What are general theorems that give sufficient criteria for a $C^{\infty}$ function to be analytic? The more general/simple the test, the better. I'm trying to understand in a more thorough way what prevents $C^{\infty}$ functions from being analytic. The canonical example of a $C^{\infty}$ function $f$ which is not analytic at a point is $f(x) = e^{-\frac{1}{x}}$ for $x \in [0, \infty]$ and $f(x) = 0$ elsewhere. This indicates to me that nonzero analytic functions cannot "flatten out" too quickly at a point where they are defined. I know this is vague, but can you give me some theorems that make it more clear just how much more general $C^{\infty}$ functions are than analytic ones, and how many more degrees of freedom they have? I would also appreciate examples of $C^{\infty}$ functions that are pathological in some sense, in a way that an analytic function cannot be. For example, is there a $C^{\infty}$ function that fails to be analytic on a dense set?
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1Look up Taylor's theorem with remainder. – treble Jan 03 '17 at 02:56
3 Answers
There's a theorem of Sergei N. Bernstein which gives a sufficient condition for a $C^\infty$ function $f:\mathbb R\rightarrow\mathbb R$ to be analytic on $[0, r)$, mainly that $f^{(k)}(x)\ge0,\forall k \in \mathbb N_0,\forall x\in [0,r]$.
This is "Theorem 12.12" in Apostol's "Calculus I" (2nd Edition). I know the question is old but in case that someone else finds this, I hope my answer to be useful to a nonempty set of people.
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I am afraid that graphics not always gives us clear visual cues about analyticity because the differences between smoothness and analyticity are somewhat subtle and technical. The following is the only difference known to me:
Proposition. Let $f: I \to \mathbb{R}$ be an analytic function defined on an open interval $ I \subseteq \mathbb{R} $. If exists $ x_0 \in I $ such that for every $ n \in \mathbb{Z}_{\ge 0}$ we have that $ f^{(n)}(x_0) = 0 $, then $ f(x) = 0 $ for every $ x \in I $.
Since the class of analytic functions is well behaved in the sense that it is closed under the operations of sum, product and composition, another thing that might help to identify non-analytic functions when dealing with combinations of elementary analytic functions is looking at points in the function maximal domain such that it could not be extended without losing its analyticity and possibly the need of a piecewise function. That is exactly the case with the function
$$ f(x) = \begin{cases} e^{-\frac{1}{x}} & x >0 \\ 0 &x \le 0 \end{cases} $$
which fails to be analytic exactly on $ x = 0 $ which in turn is the only point out of the maximal domain $ \mathbb{R} \backslash \{0\} $ of the function $ e^{-\frac{1}{x}} $ that it is not possible to assign a value such that preserves the analyticity.
Now we will state two propositions that gives sufficient criteria for a smooth function to be analytic. Let $ f: I \to \mathbb{R} $ be a smooth function defined on an open interval $ I \subseteq \mathbb{R} $.
Proposition. The function $f$ is analytic if and only if for every $ x_0 \in I $ the Taylor series of $f$ around $x_0$ converges to $f(x)$ for all $x$ in a neighbourhood of $x_0$.
Proposition. The function $f$ is analytic if and only if for every compact subset $K \subset I$ there exists a constant $ C \in \mathbb{R}_{\ge 0} $ such that for every $ n \in \mathbb{Z}_{\ge 0} $ the following innequality holds for all $x \in K$
$$ |f^{(n)}(x)| \le C^{n+1} \, n! $$
Lastly, you can find more about not so much pathological non-analytical smooth functions in the following Wikipedia articles:
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I looked at the first article you linked, and it claimed that any sequence of numbers can be the sequence of derivatives of a smooth function at a point. Is there any restriction on which sequences can be the derivatives of an analytic function at a point, other than that the power series around that point must have nonzero radius of convergence? – Vik78 Jan 03 '17 at 18:50
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It seems not as this is practically a tautology. However you can express this is terms of the sequence itself by Cauchy-Hadamard theorem. Thus, this is the same as asking for a sequence $ a_n $ such that $ \limsup_{x\to\infty} | (\frac{ a_n }{ n! })^{1/n} | \in \mathbb{R}_+ $. – mucciolo Jan 03 '17 at 20:41
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Where can I find a reference for the third proposition ($|f^{(n)}(x)| \leq C^{n+1}n!$)? – João Alves Jr. Aug 23 '21 at 17:01
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I just found a reference for that 3rd proposition: Corollary 1.2.9 and Lemma 1.2.10 of Krantz, Steven; Parks, Harold R. (2002). A Primer of Real Analytic Functions (2nd ed.). Birkhäuser. ISBN 0-8176-4264-1. – João Alves Jr. Aug 23 '21 at 17:32
One necessary and sufficient condition for a function $f: I \subset \mathbb{R} \to \mathbb{C}$ to be analytic at a point $x_0 \in I$ is that $f$ extends, in some $\mathbb{C}$-neighborhood of $x_0$, to a function complex-differentiable near $x_0$. That is, $f$ is analytic at $x_0$ iff there exists a neighborhood $U \subset \mathbb{C}$ of $x_0$ and a holomorphic function $\tilde{f}:U \to \mathbb{C}$ satisfying $\tilde{f}(x) = f(x)$ for $x \in U \cap I$. This is an immediate consequence of the fundamental result in complex analysis that holomorphic and analytic functions are the same.
With almost no work, this criteria demonstrates that your canonical example is analytic at every point other than $x_0 = 0$, and it similarly demonstrates that most elementary functions one can write down are analytic.
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