Theorem on the equality of analytic functions
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
or
), if f = g on some
, where
has an accumulation point in D, then f = g on D.[1]
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 infinitely real-differentiable functions. In comparison, analytic functions are a much more rigid notion. Informally, one sometimes summarizes the theorem by saying analytic functions are "hard" (as opposed to, say, continuous functions which are "soft").
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,
can be
on one open set, and
on another, while
is
on one, and
on another.
Lemma
If two holomorphic functions
and
on a domain D agree on a set S which has an accumulation point
in
, then
on a disk in
centered at
.
To prove this, it is enough to show that
for all
.
If this is not the case, let
be the smallest nonnegative integer with
. By holomorphy, we have the following Taylor series representation in some open neighborhood U of
:
![{\displaystyle {\begin{aligned}(f-g)(z)&{}=(z-c)^{m}\cdot \left[{\frac {(f-g)^{(m)}(c)}{m!}}+{\frac {(z-c)\cdot (f-g)^{(m+1)}(c)}{(m+1)!}}+\cdots \right]\\[6pt]&{}=(z-c)^{m}\cdot h(z).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/923abd4b57f8e6ec53edf4035f95117acff7174b)
By continuity,
is non-zero in some small open disk
around
. But then
on the punctured set
. This contradicts the assumption that
is an accumulation point of
.
This lemma shows that for a complex number
, the fiber
is a discrete (and therefore countable) set, unless
.
Proof
Define the set on which
and
have the same Taylor expansion:
![{\displaystyle S=\left\{z\in D\mid f^{(k)}(z)=g^{(k)}(z){\text{ for all }}k\geq 0\right\}=\bigcap _{k=0}^{\infty }\left\{z\in D\mid \left(f^{(k)}-g^{(k)}\right)(z)=0\right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ab1ef72ae28522fdd10f2fb8bf4cf3ded9be207)
We'll show
is nonempty, open, and closed. Then by connectedness of
,
must be all of
, which implies
on
.
By the lemma,
in a disk centered at
in
, they have the same Taylor series at
, so
,
is nonempty.
As
and
are holomorphic on
,
, the Taylor series of
and
at
have non-zero radius of convergence. Therefore, the open disk
also lies in
for some
. So
is open.
By holomorphy of
and
, they have holomorphic derivatives, so all
are continuous. This means that
is closed for all
.
is an intersection of closed sets, so it's closed.
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
. The following result can be found in.[2]
Claim
Let
denote a non-empty, connected open subset of the complex plane. For
the following are equivalent.
on
; - the set
contains an accumulation point,
; - the set
is non-empty, where
.
Proof
The directions (1
2) and (1
3) hold trivially.
For (3
1), by connectedness of
it suffices to prove that the non-empty subset,
, is clopen (since a topological space is connected if and only if it has no proper clopen subsets). Since holomorphic functions are infinitely differentiable, i.e.
, it is clear that
is closed. To show openness, consider some
. Consider an open ball
containing
, in which
has a convergent Taylor-series expansion centered on
. By virtue of
, all coefficients of this series are
, whence
on
. It follows that all
-th derivatives of
are
on
, whence
. So each
lies in the interior of
.
Towards (2
3), fix an accumulation point
. We now prove directly by induction that
for each
. To this end let
be strictly smaller than the convergence radius of the power series expansion of
around
, given by
. Fix now some
and assume that
for all
. Then for
manipulation of the power series expansion yields
![{\displaystyle h^{(n)}(z_{0})=n!{\frac {h(z)}{(z-z_{0})^{n}}}-(z-z_{0})\underbrace {n!\sum _{k=n+1}^{\infty }{\frac {h^{(k)}(z_{0})}{k!}}(z-z_{0})^{k-(n+1)}} _{=:R(z)}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96629a66444043226f087dff614af84c12b45809) | | (1) |
Note that, since
is smaller than radius of the power series, one can readily derive that the power series
is continuous and thus bounded on
.
Now, since
is an accumulation point in
, there is a sequence of points
convergent to
. Since
on
and since each
, the expression in (1) yields
![{\displaystyle h^{(n)}(z_{0})=n!{\frac {h(z^{(i)})}{(z^{(i)}-z_{0})^{n}}}-(z^{(i)}-z_{0})R(z^{(i)})=0-\underbrace {(z^{(i)}-z_{0})} _{\longrightarrow _{i}0}R(z^{(i)}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0aaa060005b9834b56e75fdc63118d81d1697a6) | | (2) |
By the boundedness of
on
, it follows that
, whence
. Via induction the claim holds. Q.E.D.
See also
References
- ^ For real functions, see Krantz, Steven G.; Parks, Harold R. (2002). A Primer of Real Analytic Functions (Second ed.). Boston: Birkhäuser. Corollary 1.2.7. ISBN 0-8176-4264-1.
- ^ Guido Walz, ed. (2017). Lexikon der Mathematik (in German). Vol. 2. Mannheim: Springer Spektrum Verlag. p. 476. ISBN 978-3-662-53503-5.
- Ablowitz, Mark J.; Fokas A. S. (1997). Complex variables: Introduction and applications. Cambridge, UK: Cambridge University Press. p. 122. ISBN 0-521-48058-2.