# What is wrong with this proof about real analytic functions?

I will give a "proof" for the following result that seems to be incorrect. Can some one tell me what is wrong with my reasoning?

RESULT: Let $$f$$, $$g$$ and $$h$$ be three real analytic functions on $$(a-R,a+R)$$ with $$R>0$$ such that $$f(x)=g(x)+h(x)$$ for all $$x\in (a-R,a+R)$$ . If $$R>0$$ is the radius of convergence of the Taylor series of two of these functions, then the radius of convergence of the Taylor series of the third function is also $$R$$.

PROOF: If I use the given equality between the functions, I get that for all $$n$$, $$f^{(n)}(x)=g^{(n)}(x)+h^{(n)}(x),$$ and in particular, $$f^{(n)}(a)=g^{(n)}(a)+h^{(n)}(a).$$ Let $$A_{n}$$, $$B_{n}$$ and $$C_{n}$$ be the general terms of the power series expansion of $$f$$, $$g$$ and $$h$$ respectively. I divide each term of the last inequality by $$n!$$, and so: $$A_{n}=\dfrac{f^{(n)}(a)}{n!}=\dfrac{g^{(n)}(a)}{n!}+\dfrac{h^{(n)}(a)}{n!} =B_{n}+C_{n}.\qquad(1)$$

Now I am going to use a theorem that says: the radius of convergence of a series $$\sum\alpha_{n}(x-a)^{n}$$ is $$R$$ if and only if for all $$0 there exists a positive constant $$k$$ such that we have for all $$n$$, $$|\alpha_{n}|\leq \dfrac{k}{r^{n}}.$$ Let $$0, and suppose for example that the radius of convergence for the Taylor's series of $$g$$ and $$h$$ is $$R$$. By the cited theorem there exists positive constants $$k_{1}$$ and $$k_{2}$$ such that for all $$n$$, $$|B_{n}|\leq \dfrac{k_{1}}{r^{n}}$$ and $$|C_{n}|\leq \dfrac{k_{2}}{r^{n}}.$$ Then according to (1) we have for all $$n$$, $$|A_{n}|\leq |B_{n}+C_{n}|\leq \dfrac{(k_{1}+k_{2})}{r^{n}}.????!!!$$

• You quote a theorem about radius of convergence: but the $r_n, \alpha_n$ there are not defined. It is not clear what role they have. – P Vanchinathan Oct 27 '18 at 0:36
• The theorem you quote uses $r$, which is never mentioned again, and the condition you put uses $r_n$, which is never defined... – Arturo Magidin Oct 27 '18 at 0:36
• What happens if $g=-h$? – Kavi Rama Murthy Oct 27 '18 at 0:38
• Of course you mean that the radius of convergence of the third function is at least $R$... – David C. Ullrich Oct 27 '18 at 0:40
• why does it seem to be incorrect? – T_M Oct 27 '18 at 1:03

Your approach is fine. We can proceed from \begin{align*} |A_{n}|\leq |B_{n}+C_{n}|\leq \dfrac{k_{1}+k_{2}}{r^{n}} \end{align*} as follows:
We set $$k=k_1+k_2$$ and obtain for $$0 \begin{align*} |A_{n}|\leq \frac{k}{r^n} \qquad\text{for all } n \end{align*} We conclude the convergence radius of $$f$$ expanded at $$0$$ is at least $$R$$.
Note the phrase at least. We have shown the claim is valid for all $$0, but we didn't say anything about $$r>R$$.
Let's consider for example $$g(x)=\frac{1}{1-x}$$ and $$h(x)=-\frac{1}{1-x}+\frac{1}{1-\frac{x}{2}}$$. Both function have convergence radius $$R=1$$ when expanded at $$0$$, but $$f(x)=g(x)+h(x)=\frac{1}{1-\frac{x}{2}}$$ has radius of convergence $$R=2$$.