered at a point will be the Taylor series. Now, $f(z)$ is real on the real axis in the set $S=\{x: a<x<b\}$ . This mean we can calculate the Taylor series of $f(z)$ by calculating the Taylor series for the real function $f(x)$ over $S$ . but $f(x)$ is zero in that interval which means all the derivatives are zero as well (it's flat). This means all the constants in the real Taylor expansion are zero. But since Taylor series are unique, this Taylor series must be the Taylor series for $f(z)$ as well, thus $f(z)=0$ throughout $\mathbb{C}$ .

That looks nice, it is also a lot nicer than what I did. We can add some rigor into your statement if it is really necessary. Say $c\in (a,b)$ and we wish to prove that $f'(c) = 0$ i.e. $f'$ vanishes on $(a,b)$ . To prove this we approach it by definition, $f'(c) = \lim_{z\to c} \frac{f(z)-f(c)}{z-c}$ . But we know that $f$ is entire so that limit (the derivative) has to exist. If the complex limit exists then it must exist along all possible paths to the limit point $c$ . By choosing the path along the real axis we get that $f'(c) = \lim_{x\to c}\frac{f(x)-f(c)}{x-c} = \lim_{x\to 0}\frac{0}{x-c} = 0$ . Therefore, $f'$ is the zero function. From here it follows that $f^{(n)}$ is a zero function for all $n\geq 1$ .