I am learning about real analytic functions on my own right now. I've been having trouble with one of the exercises, and it isn't much help that most of the resources online for analytic functions are for Complex Analysis. I am talking about real analytic functions.

For reference, here is a definition that I have been using:

A real function $f(x)$ is analytic at $x_{0}$ if there is a $r > 0$:

$$f(x) = \sum_{n = 0}^{\infty} a_{n}(x - x_{0})^{n}, |x - x_{0}| < r$$

i.e. there is some power series which converges to the function.

Using this definition, I want to solve the following exercise problem:

Find a function $f$ analytic at $x_{0} = 0$ so that $f\left(\frac{1}{n}\right) = \frac{n}{n + 1}$, $n = 1, 2, \ldots$.

Show that such a function cannot be analytic on $(-2, 0)$.

So, working backwards, I found out that $\frac{1}{1 + x}$ satisfies the property $f(\frac{1}{n}) = \frac{n}{n + 1}$. I'm really not so sure how to prove the analytic properties though. I think that now I need to show $\frac{1}{1 + x}$ is analytic, and then I need to prove the second part of the claim, which is that such a function cannot be analytic on $(-2, 0)$.

I have an example in my book which shows $1 + x + x^{2} + x^{3} + \cdots$ is analytic. Here's how they do it:

A prototypical example is the geometric series $$1 + x + x^{2} + \cdots = \lim_{n\to\infty} 1 + x + x^{2} + \ldots x^{n} = \lim_{n\to\infty}\frac{1 - x^{n + 1}}{1 - x}$$

for which it is well known equals $f(x) = \frac{1}{1 - x}$ for $|x| < 1$. To verify that the function is analytic, we need to expand about any point $x_{0} \neq 1$:

$$ \begin{align*} \frac{1}{1-x} = \frac{1}{1 - x_{0} - (x - x_{0})} = \frac{1}{1- x_{0}}\left(1 - \frac{x - x_{0}}{1 - x_{0}}\right)^{-1} \end{align*} $$

I tried outlining this example, but I couldn't make any progress. I would really appreciate some sort of help.

  • $\begingroup$ Your function is $f(z)=\frac1{1+z}$ for $z_0=0$. $\endgroup$ – xpaul Nov 12 '18 at 16:30
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    $\begingroup$ It seems the answer is already inside your own post! You know how to show that $g(x) = 1/(1-x)$ is analytic at $0$, now you only need to realize that $f(x) = 1/(1+x)$ can be related to this function $g$ by $f(x) = g(-x)$ $\endgroup$ – Vincent Nov 12 '18 at 16:33
  • $\begingroup$ @Vincent I don't see how it helps. I had also made that observation but couldn't make any progress. I also don't know how to show such a function cannot be analytic on (-2, 0). $\endgroup$ – user592402 Nov 12 '18 at 16:38
  • $\begingroup$ @joseph What can you use? Can you use the Identity Theorem for analytic functions? $\endgroup$ – Batominovski Nov 12 '18 at 17:30
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    $\begingroup$ @Vincent I think it's because it's not defined at $x_{0} = -1$ $\endgroup$ – user592402 Nov 13 '18 at 16:58

This is a solution without the Identity Theorem. It is a bit lengthy, but I have no other tricks. Combinatorial identities I use in my proof can be proven using the technique from here.

Note that $$f(x)=\dfrac{1}{1+x}\tag{*}$$ for all $x$ such that $\dfrac{1}{x}$ is a positive integer. We also see that $$f(0)=\lim_{n\to\infty}\,f\left(\dfrac1n\right)=1\,.$$ This means (*) is true when $x=0$ too.

We can try to find $f^{(k)}(0)$ for $k=1,2,3,\ldots$ by noting that $$f^{(k)}(0)=\lim_{h\to 0^+}\,\frac{1}{h^k}\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,f\big((k-r)h\big)\,.$$ Taking $h:=\dfrac{1}{m\cdot k!}$ for some positive integer $m$, we have $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,f\left(\frac{k-r}{m\cdot k!}\right)\,.$$ Since $\dfrac{m\cdot k!}{k-r}$ is a positive integer for all $r=0,1,2,\ldots,k-1$, and $f(0)=1$, we get $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{1}{1+\frac{k-r}{m\cdot k!}}\right)\,.$$ Because $\sum\limits_{r=0}^k\,\binom{k}{r}\,(-1)^r\,(k-r)^t=0$ for $t=0,1,2,\ldots,k-1$, we get $$f^{(k)}(0)=\lim_{m\to \infty}\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{1}{1+\frac{k-r}{m\cdot k!}}-\sum_{t=0}^{k-1}\,(-1)^t\,\left(\frac{k-r}{m\cdot k!}\right)^t\right)\,.$$ Using Taylor's Theorem, we have $$\frac{1}{1+\frac{k-r}{m\cdot k!}}-\sum_{t=0}^{k-1}\,(-1)^t\,\left(\frac{k-r}{m\cdot k!}\right)^t=(-1)^k\,\left(\frac{k-r}{m\cdot k!}\right)^k+\mathcal{O}\left(\frac{1}{m^{k+1}}\right)\,.$$ That is, $$f^{(k)}(0)=\lim_{m\to\infty}\,\left((-1)^k\,(m\cdot k!)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(\frac{k-r}{m\cdot k!}\right)^k+\mathcal{O}\left(\frac{1}{m}\right)\right)\,.$$ Ergo, $$f^{(k)}(0)=(-1)^k\,\sum_{r=0}^k\,\binom{k}{r}\,(-1)^r\,\left(k-r\right)^k=(-1)^k\,k!\,.$$ Therefore, $$f(x)=\sum_{k=0}^\infty\,\frac{f^{(k)}(0)}{k!}\,x^k=\sum_{k=0}^\infty\,(-1)^k\,x^k=\frac{1}{1+x}$$ for all $x\in (-1,+1)$ because the radius of convergence is $1$. This proves the existence and the uniqueness of $f$.

To show that there is no analytic function with the given property that is defined on $(-2,0)$, you need to show that such a function $f$ cannot be defined at $-1$. One way to do this is noting that $x=-1$ is a pole of $\dfrac{1}{1+x}$. Therefore, $x=-1$ is a natural boundary of $f(x)$.


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