# Proof that the inverse of an analytic function is analytic which uses only real analysis.

I would like to prove the following result:

Let $$f:R\to R$$ be such that $$f(x)=\sum\limits_{k=0}^\infty a_k(x-c)^k$$ for all $$x$$ in some open set $$O\subset R$$.

Suppose that $$f'(c)\ne 0$$.

Then there is a function $$g:T\to R$$ such that $$g(f(x))=x$$ and $$g(x)=\sum\limits_{k=0}^\infty b_k(x-f(c))^k$$, where $$T$$ is an open set of $$R$$ containing $$f(c)$$.

I know how to prove that $$f$$ is locally invertible but not how to prove that its inverse can be written as a power series on some open set.

There are many proofs for the inverse function theorem which use complex analysis but I would like to write a proof which uses only real analysis. I have been writing a book of proofs of many of the important theorems in calculus and I don't want to introduce complex analysis for the proof of one theorem.

Thanks, Andrew Murdza.

• I suggest consulting section 1.4 of A Primer of Real Analytic Functions by Krantz and Parks. Commented Mar 4, 2019 at 20:38
• This is perfect! This answers my question. If you convert your comment into an answer (with the same text) I will accept it. Commented Mar 5, 2019 at 0:26

This proof works over any Cauchy-complete field $$\mathbb F$$ with a non-trivial absolute value. The basic examples are $$\mathbb R$$ (real numbers), $$\mathbb C$$ (complex numbers), $$\mathbb Q_p$$ ($$p$$-adic numbers), and $$\mathbb F_p((X))$$ (formal Laurent series over a finite field). In fact $$\mathbb F$$ necessarily contains one of these four as a sub-field, but we don't need to know that.

After shifting/translating (which is invertible and preserves analyticity), we can assume that $$c=f(c)=0$$. Likewise, after scaling, we can assume that $$f'(c)=1$$, but let's not do that yet.

$$f(x)=\sum_{n\geq1}a_nx^n,\quad|x|\leq R$$ $$\sum_{n\geq1}|a_n|R^n<\infty$$ $$g(y)=\sum_{n\geq1}b_ny^n,\quad|y|\leq S$$ $$\sum_{n\geq1}|b_n|S^n<\infty$$

We're given $$a_n$$ and $$R>0$$, and we want to find $$b_n$$ such that $$g(f(x))=x$$ for all $$x$$ in some neighbourhood of $$0$$ and $$f(g(y))=y$$ for all $$y$$ in some neighbourhood of $$0$$.

Expanding the formal power series $$g(f(X))$$ (as shown at the bottom of this answer) and equating to the identity series $$e(X)=X$$, we can see that the required coefficients $$b_n$$ exist and are unique. Each one is a polynomial in $$a_k$$ for $$k\leq n$$, with integer coefficients, divided by some power of $$a_1\neq0$$. Thus $$f$$ has a unique formal left-inverse: $$g\circ f=e$$. By similar reasoning, $$f$$ has a unique formal right-inverse: $$f\circ h=e$$. Since composition of formal power series is associative, we have $$g=g\circ e=g\circ(f\circ h)=(g\circ f)\circ h=e\circ h=h$$, so the two inverses are in fact the same.

We only need to show that $$g(y)$$ converges for some $$y\neq0$$. That is, we need to find $$S>0$$ such that $$\sum_n|b_n|S^n<\infty$$.

Wolfram MathWorld gives this formula for the coefficients:

$$b_n=\sum_{k_2+2k_3+3k_4+\cdots=n-1}\frac{(2k_2+3k_3+4k_4+\cdots)!}{n!\;k_2!\;k_3!\;k_4!\;\cdots}\left(\frac{1}{a_1}\right)^n\left(-\frac{a_2}{a_1}\right)^{k_2}\left(-\frac{a_3}{a_1}\right)^{k_3}\left(-\frac{a_4}{a_1}\right)^{k_4}\cdots$$

I've reformulated it somewhat. I haven't verified it, but we can look for a proof of this formula elsewhere.

That factorial stuff looks like a fraction, which might be undefined if $$\mathbb F$$ has positive characteristic. But we've already seen that the coefficients are integers, so that's not a problem. We can evaluate the fraction in $$\mathbb Q$$, getting a result in $$\mathbb N$$, and then map that to $$\mathbb F$$ in the natural way.

Now let's scale $$f$$ so that $$a_1=1$$. (This simplifies the formula, but may actually make it harder to interpret and to verify, which is why I haven't done it yet.)

Applying the triangle inequality:

$$|b_n|\leq\sum_{k_2+2k_3+3k_4+\cdots=n-1}\frac{(2k_2+3k_3+4k_4+\cdots)!}{n!\;k_2!\;k_3!\;k_4!\;\cdots}|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$ $$\sum_{n\geq1}|b_n||y|^n\leq\sum_{n\geq1}\;\sum_{k_2+2k_3+3k_4+\cdots=n-1}\frac{(2k_2+3k_3+4k_4+\cdots)!}{n!\;k_2!\;k_3!\;k_4!\;\cdots}|y|^n|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$

Rearranging the series, and eliminating the variable $$n$$:

$$=\sum_{k_2,k_3,k_4,\cdots}\frac{(2k_2+3k_3+4k_4+\cdots)!}{(1+k_2+2k_3+3k_4+\cdots)!\;k_2!\;k_3!\;k_4!\;\cdots}|y|^{1+k_2+2k_3+3k_4+\cdots}|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$

Introducing a new variable $$m$$:

$$=\sum_m\sum_{k_2+k_3+k_4+\cdots=m}\frac{(2k_2+3k_3+4k_4+\cdots)!}{(1-m+2k_2+3k_3+4k_4+\cdots)!\;k_2!\;k_3!\;k_4!\;\cdots}|y|^{1-m+2k_2+3k_3+4k_4+\cdots}|a_2|^{k_2}|a_3|^{k_3}|a_4|^{k_4}\cdots$$

Distributing factors of $$|y|$$:

$$=\sum_m|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{(2k_2+3k_3+4k_4+\cdots)!}{(1-m+2k_2+3k_3+4k_4+\cdots)!\;k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$

Breaking off the $$m=0$$ term, where the inner sum reduces to a single term with $$k_2=k_3=k_4=\cdots=0$$:

$$=|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{(2k_2+3k_3+4k_4+\cdots)!}{(1-m+2k_2+3k_3+4k_4+\cdots)!\;k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$

Rewriting the factorial stuff with a binomial coefficient, and using $$\binom nk\leq\sum_k\binom nk=2^n$$:

$$=|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\binom{2k_2+3k_3+4k_4+\cdots}{m-1}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$ $$\leq|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}2^{2k_2+3k_3+4k_4+\cdots}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}|a_2y^2|^{k_2}|a_3y^3|^{k_3}|a_4y^4|^{k_4}\cdots$$

Distributing factors of $$2$$:

$$=|y|+\sum_{m\geq1}|y|^{1-m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{(m-1)!}{k_2!\;k_3!\;k_4!\;\cdots}\big(2^2|a_2y^2|\big)^{k_2}\big(2^3|a_3y^3|\big)^{k_3}\big(2^4|a_4y^4|\big)^{k_4}\cdots$$ $$=|y|+\sum_{m\geq1}\frac{|y|^{1-m}}{m}\sum_{k_2+k_3+k_4+\cdots=m}\frac{m!}{k_2!\;k_3!\;k_4!\;\cdots}\big(2^2|a_2y^2|\big)^{k_2}\big(2^3|a_3y^3|\big)^{k_3}\big(2^4|a_4y^4|\big)^{k_4}\cdots$$

Applying the multinomial formula:

$$=|y|+\sum_{m\geq1}\frac{|y|^{1-m}}{m}\Big(2^2|a_2y^2|+2^3|a_3y^3|+2^4|a_4y^4|+\cdots\Big)^m$$ $$=|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\Big(2^0|a_2y^0|+2^1|a_3y^1|+2^2|a_4y^2|+\cdots\Big)^m$$

The parenthesized series will converge if $$|y|\leq R/2$$, since we assumed $$\sum_n|a_n|R^n$$ converges:

$$=|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\left(\sum_{n\geq2}|a_n|(2|y|)^{n-2}\right)^m$$ $$\leq|y|+\sum_{m\geq1}\frac{2^2|y|^{1+m}}{m}\left(\sum_{n\geq2}|a_n|R^{n-2}\right)^m$$ $$=|y|+4|y|\sum_{m\geq1}\frac{1}{m}\left(|y|\sum_{n\geq2}|a_n|R^{n-2}\right)^m$$

The outer series will converge if $$|y|<1/\sum_{n\geq2}|a_n|R^{n-2}$$, since the Taylor series of the real logarithm is $$-\ln(1-u)=\sum_{m\geq1}u^m/m$$ for $$-1:

$$=|y|-4|y|\ln\left(1-|y|\sum_{n\geq2}|a_n|R^{n-2}\right)$$ $$<\infty$$

Thus, we've shown that $$g(y)$$ converges absolutely, with a radius at least

$$S=\min\left\{\frac{R}{2},\;\frac{0.99R^2}{\sum_{n\geq2}|a_n|R^n}\right\}.$$

• What did I do after "Applying the multinomial formula"? I factored out $2^2$ instead of $2^{2m}$... That changes the last few lines of the answer, but not too badly. The radius should be $$S=\min\left\{\frac R2,\;\frac{0.99R^2}{\mathbf4\sum_{n\geq2}|a_n|R^n}\right\}.$$ Commented Jan 11, 2023 at 19:59