Let $g:[0,\infty) \rightarrow \mathbb{R}$ be an increasing function. Is there a way to construct an entire function $f(z)$ such that $f(x)=g(|x|)$ for all real $x$?
$\begingroup$
$\endgroup$
1
-
4$\begingroup$ What about $g(x)=x$ ? There is certainly no entire $f$ s.t. $f(x)=|x|$ for all $x \in \mathbb{R}$. $\endgroup$– RalphCommented Aug 30, 2012 at 9:58
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
You should keep in mind analytic continuation. Let $f:\mathbb R\rightarrow\mathbb R$ be a $C^\infty$ function ; if there exists an entire function extending $f$, then
(1) It is unique by analytic continuation,
(2) The function $f$ must be real-analytic.
Of course (2) is not sufficient: think about the real-analytic $x\mapsto\frac{1}{1+x^2}$, which does NOT have an entire extension, since by analytic continuation, that extension should coincide with $\mathbb C\ni z\mapsto \frac{1}{1+z^2}$, which has poles at $\pm i$. For a real-analytic $f$, you can formulate a criterion dealing with the size of derivatives. Such a function has an entire extension iff $$\forall R>0,\exists C_R,\forall n\in\mathbb N,\quad \vert f^{(n)}(0)\vert\le C_R\frac{n!}{R^n}. $$ On the other hand, real-analyticity of a $C^\infty$ $f$ on the real line is equivalent to $$ \forall x\in \mathbb R,\exists r>0,\exists C>0, \exists R>0,\forall n\in\mathbb N,\quad \Vert f^{(n)}\Vert_{L^\infty(B(x,r))}\le C\frac{n!}{R^n}. $$
-
2$\begingroup$ For the entire extension of a real-analytic function, an equivalent condition is $$ \lim_{n \to \infty} |f^{(n)}(0)/n!|^{1/n} = 0$$ $\endgroup$ Commented Aug 30, 2012 at 22:24
$\begingroup$
$\endgroup$
3
You certainly need some smoothness condition for this to be possible. In fact, the condition is that $f$ be extendable to an even, real-analytic function on the line which kind of makes the result a tautology since then $f$ has the form $\sum_{n=1}^\infty a_{2n} z^{2n}$.
-
$\begingroup$ Except that it wouldn't necessarily be entire. You need an estimate on the derivatives such as Bazin's. $\endgroup$ Commented Aug 30, 2012 at 22:13
-
$\begingroup$ I propose to close this question. $\endgroup$ Commented Aug 30, 2012 at 23:34
-
$\begingroup$ Sorry. Didn't realise that I had to spell out that a power series that converges for each real number has infinite radius of convergence. $\endgroup$– jbcCommented Aug 31, 2012 at 7:53