# Composition of real-analytic functions is real-analytic

Suppose $f,g: \mathbb{R} \to \mathbb{R}$ are real analytic, i.e, locally given by convergent power series. Then $g \circ f$ is real-analytic as well.

How do I prove this? I guess the "standard" proof would be to extend $f$ and $g$ into some open subsets of $\mathbb{C}$ in the natural way via their power series, then notice that $g \circ f$ is complex differentiable, hence complex-analytic, and hence real-analytic when restricted to $\mathbb{R}$.

But is there another proof of this, one that doesn't use complex-analytic extensions?

I want a proof which can be extended to the multivariate case, i.e, if $f: \mathbb{R}^n \to \mathbb{R}^m$ and $g:\mathbb{R}^m \to \mathbb{R}^p$ are both analytic (i.e, their components are locally given by multivariate power series), then $g \circ f$ is also real-analytic. Is the standard proof for the multivariate case via complex-analytic extensions as well? Does anyone know a good book for this subject?

A good reference for this stuff is A Primer of Real Analytic Functions by Steven Krantz and Harold Parks.

• Great! I hoped that the calculation would be less involved, but it's still nice. Apr 30, 2017 at 21:22

Here's a proof for the single-variable case. This works as well for $$\mathbb R$$ as for $$\mathbb C$$.

Let $$c$$ be any point in the domain of $$g\circ f$$; so $$c$$ is in the domain of $$f$$ and $$f(c)$$ is in the domain of $$g$$. Expand $$f$$ and $$g$$ as power series, converging absolutely on closed disks/intervals:

$$f(x)=\sum_ma_m(x-c)^m,\quad|x-c|\leq R;\qquad\sum_m|a_m|R^m<\infty$$

$$g(y)=\sum_nb_n(y-f(c))^n,\quad|y-f(c)|\leq S;\qquad\sum_n|b_n|S^n<\infty$$

($$R$$ and $$S$$ may be smaller than the radii of convergence. Any open disk contains a closed disk.)

Since $$f$$ is continuous (meaning $$f(x)\to f(c)=a_0$$ when $$x\to c$$), there is some $$r\leq R$$ such that $$|f(x)-f(c)|\leq S$$ when $$|x-c|\leq r$$. Thus the two power series involved in $$g(f(x))$$ are both absolutely convergent when $$|x-c|\leq r$$.

$$g(f(x))=\sum_{n\geq0}b_n\left(\sum_{m\geq1}a_m(x-c)^m\right)^n$$

But that's not quite what we need, in order to validly rearrange the terms here. We need something stronger:

$$\sum_{m\geq1}|a_m||x-c|^m\leq|x-c|\sum_{m\geq1}|a_m|R^{m-1}=M|x-c|$$

$$M=\frac1R\sum_{m\geq1}|a_m|R^m\leq\frac1R\sum_{m\geq0}|a_m|R^m<\infty$$

We can take $$r=\min\{S/M,R\}$$, so that

$$\sum_{m\geq1}|a_m||x-c|^m\leq M|x-c|\leq Mr\leq S$$

when $$|x-c|\leq r$$. Now we get the required absolute convergence:

$$\sum_{n\geq0}|b_n|\left(\sum_{m\geq1}|a_m||x-c|^m\right)^n\leq\sum_{n\geq0}|b_n|S^n<\infty$$

It follows that the terms in $$g(f(x))$$ can be rearranged in any way, and we can write it in the standard form (with no nested or repeated terms) as a power series in $$(x-c)$$.

I suppose I should also note that the coefficients are finite sums:

$$g(f(x))=\sum_{n\geq0}b_n\left(\sum_{m\geq1}a_m(x-c)^m\right)^n$$

$$=\sum_{n\geq0}b_n(x-c)^n\left(\sum_{m\geq1}a_m(x-c)^{m-1}\right)^n$$

$$=b_0+b_1(x-c)\Big(a_1+a_2(x-c)+a_3(x-c)^2+a_4(x-c)^3+\cdots\Big)\\+b_2(x-c)^2\Big(a_1+a_2(x-c)+a_3(x-c)^2+a_4(x-c)^3+\cdots\Big)^2\\+b_3(x-c)^3\Big(a_1+a_2(x-c)+a_3(x-c)^2+a_4(x-c)^3+\cdots\Big)^3\\+\cdots$$

$$=b_0+b_1a_1(x-c)+(b_1a_2+b_2a_1^2)(x-c)^2+(b_1a_3+2b_2a_1a_2+b_3a_1^3)(x-c)^3+\cdots$$

$$=\sum_{p\geq0}(x-c)^p\sum_{n\geq0\\n\leq p}b_n\sum_{k_1,k_2,k_3,\cdots\geq0\\k_1+2k_2+3k_3+\cdots=p\\k_1+k_2+k_3+\cdots=n}\frac{n!}{k_1!\,k_2!\,k_3!\,\cdots\,}a_1^{k_1}a_2^{k_2}a_3^{k_3}\cdots$$

• Yes, this is what is needed to do. Surprising how messier this is, compared with the one-line proof that comes from complex analysis. However this is also much more robust than the complex analytic proof. It also gives the coefficients of $f\circ g$. Jun 16, 2022 at 21:57

You may use the idea of majorants as in the proof of Cauchy-Kowalevski Theorem, where the functions real analytic on open subsets of $\mathbb R^m$.