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Let us assume that the zeros of $f = \{Z_1,\ldots,Z_n,a\}$ are infinite and converge towards $a$.

The book which I am reading says that any neighborhood of $a$ will contain infinite zeros. Since $f$ is continuous, $f$ must be identically zero in this neighborhood. .....(A)

Then as the neighborhood grows, it will be identically $0$ in that neighborhood too until it engulfs the whole of this region. Hence, the function becomes identically zero in the whole region.

I don't quite understand (A). I am confused over (A) with this analogy. Consider $f(z) = z-1$. This has a zero at $z=1$. Since $f$ is continuous in the $z$ plane, this means that as per (A) points in the neighborhood of $z =1$ should also have mapped value $= 0$.

I am confused; where could I be making a mistake? Thanks

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    $\begingroup$ Your exampel is not valid because in the proof it is assumed that $f$ has infinitely many zeros, but your $f$ has only one zero. $\endgroup$ May 12, 2013 at 10:20
  • $\begingroup$ Like fgp I'm trying to find a smart counter-example; I don't see why the continuity of $f$ would imply that it is identically $0$ in a neighborhood of $a$. I think an additional information about the density of zeros is necessary to conclude, infinite doesn't seem sufficient to me yet... $\endgroup$
    – Jonathan H
    May 12, 2013 at 10:51
  • $\begingroup$ yeah ; I find the proof quite vague from this perspective . How do i make a more meaningful proof statement ? $\endgroup$
    – MathMan
    May 12, 2013 at 10:55

3 Answers 3

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The statement is wrong the way you quoted it. Take $f \,:\, \mathbb{R} \to \mathbb{R}$, $$ f(x) = \begin{cases} x\sin \frac{1}{x} &\textrm{if } x > 0 \\ -x &\textrm{if } x \leq 0 \text{.} \end{cases} $$ Since $\lim_{x\to 0^+} f(x) = 0$ (and obviously also $\lim_{x\to 0^-} f(x) = 0$), that function is continuous. It has zeros at $$ \left\{\frac{1}{n\pi} \,:\, n \in \mathbb{N}\right\} \text{,} $$ which obviously have $0$ as a cluster point. Yet $f$ isn't identically zero on any neighbourhood of $0$.

For analytic functions, however, you also have that they are $C^\infty$. Your probably have to use that. Note that that only requiring $C^\infty$ isn't enough. Take for example $$ f(x) = \begin{cases} e^{-\frac{1}{x}} \sin \frac{1}{x} &\textrm{if } x > 0 \\ -x &\textrm{if } x \leq 0 \text{.} \end{cases} $$ That $e^{-\frac{1}{x}}$ should damp things strong enough to ensure the existance of all derivatives of $f$ at zero, i.e. $f$ should be $C^\infty$, yet the theorem still doesn't hold since $f$ still isn't identically zero on a neighbourhood of zero.


So from the above it's clear the we need to use that $f$ is analytic to prove the theorem. We'll wlog assume that $a = 0$ (simply look at $f'(x)=f(x+a)$ if it isn't). Let $(x_n)$ be a sequence of zeros of $f$, with $x_n \to 0$. Since $f$ is analytic at $0$, there's a $r \geq 0$ such that for all $|x| < r$ we have $$ f(x) = \sum_{k=0}^\infty c_k x^k \text{.} $$ Let's further wlog assume that $|x_n| < r$ for all $n$ and that $|x_{n+1}| < |x_n|$ (Simply remove those $x_n$ which violate one of the conditions. Since the sequence converges to 0, infinitely many $x_n$ will remain). From the continuity of $f$ it follows that $f(0)=0$, i.e. that $c_0 = 0$. For $f(x_n)$ to be zero, it must hold that \begin{align} &\sum_{k=0}^\infty c_k x_n^k = 0 \implies \underbrace{c_0}_{=0} + c_1 x_n = -x_n^2\sum_{k=2}^\infty c_k x_n^{k-2} \implies c_1 = -x_n\sum_{k=2}^\infty c_k x_n^{k-2} \\ \implies &|c_1| \leq |x_n|\underbrace{\sum_{k=2}^\infty |c_k| |x_n|^{k-2}}_{:=M_n} \text{.} \end{align} (The last step uses that the series is absolutely convergent, which you have for series expansions of analytic functions)

Now observe that since $|x_{n+1} < x_n|$ you have $M_{n+1} \leq M_n$, and hence for arbitrary $n$ that $|c_1| \leq |x_n|M_n$. But since $x_n \to 0$, it follows that $c_1=0$.

Knowing that $c_1=0$, you can use the same method to show that, in fact $c_2=0$, and from that $c_3=$, and so on. You thus get that all the $c_n$ are zero, and hence that $f$ is identically zero on the ball with radius $r$ around $0$.

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  • $\begingroup$ Sorry, I might be wrong, but I think $f(0^+)$ is not $0$ here. $\endgroup$
    – Jonathan H
    May 12, 2013 at 10:44
  • $\begingroup$ @Sh3ljohn Ah, yeah, I made a typo. Should be $x\sin(1/x)$! Will fix! $\endgroup$
    – fgp
    May 12, 2013 at 10:49
  • $\begingroup$ @Sh3ljohn Better? ;-) $\endgroup$
    – fgp
    May 12, 2013 at 10:50
  • $\begingroup$ :) Yep! Now it's an interesting counter-example! $\endgroup$
    – Jonathan H
    May 12, 2013 at 10:52
  • $\begingroup$ @fgp So, How do i make a more meaningful proof statement . Only if i was able to prove that the inclusion of infinite points has made a difference than if a single point was present ? What is that difference, i am still confused $\endgroup$
    – MathMan
    May 12, 2013 at 11:26
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I think that problem is that continuity is not enough. (Although I can't think of a counter-example off-hand). What you really need is for the function to be holomorphic in some neighbourhood of it's zeroes.

I know of the related result (known as the "principle of isolated zeroes"):

Suppose $f:B_a(r)\rightarrow \mathbb{C}$ is a non-zero holomorphic function on $B_a(r)$with $f(a) = 0$. Then $\exists 0 <\rho<r$ such that $\forall z \in B_a(\rho)\text{ \ }\{a\}, f(z) \not = 0$.

To prove this, we need Taylor's theorem. This says that on $B_a(r)$, $f$ can be written as

$$f(z) =\sum_{n=0}^\infty a_n(z-a)^n$$ If $N$ is the largest integer such that $a_n = 0$ for all $n \leq N$ then we can write $f(z) = (z-a)^Ng(z)$ where $g$ is a holomorphic function and $g(a)$ is not zero. By the continuity of $g$ there is some open ball around $a$ (call it $B_a(\rho)$) where $g$ is not zero, so since $(z-a)^N$ is non-zero in $B_a(\rho)\text{ \ }\{a\}$, $f$ is also non-zero there.

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  • $\begingroup$ Thanks . i got one query though : ( May be i still don't understand the proof enough, thats why ) how do we account for the presence of finite/ infinite zeros of f ? $\endgroup$
    – MathMan
    May 12, 2013 at 18:23
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    $\begingroup$ @VishakHPillai This theorem is true regardless of the number of zeroes that $f$ has. If it has finitely many though, the theorem is much easier to prove because if you take $d$ to be the smallest distance between the zeroes, the open disc of radius $\frac{d}{2}$ centred on any of the zeroes cannot contain any other zero. Is this what you mean? $\endgroup$ May 12, 2013 at 19:42
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    $\begingroup$ WHAT IF THERE IS NO SUCH LARGEST N? I MEAN WHAT IF THE ZERO HAS INFINITE ORDER ? HOW CAN YOU PROVE THIS IN THAT CASE . $\endgroup$ May 30, 2019 at 12:03
  • $\begingroup$ @AaqibIqbal Consider $\mathcal{U}_0=\{z : f^{(i)}(z)=0 \ \forall \ i \geq 0\}$. This is the collection of zeroes of infinite order. This set is closed. Now, try proving that this set is also open. However, our domain is connected, so we end up getting a clopen set that is not $D$ or the $\emptyset$. Hence this is a contradiction to the connectedness of our domain and so $\mathcal{U}_0$ must be empty. $\endgroup$ Dec 8, 2020 at 5:49
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The set of Zeros of an analytic function may be countably infinite. For example $f(z)=\sin\left(\frac1z\right)$ is analytic on $C\setminus\{0\}$. It is true that The set of Zeros of an analytic function at most countably infinite.

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