I'm trying to prove a version of the Identity Theorem for single variable real functions:

Let $J\subset I \subset \mathbb{R}$ be non-empty, open intervals. Show that, if functions $f, g: I \to \mathbb{R}$ are real analytic on the interval $I$ such that $$f(x)=g(x) \ \forall x \in J,$$ then $$f(x)=g(x) \ \forall x \in I.$$

I have read this question, but unfortunately I'm only a beginner in real analysis and I don't know anything about topology etc, so the answer for that question goes over my head.

My idea for proving this was to expand the set for which the functions are equal step by step. Let $J=\left]a,b\right[$ and $I=\left]c,d\right[$. First, consider the function $f-g$ at the point $a$. To the right of $a$, the function $f-g$ must be zero, and using the power series expansion for $f-g$, one can conclude that $f-g$ must also be zero left of $a$, but only for some distance $\delta_1>0$ as given by the analyticity of $f$ and $g$. Now just repeat at each new endpoint until the set $I$ is reached.

However, what guarantees that the sequence $\{\delta_i\}$ doesn't approach zero too fast? Too fast in the sense that the endpoint of $I$ is never reached using this method. Is there a way to restrict the size of $\delta_i$ or is this approach simply not viable?

  • $\begingroup$ Look at the set of points $x$ such that the functions coincide on $(x,b)$. Find a contradiction if the infimum of this set is not $c$. $\endgroup$ – hmakholm left over Monica May 7 '19 at 11:02
  • $\begingroup$ If $f$ is analytic on $[A,B]$ then for every $x_0 \in [A,B]$ it is given by a power series on $x \in [A,B],|x-x_0| < r(x_0)$, check that $r(x_0)$ is continuous in $x_0$ so that $\inf_{x_0\in [A,B]} r(x_0)$ needs to be $> 0$. Thus equality of $f,g$ on one of those disks implies equality everywhere else. $\endgroup$ – reuns May 7 '19 at 11:43

The hypotheses of your statement are stronger than necessary and I think that potentially obscures why the theorem holds, so let me outline (the missing details are easy to fill in) the following:

If $\sum_{n=0}^{\infty}a_nx^n$ and $\sum_{n=0}^{\infty}b_nx^n$ are two real power series with positive radius of convergence and they coincide on a zero sequence $(x_n)_n\in\left(\mathbb{R}\setminus\{0\}\right)^{\mathbb{N}}$, then they are equal.

To prove this, denote $f_0(x)=\sum_{n=0}^{\infty}a_nx^n$ and $g_0(x)=\sum_{n=0}^{\infty}b_nx^n$. Since power series are continuous in their radius of convergence, we obtain $$a_0=f_0(0)=\lim_{n\rightarrow\infty}f_0(x_n)=\lim_{n\rightarrow\infty}g_0(x_n)=g_0(0)=b_0.$$ Now, we define $$f_1(x)=\begin{cases}\frac{f_0(x)-a_0}{x},&x\neq0\\a_1,&x=0\end{cases},\qquad g_1(x)=\begin{cases}\frac{g_0(x)-b_0}{x},&x\neq0\\b_1,&x=0\end{cases}.$$ Note that $f_1$ and $g_1$ are still power series with the same radius of convergence as $f_0$ and $g_0$ respectively. Note furthermore that they still coincide on the same sequence $(x_n)_n$ (and also that this is the step where it is crucial that the sequence $(x_n)_n$ takes values in $\mathbb{R}\setminus\{0\}$). The same calculation as above now yields $a_1=b_1$. Continuing the process inductively yields $a_n=b_n$ for all $n\in\mathbb{N}$, which is the desired statement.

The statement you are trying to prove is just a (if necessary local) application of this. As a bonus, the above theorem also holds for complex power series and the proof is virtually the same.

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