Identity theorem

 ( Theorems In Real Analysis ) 

In real analysis and complex analysis, branches of mathematics, the identity theorem for analytic functions states: given functions f and g analytic on a domain D (open and connected subset of $\backslash mathbb\{R\}$ or $\backslash mathbb\{C\}$), if f = g on some $S\; \backslash subseteq\; D$, where $S$ has an accumulation point in D, then f = g on D.For real functions, see

Thus an analytic function is completely determined by its values on a single open neighborhood in D, or even a countable subset of D (provided this contains a converging sequence together with its limit). This is not true in general for real-differentiable functions, even smoothness. In comparison, analytic functions are a much more rigid notion.

The underpinning fact from which the theorem is established is the expandability of a holomorphic function into its Taylor series.

The connectedness assumption on the domain D is necessary. For example, if D consists of two disjoint open sets, $f$ can be $0$ on one open set, and $1$ on another, while $g$ is $0$ on one, and $2$ on another.

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Lemma If two holomorphic functions $f$ and $g$ on a domain D agree on a set S which has an accumulation point $c$ in $D$, then $f\; =\; g$ on a disk in $D$ centered at $c$.

To prove this, it is enough to show that $f^\{(n)\}(c)=\; g^\{(n)\}(c)$ for all $n\backslash geq\; 0$, since both functions are analytic.

If this is not the case, let $m$ be the smallest nonnegative integer with $f^\{(m)\}(c)\backslash ne\; g^\{(m)\}(c)$. By holomorphy, we have the following Taylor series representation in some open neighborhood U of $c$:

$$

\begin{align} (f - g)(z) &{}=(z - c)^m \cdot \left\frac{(f \\6pt

          &{}=(z - c)^m  \cdot h(z).
     

\end{align}

By continuity, $h$ is non-zero in some small open disk $B$ around $c$. But then $f-g\backslash neq\; 0$ on the punctured set $B-\backslash \{c\backslash \}$. This contradicts the assumption that $c$ is an accumulation point of $\backslash \{f\; =\; g\backslash \}$.

This lemma shows that for a complex number $a\; \backslash in\; \backslash mathbb\{C\}$, the fiber $f^\{-1\}(a)$ is a discrete (and therefore countable) set, unless $f\; \backslash equiv\; a$.

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Proof Define the set on which $f$ and $g$ have the same Taylor series: $$S\; =\; \backslash left\backslash \{\; z\; \backslash in\; D\; \backslash mathrel\{\backslash Big\backslash vert\}\; f^\{(k)\}(z)\; =\; g^\{(k)\}(z)\; \backslash text\{\; for\; all\; \}\; k\; \backslash geq\; 0\backslash right\backslash \}\; =\; \backslash bigcap\_\{k=0\}^\backslash infty\; \backslash left\backslash \{\; z\; \backslash in\; D\; \backslash mathrel\{\backslash Big\backslash vert\}\; \backslash bigl(f^\{(k)\}-\; g^\{(k)\}\backslash bigr)(z)\; =\; 0\backslash right\backslash \}.$$

We'll show $S$ is nonempty, Clopen set. Then by Connected space of $D$, $S$ must be all of $D$, which implies $f=g$ on $S=D$.

By the lemma, $f\; =\; g$ in a disk centered at $c$ in $D$, they have the same Taylor series at $c$, so $c\backslash in\; S$, $S$ is nonempty.

As $f$ and $g$ are holomorphic on $D$, $\backslash forall\; w\backslash in\; S$, the Taylor series of $f$ and $g$ at $w$ have non-zero radius of convergence. Therefore, the open disk $B\_r(w)$ also lies in $S$ for some $r$. So $S$ is open.

By holomorphy of $f$ and $g$, they have holomorphic derivatives, so all $\backslash textstyle\; f^\{(n)\},\; g^\{(n)\}$ are continuous. This means that $\backslash textstyle\; \backslash bigl\backslash \{z\; \backslash in\; D\; \backslash mathrel\{\backslash big\backslash vert\}\; \backslash bigl(f^\{(k)\}\; -\; g^\{(k)\}\backslash bigr)(z)\; =\; 0\backslash bigr\backslash \}$ is closed for all $k$. $S$ is an intersection of closed sets, so it's closed.

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Full characterisation Since the identity theorem is concerned with the equality of two holomorphic functions, we can simply consider the difference (which remains holomorphic) and can simply characterise when a holomorphic function is identically $0$. The following result can be found in.

(2025). 9783662535035, Springer Spektrum Verlag. ISBN 9783662535035

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Claim Let $G\backslash subseteq\backslash mathbb\{C\}$ denote a non-empty, connectedness open subset of the complex plane. For analytic $h:G\backslash to\backslash mathbb\{C\}$ the following are equivalent.

1. $h\backslash equiv\; 0$ on $G$;
2. the set $G\_\{0\}=\backslash \{z\backslash in\; G\backslash mid\; h(z)=0\backslash \}$ contains an accumulation point, $z\_\{0\}$;
3. the set $G\_\{\backslash ast\}=\backslash bigcap\_\{n\backslash in\backslash N\_0\}\; G\_\{n\}$ is non-empty, where $G\_\{n\}\; :=\; \backslash \{z\backslash in\; G\backslash mid\; h^\{(n)\}(z)=0\backslash \}$.

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Proof (1 $\backslash Rightarrow$ 2) holds trivially.

(2 $\backslash Rightarrow$ 3) is shown in section Lemma in part with Taylor expansion at accumulation point, just substitute g=0.

(3 $\backslash Rightarrow$ 1) is shown in section Proof with set where all derivatives of f-g vanishes, just substitute g=0.

Q.E.D.

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See also

• Analytic continuation
• Identity theorem for Riemann surfaces

• (1997). 9780521480581, Cambridge University Press. ISBN 9780521480581

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