Quasi-analytic Classes (Complex Analysis)

Motivation

There are a lot of nice properties of analytic functions, whose class is denoted by Cω. Formally we have the following definition:

If f∈Cω and x0∈R, one can write f=a0+a1(x−x0)+a2(x−x0)2+⋯.

Obviously f∈C∞ (and hence Cω⊂C∞) and alternatively we have the Taylor series converges to f for any x0∈R: T(x)=∞∑n=0Dnf(x0)n!(x−x0)n. One interesting thing is, every f∈Cω is uniquely determined by a sequence D0f(x0),Df(x0),D2f(x0),⋯.

Unfortunately, this property is not generally true on C∞. For example, we can consider the bump function φ (a simple example can be found on wikipedia). In brief, φ=0 for all x∈(−∞,−1]∪[1,+∞) but φ>0 on (−1,1). And more importantly, φ∈C∞. However, if we take f=φ and g=2φ, then f≠g, but Dnf(−2)=Dng(−2)=0 for all n≥0. We get a sequence of derivatives of different orders, but this sequence does not determine a unique C∞ function.

The term "uniquely determined" can also be described in an alternative way: If f∈Cω and Dk(x0)=0 for all k≥0, then f=0 everywhere.

So a question comes up naturally: how many functions can be determined by its derivatives of all orders? Does Cω 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 Ω be the set of all z=x+iy such that |y|<δ and suppose f∈H(Ω) and |f(z)|<β for all z∈Ω. By Cauchy's Estimate, we get |Dnf(x)|≤βδ−nn!n∈N,x∈R. Also the restriction of f on R is real-analytic. Here comes the interesting part: β and 1δ 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{Mn}.

The classes C{Mn}

Let {Mn} be a sequence of positive numbers, we let C{Mn} denote the class of all f∈C∞ such that ‖Dnf‖∞≤βfBnfMn, where ‖⋅‖∞ is the supremum norm defined on R, and βf,Bf are constants only determined by f but not n.

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

The algebraic structure of C{Mn}

Indeed, Bf plays an much more important rule, since we have lim sup 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.