# Quasi-analytic Classes

## Motivation

There are a lot of nice properties of analytic functions, whose class is denoted by $C^\omega$. Formally we have the following definition:

If $f \in C^\omega$ and $x_0 \in \mathbb{R}$, one can write $f = a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots.$

Obviously $f \in C^\infty$ (and hence $C^\omega \subset C^\infty$) and alternatively we have the Taylor series converges to $f$ for any $x_0 \in \mathbb{R}$: $T(x) = \sum_{n=0}^{\infty}\frac{D^nf(x_0)}{n!}(x-x_0)^n.$ One interesting thing is, every $f \in C^\omega$ is uniquely determined by a sequence $D^0f(x_0), Df(x_0),D^2f(x_0),\cdots$.

Unfortunately, this property is not generally true on $C^\infty$. For example, we can consider the bump function $\varphi$ (a simple example can be found on wikipedia). In brief, $\varphi=0$ for all $x \in (-\infty,-1] \cup [1,+\infty)$ but $\varphi>0$ on $(-1,1)$. And more importantly, $\varphi \in C^\infty$. However, if we take $f = \varphi$ and $g = 2\varphi$, then $f \neq g$, but $D^nf(-2)=D^ng(-2)=0$ for all $n \geq 0$. We get a sequence of derivatives of different orders, but this sequence does not determine a unique $C^\infty$ function.

The term "uniquely determined" can also be described in an alternative way: If $f \in C^\omega$ and $D^k(x_0)=0$ for all $k \geq 0$, then $f=0$ everywhere.

So a question comes up naturally: how many functions can be determined by its derivatives of all orders? Does $C^\omega$ contain all we can get? If not, how can we describe them?

The class of analytics functions is our source of motivation, so it makes sense to dig into its properties to find more. For an analytic function it is natural to consider the restriction of a holomorphic function on the complex plane. Let $\Omega$ be the set of all $z=x+iy$ such that $|y| < \delta$ and suppose $f \in H(\Omega)$ and $|f(z)|<\beta$ for all $z \in \Omega$. By Cauchy's Estimate, we get $|D^n f(x)| \leq \beta \delta^{-n}n!\quad n \in \mathbb{N},x\in \mathbb{R}.$ Also the restriction of $f$ on $\mathbb{R}$ is real-analytic. Here comes the interesting part: $\beta$ and $\frac{1}{\delta}$ is determined only by $f$ and have nothing to do with $n$, meanwhile $n!$ is a special sequence that dominated $f$ to some extent.

This motivates us to define a special class of functions, which is called the class $C\{M_n\}$.

## The classes $C\{M_n\}$

Let $\{M_n\}$ be a sequence of positive numbers, we let $C\{M_n\}$ denote the class of all $f \in C^\infty$ such that $\lVert D^nf\rVert_\infty \leq \beta_f B^n_f M_n,$ where $\lVert \cdot \rVert_\infty$ is the supremum norm defined on $\mathbb{R}$, and $\beta_f,B_f$ are constants only determined by $f$ but not $n$.

In order to equip $C\{M_n\}$ with some satisfying algebraic structures, which can simplify our work, we need some restrictions.

### The algebraic structure of $C\{M_n\}$

Indeed, $B_f$ plays an much more important rule, since we have $\limsup_{n \to \infty}\left(\frac{\lVert D^n f\rVert_\infty}{M_n}\right)^{1/n} \leq B_f$ while $\beta_f$ was eliminated to $1$ in this limit. However, if we eliminate $\beta_f$ at the beginning, i.e. put $\beta_f = 1$ for all $f \in C\{M_n\}$, then when $n=0$, we have $\lVert f \rVert_\infty \leq M_0,$ which prevents $C\{M_n\}$ to be a vector space. For example, if $\lVert f \rVert_\infty = M_0$, then $\lVert 2f \rVert_\infty = 2M_0 > M_0$, hence $2f \not\in C\{M_n\}$. However, if we add $\beta_f$ no matter what, say $\lVert f \rVert_\infty \leq \beta_f M_0$, then whenever we do addition and scalar multiplication, there is a different constant with respect to the function, which makes sure that $C\{M_n\}$ is closed under addition and scalar multiplication, i.e. is a vector space. If we don't add such a constant, our class contains way too few functions.

Further, we have some restriction on the sequence $\{M_n\}$:

1. $M_0=1$.
2. $M_n^2 \leq M_{n-1}M_{n+1}$ ($\{\log M_n\}$ is a convex sequence).

As we will see soon, this makes $C\{M_n\}$ an algebra over $\mathbb{R}$, where multiplication is defined pointwise.

Proof. If $f,g \in C\{M_n\}$, then we need to show that $fg \in C\{M_n\}$. We have the product rule for differentiation: $D^n(fg) = \sum_{j=0}^{n}{n \choose k}(D^jf)(D^{n-j}g).$ Since $f,g \in C\{M_n\}$, we have $|D^n(fg)| \leq \sum_{j=0}^{n}{n \choose k}\beta_fB_f^jM_j\beta_gB_g^{n-j}M_{n-j} = \beta_f\beta_g\sum_{j=0}^{n}{n \choose k}B_f^jB_g^{n-j}M_jM_{n-j}.$ Of course we want to eliminate $M_jM_{n-j}$ to obtain a binomial expansion. To do this we need the convexity of the sequence $\{\log M_n\}$. Note $M_n^2 \leq M_{n-1}M_{n+1}$ implies $\log M_n - \log M_{n-1} \leq \log M_{n+1} - \log M_n.$ As a result, the line segment connecting $(n,\log M_n)$ and $(n-1,\log M_{n-1})$ is steeper and steeper as $n$ grows. By connecting these points, we actually gets a convex function but we will be more rigorous. For $0 < j < n$, we have \begin{aligned} \log M_n - \log M_j &= \sum_{k=j+1}^{n}\left(\log M_k - \log M_{k-1}\right) \\ &\geq \sum_{k = j}^{n-1}\left(\log M_{k} - \log M_{k-1}\right) \\ &\geq \sum_{k=1}^{n-j}(\log M_k - \log M_{k-1}) \quad\text{(note \log M_0=0)} \\ &= \log M_{n-j}. \end{aligned} Hence $M_n \geq M_jM_{n-j}$ for $0<j<n$. It also hold when $j=0$ or $j=n$, hence we get $|D^n(fg)|= \beta_f\beta_g\sum_{j=0}^{n}{n \choose k}B_f^jB_g^{n-j}M_jM_{n-j} \leq \beta_f\beta_g\sum_{j=0}^{n}{n \choose k}B_f^jB_g^{n-j}M_n = \beta_f\beta_g(B_f+B_g)^nM_n.$ Hence $fg \in C\{M_n\}$. The reason why $C\{M_n\}$ is a vector space has been stated already. $\square$

This restriction does not hurt the generality. In fact whenever we are given a positive sequence $\{M_n\}$, we have another sequence $\{M'_n\}$ satisfying the two restrictions such that $C\{M_n\}=C\{M'_n\}$.

### The Quasi-analytic class

A class $C\{M_n\}$ is said to be quasi-analytic if the condition $f \in C\{M_n\},\quad (D^nf)(0)=0$ for all $n \in \mathbb{N}$ implies that $f = 0$ for all $x \in \mathbb{R}$.

The reason we try to check whether it's equal to $0$ everywhere, instead of check whether it is 'uniquely determined' by a sequence of derivative of different order is, this one is much simpler to work with. If a sequence of derivative of different order determines two functions, then their difference is always $0$.

#### $C\{n!\}$ as an example

We have seen that $C\{n!\}$ contains all functions which is a restriction of a holomorphic function in the strip defined by $|\Im(z)|<\delta$. Conversely, we show that any function in $C\{n!\}$ defined on the real axis can be extended to a holomorphic function with the same property. As a result, $C\{n!\}$ is a quasi-analytics class (which contains all bounded function of $C^\omega$). If we only consider functions defined on a closed and bounded interval $[a,b]$, then $C\{n!\}$ is exactly $C^\omega$.

Suppose $f \in C\{n!\}$. First of all we have $\lVert D^nf \rVert_\infty \leq \beta B^nn!$ for $n \in \mathbb{N}$. By Taylor's formulae $f(x) = \sum_{j=0}^{n-1}\frac{D^jf(a)}{j!}+\frac{1}{(n-1)!}\int_a^x(x-t)^{n-1}D^nf(t)dt.$ The remainder is therefore dominated by $\frac{n!}{(n-1)!}\beta B^n\left\vert\int_a^x(x-t)^{n-1}dt\right\vert = \beta|B(x-a)|^n.$ If $|B(x-a)|<1$, then $\lim_{n \to \infty}|B(x-a)|^n = 0$, and we can safely write the expansion $f(x) = \sum_{n=0}^{\infty}\frac{D^nf(a)}{n!}(x-a)^n.$ Pick $0<\delta<\frac{1}{B}$, we can replace $x$ in the expansion above with $z$ such that $|z-a|<\delta$. This defines a holomorphic function $F_a$ on $D(a,\delta)$ (the open disk centred at $a$ with radius $\delta$). If $x \in D(a,\delta)$ is real, then $F_a(x)=f(x)$. Therefore $F_a$ is the analytic continuation of $f$; all $F_a$ form a holomorphic extension $F$ of $f$ in the strip $|\Im(z)|<\delta$. As a result, for $z = a+iy$ with $|y|<\delta$, we have $|F(z)|=|F_a(z)| = \left\vert\sum_{n=0}^{\infty}\frac{D^nf(a)}{n!}(iy)^n\right\vert \leq \beta \sum_{n=0}^{\infty}(B\delta)^n = \frac{1}{1-B\delta}$ Hence $F$ is bounded in such a region.

## The fundamental theorem about quasi-analytic classes

In general, if $M_n \to \infty$ way too fast (at least faster than $n!$) as $n \to \infty$, then $C\{M_n\}$ is quasi-analytic. There are several equivalent statements on whether $C\{M_n\}$ is a quasi-analytic class, which is given by the Denjoy-Carleman theorem. Here I collect all conditions that I have found:

(Denjoy-Carleman theorem) The following conditions are equivalent:

1. $C\{M_n\}$ is not quasi-analytic.
2. $\int_0^\infty \log Q(x)\frac{dx}{1+x^2}<\infty$, where $Q(x)=\sum_{n=0}^{\infty}\frac{x^n}{M_n}$.
3. $\int_0^\infty \log q(x) \frac{dx}{1+x^2}<\infty$, where $q(x) = \sup \frac{x^n}{M_n}$.
4. $\sum_{n=1}^{\infty}\left(\frac{1}{M_n}\right)^{1/n}<\infty$.
5. $\sum_{n=1}^{\infty}\frac{M_{n-1}}{M_n}<\infty$
6. $C\{M_n\}$ contains nontrivial function with compact support.
7. $\sum_{n=1}^{\infty}\frac{1}{\lambda_n}<\infty$ where $\lambda_n = \inf_{k \geq n}M_k^{\frac{1}{k}}$.

You may find condition 7 is ridiculous. In fact, in this condition $\{M_n\}$ is not required to satisfy the two restriction. This one is what Denjoy and Carleman found initially. Later, mathematicians find that for a sequence $\{M_n\}$ we can obtain its convex minorant ${M_n'}$ such that

1. $M_n \geq M_n'$ for all $n$.
2. $\{\log M_n'\}$ is convex.
3. There is a sequence $0=n_0<n_1<\cdots$ such that $M_{n_0} = M'_{n_0}$ and $\log M_k$ is linear for $n_i \leq k \leq n_{i+1}$.

And as you may guess, the convex minorant $\{M_n'\}$ is what we are using today.

The proof of the Denjoy-Carleman theorem will come out in my next blog post. There are quite a lot of work to do to finish the proof, and it cannot be done within hours. We will be using many complex analysis theories. Also, I will try to cover some extra properties of quasi-analytic classes as well as why convex minorant is sufficient.

Desvl

2021-03-30

2021-10-11