About Approximation theory in Functional analysis

July 14, 2023

Abstract

This article investigates the approximation theory in the sense of functional analysis, to study the inner logics of approximation. And then the condition of formal series space normability is discussed.

Introduction

We have once learnt the Weierstrass Approximation Theorem in Calculus courses, which indicates that $\forall f\in\mathcal{C}[a,b]$ , there exists polynomial series capable of converging to $f$ uniformly. In the least square approximation problem, we also studied the method of Chebyshev polynomial base to approximate continuous functions. To analyze those issues, it is natural to introduce normed spaces.

To represent an element of some normed space, there is a concept called base, whose general idea is to destruct every element into combinations of some elements. In finite dimension normed spaces, base are widely used to simplify situations, while it’s also a definition of the finite dimension property of finite dimension normed spaces. With the aid of base theory, we can conclude that some sort of methods of base cannot represent banach spaces, thus cannot represent $\mathcal{C}[a,b]$ or $L^p(E)$ .

The text is based on [1,2].

Two types of basis for infinite dimension normed spaces

For a finite normed space $X$ , there exists $n\in \mathbb N$ , such that exists linear irrelevant $\{e_1,\cdots, e_n\}$ where every element $x\in X$ can be represented as $x=\sum_{i=1}^{n}k_ie_i$ . For infinite dimension normed spaces, the infinity term must exists, either in the sum or in the cardinality of base.

The concern of base is from the naive thought that every polynomial can be represented using the linear combination of $\{x^i|i\in \mathbb N\}$ . If elements in a normed space can be approximated in polynomial, then the base of normed space should be approximated in polynomial, or more precise, to be represented in some sort of polynomial base. Thus we need to learn something about base.

Hamel basis

Firstly we define $\operatorname{span}\{e_i\}_{i\in I}$ as

\operatorname{span}\{e_i\}_{i\in I} = \left\{\sum_{i=1}^Nk_ie_i|N\in \mathbb N, k_i\in \mathbb K\right\}

If we let the base have infinite cardinity, whether countable or uncountable, and let the whole space be

X=\operatorname{span}\{e_i\}_{i\in I}

, where

\{e_i\}_{i\in I}

are linear independent, then we got so called Hamel basis.

From the definition we instantly get that every element in a space with Hamel basis can be represented as finite sum of Hamel base.

Actually from Zorn’s lemma it’s easy to see that every linear space has its Hamel base.

Lemma 1 (Zorn’s lemma). Let $(\mathcal F,\prec)$ is a partial order set, if every order complete subset has its upper bound, then there is maximal element in $(\mathcal F,\prec)$ .

And then we can prove the statement above.

Proposition 1. Every linear space has its Hamel base.

Proof. Let $\mathcal F$ be the set containing all linear independent sets, then we define the $\prec$ : $\{e_i\}_{i\in I}\prec \{e_j\}_{j\in J} \iff \{e_i\}_{i\in I}\subset \{e_j\}_{j\in J}$ .

For any order complete subset $\mathcal A$ , consider the set $B = \bigcup_{E\in \mathcal A} E$ . As $\mathcal A$ is order complete, then for any finite subset $F\in G$ , exists $E\in \mathcal A$ such that $F\subseteq E$ . Thus the elements in $B$ are linear independent, so $B\in \mathcal F$ and for any $E\in \mathcal A$ , $E\subseteq B$ . That is, $B$ is an upper bound of $\mathcal A$ .

Then according to Zorn’s lemma, there exists a maximum element $M$ , which is a Hamel base of the space. ◻

Actually there are many base that are Hamel base, such as $\{x^i|i\in \mathbb N\}$ . However, the next subsection showed that those (countable) Hamel base are not much helpful as they are not fit in Banach spaces.

The inability of countable Hamel base to represent a infinite dimension Banach space

Although Hamel base exists in all linear spaces, What we are aware of are countable base, which are incompatible with infinite dimension Banach spaces. The incompatibility can then be derived from Baire Category Theorem.

Theorem 2 (Baire Category Theorem). A banach space cannot be a countable union of nowhere dense sets.

Then we can get the following theorem.

Theorem 3. A Banach space with infinity dimension cannot have a countable Hamel base.

Proof. If a Banach space $(X,\lVert \cdot \rVert )$ have a countable Hamel base, denoting $\{e_i\}_{i=1}^{\infty}$ , then we can decompose the space into union of closed sets. Denote $F_n$ as follows,

F_n = \operatorname{span}\{e_i\}_{i=1}^{n},

then we get that

X=\bigcup_{i=1}^{\infty}F_{n}

, and

F_n

are closed (as components of every base are closed, which is a conclusion of that

\mathbb K

is banach). To prove that

F_n

are nowhere dense, we need to show that the internal of

F_n

is empty. Denote the internal of

F_n

to be

\mathop{\mathrm{int}}F_n

. If

\mathop{\mathrm{int}}F_n\neq \emptyset

, then there exists closed ball

\overline{B(x,r)}\subset F_n

. Then we can move

e_{n+1}

into

\overline{B(x,r)}

, and let y be follows,

y = \frac{rx_{n+1}}{\lVert x_{n+1} \rVert } + x,

Thus

y\in \overline{B(x,r)}\in F_n

, Thus

e_{n+1}\in F_{n}

. Contradict.

Thus $X$ is the countable union of nowhere dense sets $\{F_n\}_{n=1}^{\infty}$ , contradict. ◻

Such theorem shows that any polynomial space cannot be banach with any norm.

Schauder basis

The shortcomings of the Hamel base is shown before, and seems that we need some other basis difinition for better dealing with infinite dimension spaces like Banach spaces. the Schauder base is one of the suitable base.

Let $\{e_n\}_{n=1}^{\infty}\subset X$ . If for every $x\in X$ , there exists $x=\sum_{i=1}^{\infty}k_ie_i, k_i\in \mathbb K$ , then we call $\{e_n\}_{n=1}^{\infty}\subset X$ a Schauder base of $X$ .

Worth attention that power basis is not a Schauder basis, although according to the Stone-Weierstrass Theorem, the span of power basis is dense in the space $\mathcal C[a,b]$ . For example, let $f\in \mathcal C[0,1]$ , defined as follows,

f(x)=\begin{cases} 0,& x\in[0,1/2),\\ 2x-1,& \text{otherwise.} \end{cases}

If power basis is a Schauder basis, then it should converge uniformly on the interval

[0,1]

, thus it converges to

0

uniformly in

[0,1/2)

, from which we concluded that

f\equiv 0

, contradict.

$\mathcal{C}[a,b]$ and Stone-Weierstrass Theorem

As is known in the Functional analysis, the norm of $\mathcal C[a,b]$ means that convergence in the $\mathcal C[a,b]$ is meant uniform convergence to some continuous function in the interval $\mathcal C[a,b]$ . From Stone-Weierstrass Theorem, we get that every element from $\mathcal C[a,b]$ , there is polynomial series that converges uniformly to the element. Stone-Weierstrass Theorem is not limited on the interval, but on the compact set.

Consider an algebra, which is meant an vector space $X$ on $\mathbb K=\mathbb C$ or $\mathbb K=\mathbb R$ , with multiplication arithmetic

(x,y)\in X\times X\rightarrow xy\in X,

meeting the following properties: for any

x,y,z\in X

\alpha,\beta\in \mathbb K

(Associativity) $(xy)z=x(yz)$ ;
(Distributivity) $x(y+z)=xy+xz$ , $(x+y)z=xz+yz$ ;
$(\alpha x)(\beta y)=(\alpha\beta)(xy)$ .

And subalgebra means subspace itself is an algebra.

Now we can introduce the Stone-Weierstrass Theorem. In Calculus courses, the Weierstrass Approximation Theorem is usually conducted with Bernstein polynomials. However, the generalized theorem only needs some basic assumption on a compact space $K$ and its normed space $\mathcal C(K)$ .

Theorem 4 (Stone-Weierstrass Theorem). Denote $K$ to be a compact metric space. Let $\mathcal A$ be the subalgebra of real space $\mathcal C(K)$ , with following properties,

Exists constant function $c\in \mathcal A$ ;
For any different point $\xi,\eta\in K$ , exists function $g\in \mathcal{A}$ , such that $g(\xi)\neq g(\eta)$ .

Then $\mathcal{A}$ is dense in $\mathcal{C}(K)$ .

Using the Stone-Weierstrass Theorem, it’s easy to obtain the Weierstrass Approximation Theorem.

Corollary 1 (Weierstrass Approximation Theorem). Denote $K$ to be compact subset of $\mathbb{R}^{n}$ , $\mathcal{P}(K)$ is all the n-variate polynomials restricted on $K$ , then $\mathcal{P}(K)$ is dense in $\mathcal{C}(K)$ .

Proof. It’s obvious that constant function is polynomials, thus the first requirement is met.

For any different point $\xi,\eta\in K$ , exists $1\le i\le n$ , such that $\xi_{i}\ne \eta_{i}$ , thus we may let $g(x)=x_i$ , where $x_i$ denotes the ith component of $x$ . Thus the second requirement is met.

According to the Stone-Weierstrass Theorem, $\mathcal{P}(K)$ is dense in $\mathcal{C}(K)$ . ◻

The proof is quite simple, owing to the strength of Stone-Weierstrass Theorem.

There is also a complex space version of Stone-Weierstrass Theorem, using which we can conclude the trigonometric version of Weierstrass Approximation Theorem.

Theorem 5 (Complex Stone-Weierstrass Theorem). Denote $K$ to be a compact metric space. Let $\mathcal A$ be the subalgebra of complex space $\mathcal C(K)$ , with following properties,

Exists constant function $c\in \mathcal A$ ;
For any different point $\xi,\eta\in K$ , exists function $g\in \mathcal{A}$ , such that $g(\xi)\neq g(\eta)$ .
If $g\in \mathcal{A}$ , then $\overline{g}\in \mathcal{A}$ .

Then $\mathcal{A}$ is dense in $\mathcal{C}(K)$ .

The Weierstrass Approximation Theorems told us that $\mathcal{P}(K)$ is dense in $\mathcal{C}(K)$ , in another word, that is $\mathcal{C}(K)\subseteq \overline{\mathcal{P}(K)}$ . The latter is then a restrict of $\mathcal{P}\lBrack X_i\rBrack_{i=1}^{n}$ .

$P[x]$ and its completion $P\lBrack x\rBrack$ , Normability problem

As is shown in the sections above, $\mathcal{C}(K)\subseteq \overline{\mathcal{P}(K)}$ . What’s more, there is elements such that $\overline{\mathcal{P}(K)}-\mathcal{C}(K)\neq \emptyset$ . For example, let $K=[0,1]$ , then $\sum_{n=1}^{\infty}(2x)^{n}$ diverges on the $\operatorname{sup}$ norm.

However, we can find subspace of $\overline{\mathcal{P}(K)}$ . For example, we may define weighted Hardy space on subspace of the formal series, which is even Hilbert spaces.

Actually, there is a theorem to provide a necessary and sufficient condition for a topological vector space to be normable.

Theorem 6 (Kolmogorov’s normability criterion). A topological vector space is normable if and only if it is a $T_1$ space and admits a bounded convex neighborhood of the origin.

Considering that the formal power series lacks topological properties, it may be normable under some constructed topologies.

Acknowledgements

The topic originates from a short conversation among me, my classmate Yue Wu, and Prof. Yi, which concenterated on the question whether a continuous function can be approximated with a series of rational polynomials. Further, the question is rewritten in a more precise form, “is rational polynomial space dense in $\mathcal{C}[a,b]$ or some sort of other normed space?” Surely the question needs more conditions on the rational polynomial space, for example, the rational polynomial space needs some sort of norms.

Furthermore, I come about some thought on the problem in the opposite direction that, as polynomial space $P[x]$ can be completed into formal power series space $P\lBrack x \rBrack$ , can the $P\lBrack x\rBrack$ be normable? Surely according to Stone-Weierstrass Theorem, we get that $\mathcal C[a,b]\subset P\lBrack x\rBrack$ . However, for the cause of unconvergence for some element of the space $P\lBrack x\rBrack$ , for example, $\sum_{n=1}^{\infty}x^{n}$ , we cannot use that norm. Then under some investigation and organization, The article is done.

Huge thanks to xxx, and Prof. yyy!

CIARLET, P.G. Linear and Nonlinear Functional Analysis with Applications: With 401 Problems and 52 Figures; Society for Industrial; Applied Mathematics, 2013; ISBN 9781611972580.

刘炳初 泛函分析; 南开大学数学教学丛书; 科学出版社, 1998; ISBN 9787030064851.