# Identity theorem

Identity theorem 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 {displaystyle mathbb {R} } or {displaystyle mathbb {C} } ), if f = g on some {displaystyle Ssubseteq D} , where {displaystyle S} has an accumulation point, then f = g on D.

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). 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, {displaystyle f} can be {displaystyle 0} on one open set, and {displaystyle 1} on another, while {displaystyle g} is {displaystyle 0} on one, and {displaystyle 2} on another.

Contents 1 Lemma 2 Proof 3 Full characterisation 3.1 Claim 3.2 Proof 4 See also 5 References Lemma If two holomorphic functions {displaystyle f} and {displaystyle g} on a domain D agree on a set S which has an accumulation point {displaystyle c} in {displaystyle D} , then {displaystyle f=g} on a disk in {displaystyle D} centered at {displaystyle c} .

To prove this, it is enough to show that {displaystyle f^{(n)}(c)=g^{(n)}(c)} for all {displaystyle ngeq 0} .

If this is not the case, let {displaystyle m} be the smallest nonnegative integer with {displaystyle f^{(m)}(c)neq g^{(m)}(c)} . By holomorphy, we have the following Taylor series representation in some open neighborhood U of {displaystyle c} : {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}}} By continuity, {displaystyle h} is non-zero in some small open disk {displaystyle B} around {displaystyle c} . But then {displaystyle f-gneq 0} on the punctured set {displaystyle B-{c}} . This contradicts the assumption that {displaystyle c} is an accumulation point of {displaystyle {f=g}} .

This lemma shows that for a complex number {displaystyle ain mathbb {C} } , the fiber {displaystyle f^{-1}(a)} is a discrete (and therefore countable) set, unless {displaystyle fequiv a} .

Proof Define the set on which {displaystyle f} and {displaystyle g} have the same Taylor expansion: {displaystyle S=left{zin Dmid f^{(k)}(z)=g^{(k)}(z){text{ for all }}kgeq 0right}=bigcap _{k=0}^{infty }left{zin Dmid left(f^{(k)}-g^{(k)}right)(z)=0right}.} We'll show {displaystyle S} is nonempty, open, and closed. Then by connectedness of {displaystyle D} , {displaystyle S} must be all of {displaystyle D} , which implies {displaystyle f=g} on {displaystyle S=D} .

By the lemma, {displaystyle f=g} in a disk centered at {displaystyle c} in {displaystyle D} , they have the same Taylor series at {displaystyle c} , so {displaystyle cin S} , {displaystyle S} is nonempty.

As {displaystyle f} and {displaystyle g} are holomorphic on {displaystyle D} , {displaystyle forall win S} , the Taylor series of {displaystyle f} and {displaystyle g} at {displaystyle w} have non-zero radius of convergence. Therefore, the open disk {displaystyle B_{r}(w)} also lies in {displaystyle S} for some {displaystyle r} . So {displaystyle S} is open.

By holomorphy of {displaystyle f} and {displaystyle g} , they have holomorphic derivatives, so all {displaystyle f^{(n)},g^{(n)}} are continuous. This means that {displaystyle {zin Dmid (f^{(k)}-g^{(k)})(z)=0}} is closed for all {displaystyle k} . {displaystyle S} 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 {textstyle 0} . The following result can be found in.[1] Claim Let {textstyle Gsubseteq mathbb {C} } denote a non-empty, connected open subset of the complex plane. For {textstyle h:Gto mathbb {C} } the following are equivalent.

{textstyle hequiv 0} on {textstyle G} ; the set {textstyle G_{0}={zin Gmid h(z)=0}} contains an accumulation point, {textstyle z_{0}} ; the set {textstyle G_{ast }=bigcap _{nin mathbb {N} _{0}}G_{n}} is non-empty, where {textstyle G_{n}:={zin Gmid h^{(n)}(z)=0}} . Proof The directions (1 {textstyle Rightarrow } 2) and (1 {textstyle Rightarrow } 3) hold trivially.

For (3 {textstyle Rightarrow } 1), by connectedness of {textstyle G} it suffices to prove that the non-empty subset, {textstyle G_{ast }subseteq G} , 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. {textstyle hin C^{infty }(G)} , it is clear that {textstyle G_{ast }} is closed. To show openness, consider some {textstyle uin G_{ast }} . Consider an open ball {textstyle Usubseteq G} containing {textstyle u} , in which {textstyle h} has a convergent Taylor-series expansion centered on {textstyle u} . By virtue of {textstyle uin G_{ast }} , all coefficients of this series are {textstyle 0} , whence {textstyle hequiv 0} on {textstyle U} . It follows that all {textstyle n} -th derivatives of {textstyle h} are {textstyle 0} on {textstyle U} , whence {textstyle Usubseteq G_{ast }} . So each {textstyle uin G_{ast }} lies in the interior of {textstyle G_{ast }} .

Towards (2 {textstyle Rightarrow } 3), fix an accumulation point {textstyle z_{0}in G_{0}} . We now prove directly by induction that {textstyle z_{0}in G_{n}} for each {textstyle nin mathbb {N} _{0}} . To this end let {textstyle rin (0,infty )} be strictly smaller than the convergence radius of the power series expansion of {textstyle h} around {textstyle z_{0}} , given by {textstyle sum _{kin mathbb {N} _{0}}{frac {h^{(k)}(z_{0})}{k!}}(z-z_{0})^{k}} . Fix now some {textstyle ngeq 0} and assume that {textstyle z_{0}in G_{k}} for all {textstyle k

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