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I think there can't be a positive radius of convergence. I don't know what happens now...

My reasoning is was as follows:

Basically suppose there was a positive radius of convergence and call it $R>0$.
Now for $1>\lambda> 0$ set

$g_\lambda(x)=\frac{1}{\lambda} f(\lambda x)$. Clearly, $g_\lambda$ is analytic on $|x|<\frac{R}{\lambda}$. Moreover, as $\lambda\to 0$ we see that $g_\lambda\to c_1=1$ $g_\lambda(x) \to x$ (NOT $g_\lambda\to c_1=1$) uniformly on compact subsets (by Arzela-Ascoli).

Now it is clear that $g_\lambda(g_\lambda(x))=\frac{1}{\lambda} f(f(\lambda x))$

but

$\frac{1}{\lambda} f(f(\lambda x))=\frac{1}{\lambda} (\lambda x+f(\lambda x) ^2)=x+\lambda g_\lambda(x)^2$.

That is $g_\lambda$ satisfies the equation

$g_\lambda(g_\lambda(x))=x+\lambda(g_\lambda(x))^2$

Now let $\lambda\to 0$. For any $x$ with $|x|<2$ the left hand side tends to $c_1=1$$x$ (NOT $c_1=1$) while the right hand side tends to $x$. This is clearly impossible. So everything is consistent.

I think this works. Obviously not.

I think there can't be a positive radius of convergence. I don't know what happens now...

My reasoning is was as follows:

Basically suppose there was a positive radius of convergence and call it $R>0$.
Now for $1>\lambda> 0$ set

$g_\lambda(x)=\frac{1}{\lambda} f(\lambda x)$. Clearly, $g_\lambda$ is analytic on $|x|<\frac{R}{\lambda}$. Moreover, as $\lambda\to 0$ we see that $g_\lambda\to c_1=1$ $g_\lambda(x) \to x$ uniformly on compact subsets (by Arzela-Ascoli).

Now it is clear that $g_\lambda(g_\lambda(x))=\frac{1}{\lambda} f(f(\lambda x))$

but

$\frac{1}{\lambda} f(f(\lambda x))=\frac{1}{\lambda} (\lambda x+f(\lambda x) ^2)=x+\lambda g_\lambda(x)^2$.

That is $g_\lambda$ satisfies the equation

$g_\lambda(g_\lambda(x))=x+\lambda(g_\lambda(x))^2$

Now let $\lambda\to 0$. For any $x$ with $|x|<2$ the left hand side tends to $c_1=1$ $x$ while the right hand side tends to $x$. This is clearly impossible. So everything is consistent.

I think this works. Obviously not.

2 More typos

I think there can't be a positive radius of convergence.

My reasoning is as follows:

Basically suppose there was a positive radius of convergence and call it $R>0$.
Now for $1>\lambda> 0$ set

$g_\lambda(x)=\frac{1}{\lambda} f(\lambda x)$. Clearly, $g_\lambda$ is analytic on $|x|<\frac{R}{\lambda}$. Moreover, as $\lambda\to 0$ we see that $g_\lambda\to c_1=1$ uniformly on compact subsets (by Arzela-Ascoli).

Now it is clear that $g_\lambda(g_\lambda(x))=\frac{1}{\lambda} f(f(\lambda(x))$f(f(\lambda x))$but$\frac{1}{\lambda} f(f(\lambda(x))=\frac{1}{\lambdaf(f(\lambda x))=\frac{1}{\lambda} (\lambda x+f(\lambda x) ^2)=x+\lambda g_\lambda(x)^2$. That is$g_\lambda$satisfies the equation$g_\lambda(g_\lambda(x))=x+\lambda(g_\lambda(x))^2$Now let$\lambda\to 0$. For any$x$with$|x|<2$the left hand side tends to$c_1=1$while the right hand side tends to$x\$. This is clearly impossible.

I think this works.

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