Arzelà–Ascoli theorem

From Wikipedia, the free encyclopedia

In mathematics, the Arzelà–Ascoli theorem of functional analysis gives necessary and sufficient conditions to decide whether a set of continuous functions from a compact metric space into a metric space is compact in the topology of uniform convergence. The main condition is equicontinuity of the set of functions.

The notion of equicontinuity was introduced at around the same time by Ascoli (1883–1884) and Arzelà (1882–1883). A weak form of the theorem was proven by Ascoli (1883–1884), who established the sufficient condition for compactness, and by Arzelà (1895), who established the necessary condition and gave the first clear presentation of the result. A further generalization of the theorem was proven by Fréchet (1906) for any space in which the notion of a limit makes sense (such as a metric space or Hausdorff space).

The theorem is a fundamental result in mathematics. In particular, it forms the basis for the proof of the Peano existence theorem in the theory of ordinary differential equations and Montel's theorem in complex analysis. It also plays a decisive role in the proof of the Peter-Weyl theorem.

1 Relevant definitions
2 Statements
- 2.1 Real line
- 2.2 Compact metric spaces and compact Hausdorff spaces
3 Proofs
- 3.1 Proof of the sufficient condition for compactness
  - 3.1.1 Alternative proof for sufficiency
- 3.2 Proof of the necessary condition for compactness
4 Examples
5 References
6 See also

[edit] Relevant definitions

Let K be a compact Hausdorff space. A subset $\mathcal{F} \subset C(K, \mathbf{R})$ is said to be equicontinuous if for every x ∈ K and every ε > 0, x has a neighborhood $U x$ such that

| f (y) - f (x) | < ε

for all $y \in U_x$ and ƒ ∈ F.

A set $\mathcal{F} \subset C(K, \mathbf{R})$ is said to be pointwise bounded if for every x ∈ K,

$\sup \{ | f(x) | : f \in \mathcal{F} \} < \infty.$

To avoid potential confusion, one sometimes says that a set $\mathcal{F} \subset C(K, \mathbf{R})$ is uniformly bounded when it is bounded in the usual sense, i.e., bounded in the Banach space C(K) equipped with the uniform norm.

[edit] Statements

[edit] Real line

In simplest terms, the theorem can be stated as follows :

Consider a sequence of continuous functions (f_n)_n∈N defined on a closed and bounded interval [a, b] of the real line with real values. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence (f_{n_k}) that converges uniformly.

For example, the theorem's hypotheses are satisfied by a uniformly bounded sequence of differentiable functions with uniformly bounded derivatives. If the sequence of second derivatives is also uniformly bounded, then the derivatives also converge uniformly (up to a subsequence) , and so on.

Another generalization holds for continuously differentiable functions. Suppose that the functions f_n are continuously differentiable with derivatives f_n′. Suppose that f_n′ are uniformly equicontinuous and uniformly bounded, and that the sequence f_n is pointwise bounded (or just bounded at a single point). Then there is subsequence of the f_n converging uniformly to a continuously differentiable function.

The above theorem also holds if the functions f_n take values in d-dimensional Euclidean space R^d, and the proof is very simple: just apply the R-valued version of the Arzelà–Ascoli theorem d times to extract a subsequence that converges uniformly in the first coordinate, then a sub-subsequence that converges uniformly in the first two coordinates, and so on. More generally, the theorem holds for mappings between two Euclidean spaces.

[edit] Compact metric spaces and compact Hausdorff spaces

The general version of this theorem for metric spaces is as follows :

Let

X

be a compact metric space,

Y

a metric space. Then a subset

F

of

C (X, Y)

is compact in the compact-open topology if and only if it is equicontinuous, pointwise relatively compact and closed.

Here, $C (X, Y)$ denotes the set of all continuous functions from $X$ to $Y$ , and a subset $F$ is pointwise relatively compact if and only if $\forall x \in X$ , the set ${f (x): f$ is in $F}$ is relatively compact in $Y$ . If a set is compact in the compact-open topology, then in particular every sequence from the set has a subsequence which converges uniformly on compact subsets.

More generally, this holds if X is a compact Hausdorff space (Dunford & Schwartz 1958, §IV.6.7):

Let

X

be a compact Hausdorff space. Then a subset

F

of

C (X)

is compact in the compact-open topology if and only if it is equicontinuous, pointwise relatively compact and closed.

The Arzelà–Ascoli theorem is thus a fundamental result in the study of the algebra of continuous functions on a compact Hausdorff space. Various generalizations of the above quoted result are possible. For instance, the functions can assume values in a metric space or (Hausdorff) topological vector space with only minimal changes to the statement (see, for instance, Kelley & Namioka (1982, §8), Kelley (1975, Chapter 7)):

Let

X

be a compact Hausdorff space and

Y

a metric space. Then a subset

F

of

C (X, Y)

is compact in the compact-open topology if and only if it is equicontinuous, pointwise relatively compact and closed.

[edit] Proofs

[edit] Proof of the sufficient condition for compactness

This section gives two versions of the proof of sufficiency.

The first one is a proof of the following version of the theorem, valid for real-valued functions on closed and bounded intervals in R. Proofs of other versions of the theorem are quite similar, provided the necessary parts of the proof are abstracted to the more general situation.

Let I ⊂ R be a closed and bounded interval. If F = {ƒ} is an infinite set of functions ƒ : I → R which is uniformly bounded and equicontinuous, then there is a sequence ƒ_n of elements of F such that ƒ_n converges uniformly on I.

Fix an enumeration {x_i}_i=1,2,3,... of rational numbers in I. Since F is uniformly bounded, the set of points {ƒ(x₁)}_ƒ∈F is bounded, and hence by the Bolzano-Weierstrass theorem, there is a sequence {ƒ_n₁} of distinct functions in F such that {ƒ_n₁(x₁)} converges. Repeating the same argument for the sequence of points {ƒ_n₁(x₂)}, there is a subsequence {ƒ_n₂} of {ƒ_n₁} such that {ƒ_n₂(x₂)} converges.

By mathematical induction this process can be continued, and so there is a chain of subsequences

$\{f_{n_1}\}\supset \{f_{n_2}\}\supset \cdots$

such that, for each k = 1, 2, 3, …, the subsequence {ƒ_{n_k}} converges at x₁,...,x_k. Now form the diagonal subsequence ${f m}$ whose mth term $f m$ is the mth term in the mth subsequence $\{f_{n_m}\}.$ By construction, ƒ_m converges at every rational point of I.

Therefore, given any ε > 0 and rational x_k in I, there is an integer N = N(ε,x_k) such that

$|f_n(x_k) - f_m(x_k)| < \varepsilon/3,\quad n,m \ge N.\,$

Since the family F is equicontinuous, for this fixed ε and for every x in I, there is an open interval U_x containing x such that

$|f(s)-f(t)| < \varepsilon/3\,$

for all ƒ ∈ F and all s, t in I such that s, t ∈ U_x.

The collection of intervals U_x, x ∈ I, forms an open cover of I. Since I is compact, this covering admits a finite subcover U₁, ..., U_J. There exists an integer K such that each open interval U_j, 1 ≤ j ≤ J, contains a rational x_k with 1 ≤ k ≤ K. Finally, for any t ∈ I, there are j and k so that t and x_k belong to the same interval U_j. For this choice of k,

$\begin{align} |f_n(t)-f_m(t)| & {} \le |f_n(t) - f_n(x_k)| + |f_n(x_k) - f_m(x_k)| + |f_m(x_k) - f_m(t)| \\ & {} < \varepsilon/3 + \varepsilon/3 +\varepsilon/3 \end{align}$

for all n, m > N = max{N(ε,x₁), ..., N(ε,x_K)}. Consequently, the sequence {ƒ_n} is uniformly Cauchy, and therefore converges to a continuous function, as claimed. This completes the proof.

[edit] Alternative proof for sufficiency

The following proves that every closed, pointwise bounded, equicontinuous subset $\mathcal{F}$ of $C(K, \mathbf{R})$ is compact when K is a compact space (not necessarily a metric space). Unlike the proof above, separability is not needed in this proof.

Since $C(K, \mathbf{R})$ is complete, $\mathcal{F}$ is complete and thus it suffices to prove that $\mathcal{F}$ is totally bounded. For that, let $ε > 0$ be given. By equicontinuity, each $x \in K$ has a neighborhood $U x$ such that

| f (y) - f (x) | < ε / 3

for all $f \in \mathcal{F}$ and $y \in U_x$ . Since $K$ is compact, there are $z_1, \ldots, z_n \in K$ such that $K$ is the union of $U_{z_1}, \ldots, U_{z_n}$ . Let

$A = \{ (f(z_1), \ldots, f(z_n)) | f \in \mathcal{F} \}$ .

By hypothesis, $A$ is a bounded, thus, totally bounded, subset of $\mathbf{R}^n$ . Thus, we can find $g_1, \ldots, g_m \in \mathcal{F}$ such that:

$A \subset \bigcup_{k=1}^m B( (g_k(z_1), \ldots, g_k(z_n)), \epsilon / 3)$

where B(x, r) denotes an open ball of radius r with center x ∈ Rⁿ. The proof is complete if we show

$\mathcal{F} \subset \bigcup_{k=1}^m B(g_k, \epsilon)$ .

If $f \in \mathcal{F}$ , then we can find $k$ such that

$\|(f(z_1), \ldots, f(z_n)) - (g_k(z_1), \ldots, g_k(z_n))\| < \epsilon / 3$

It follows that $\sup_K |f - g_k| < \epsilon$ . Indeed, if $x \in K$ , then $x \in U_{z_j}$ for some j and so:

$|f(x) - g_k(x)| \le |f(x) - f(z_j)| + |f(z_j) - g_k(z_j)| + |g_k(z_j) - g_k(x)| < \epsilon$ .

$\square$

[edit] Proof of the necessary condition for compactness

Here is the proof that every compact subset $\mathcal{F}$ of $C(K, \mathbf{R})$ is closed, pointwise bounded, and equicontinuous.

If $\mathcal{F}$ is compact, then it is closed, and it is also bounded; a fortiori, pointwise bounded. For equicontinuity, let ε > 0 and x ∈ K be given. For each $f \in \mathcal{F}$ , by continuity, we can find a neighborhood $U f$ of x such that:

$|f(y) - f(x)| < \varepsilon$ for every $y \in U_f$ .

Since $\mathcal{F}$ is compact, $\mathcal{F}$ contains a finite subset E such that $\mathcal{F}$ is the union of the sets of the form

$\{ f; f \in \mathcal{F}, \, \sup_K |f - g| < \varepsilon / 3 \}$

over $g \in E$ . Let U be the intersection of $U g$ over $g \in E$ . (U is a neighborhood of x.) If $f \in \mathcal{F}$ , then there is $g \in E$ with $\sup_K |f - g| < \varepsilon / 3$ , and, for every y ∈ U, one has that:

$|f(y) - f(x)| \le |f(y) - g(y)| + |g(y) - g(x)| + |g(x) - f(x)| < \varepsilon.$

This proves that $\mathcal{F}$ is equicontinuous at x.

[edit] Examples

The set F of functions ƒ on [0, 1] that are bounded by 1 and satisfy a Hölder condition of order α, 0 < α ≤ 1, with a fixed constant M,

$|f(x) - f(y)] \le M \, |x - y|^\alpha, \quad x, y \in [0, 1]$

is compact in C([0, 1]). In other words, the unit ball of the Hölder space C^0, α([0, 1]) is compact in C([0, 1]).

This holds more generally for scalar functions on a compact metric space X satisfying a Hölder condition with respect to the metric on X.

To every function g that is p-integrable on [0, 1], 1 < p ≤ ∞, associate the function G defined on [0, 1] by

$G(x) = \int_0^x g(t) \, \mathrm{d}t.$

Let F be the set of functions G corresponding to functions g in the unit ball of the space L^p([0, 1]). If q is the Hölder conjugate of p, defined by 1/p + 1/q = 1, then Hölder's inequality implies that all functions in F satisfy a Hölder condition with α = 1/q and constant M = 1.

It follows that F is compact in C([0, 1]). This means that the correspondence g → G defines a compact linear operator T between the Banach spaces L^p([0, 1]) and C([0, 1]). Composing with the injection of C([0, 1]) into L^p([0, 1]), one sees that T acts compactly from L^p([0, 1]) to itself. The case p = 2 can be seen as a simple instance of the fact that the injection from the Sobolev space $H^1_0(\Omega)$ into L²(Ω), for Ω a bounded open set in R^d, is compact.

When T is a compact linear operator from a Banach space X to a Banach space Y, its transpose T^∗ is compact from the (continuous) dual Y^∗ to X^∗. This can be checked by the Arzelà–Ascoli theorem.

Indeed, the image T(B) of the closed unit ball B of X is contained in a compact subset K of Y. The unit ball B^∗ of Y^∗ defines, by restricting from Y to K, a set F of (linear) continuous functions on K that is bounded and equicontinuous. By Arzelà–Ascoli, for every sequence {y^∗_n} in B^∗, there is a subsequence that converges uniformly on K, and this implies that the image $T^*(y^*_{n_k})$ of that subsequence is Cauchy in X^∗.

When ƒ is holomorphic in an open disk D₁ = B(z₀, r), with modulus bounded by M, then (for example by Cauchy's formula) its derivative ƒ′ has modulus bounded by 4M ⁄ r in the smaller disk D₂ = B(z₀, r ⁄ 2). If a family of holomorphic functions on D₁ is bounded by M on D₁, it follows that the family F of restrictions to D₂ is equicontinuous on D₂. Therefore, a sequence converging uniformly on D₂ can be extracted. This is a first step in the direction of Montel's theorem.

[edit] References

Arzelà, Cesare (1895), "Sulle funzioni di linee", Mem. Accad. Sci. Ist. Bologna Cl. Sci. Fis. Mat. 5 (5): 55–74 .
Arzelà, Cesare (1882–1883), "Un'osservazione intorno alle serie di funzioni", Rend. Dell' Accad. R. Delle Sci. Dell'Istituto di Bologna: 142–159 .
Ascoli, G. (1883–1884), "Le curve limiti di una varietà data di curve", Atti della R. Accad. Dei Lincei Memorie della Cl. Sci. Fis. Mat. Nat. 18 (3): 521–586 .
Bourbaki, Nicolas (1966), Elements of mathematics, General topology, Hermann .
Dieudonné, Jean (1988), Foundations of modern analysis, Academic Press, ISBN 978-0122155079
Dunford, Nelson; Schwartz, Jacob T. (1958), Linear operators, volume 1, Wiley-Interscience .
Fréchet, Maurice (1906), "Sur quelques points du calcul fonctionnel", Rend. Circ. Mat. Palermo 22: 1–74, doi:10.1007/BF03018603 .
Kelley, J. L. (1975), General topology, Springer-Verlag, ISBN 978-0387901251
Kelley, J. L.; Namioka, I. (1982), Linear Topological Spaces, Springer-Verlag, ISBN 978-0387901695
Rudin, Walter (1976), Principles of mathematical analysis, McGraw-Hill, ISBN 978-0070542358

This article incorporates material from Ascoli–Arzelà theorem on PlanetMath, which is licensed under the GFDL.

[edit] See also

Helly's selection theorem

Arzelà–Ascoli theorem

From Wikipedia, the free encyclopedia

Contents

[edit] Relevant definitions

[edit] Statements

[edit] Real line

[edit] Compact metric spaces and compact Hausdorff spaces

[edit] Proofs

[edit] Proof of the sufficient condition for compactness

[edit] Alternative proof for sufficiency

[edit] Proof of the necessary condition for compactness

[edit] Examples

[edit] References

[edit] See also

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