User:Bdmy/Arzela

Proof of a version of Arzelà–Ascoli theorem

We will prove here 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_nk} 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

$|f_{n}(x_{k})-f_{m}(x_{k})|<\varepsilon /3,\quad n,m\geq N.\,$

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,

${\begin{aligned}|f_{n}(t)-f_{m}(t)|&{}\leq |f_{n}(t)-f_{n}(x_{k})|+|f_{n}(x_{k})-f_{m}(x_{k})|+|f_{m}(x_{k})-f_{m}(t)|\\&{}<\varepsilon /3+\varepsilon /3+\varepsilon /3\end{aligned}}$

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.

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,

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

is compact in C([0, 1]). In other words, the unit ball of the Hölder space  $C^{0,\alpha }([0,1])$ 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]).

• 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 to K, a set F  of (linear) continuous functions on K that is bounded and equicontinuous. 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 a 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. This is a first step in the direction of Montel's theorem.

Compact operators

Let T be a compact operator on a Hilbert space H. A finite (possibly empty) or countably infinite orthonormal sequence {e_n} of eigenvectors of T, with corresponding non-zero eigenvalues, will be constructed by induction as follows. Let H₀ = H and T₀ = T. If m(T₀) = 0, then T = 0 and the construction stops without producing any eigenvector e_n. Suppose that orthonormal eigenvectors e₀, …, e_n−1 of T have been found. Then span(e₀, …, e_n−1) is invariant under T, and by self-adjointness, the orthogonal complement H_n of span(e₀, …, e_n−1) is an invariant subspace of T. Let T_n denote the restriction of T to H_n. If m(T_n) = 0, then T_n = 0, and the construction stops. Otherwise, by the claim above, applied to T_n, there is a norm one eigenvector e_n of T in H_n, with corresponding non-zero eigenvalue λ_n = ± m(T_n).

At this point, a finite or infinite orthonormal sequence {e_n} of eigenvectors of T, with non-zero eigenvalues, has been constructed. Let F = (span{e_n})^⊥; by self-adjointness, F is invariant under T. Let S denote the restriction of T to F. If the process was stopped after finitely many steps, with a last vector e_m−1, then F = H_m and S = T_m = 0 by construction.

In the infinite case, compactness of T and the weak-convergence of e_n to 0 imply that T e_n = λ_n e_n → 0, therefore λ_n → 0. Since F is contained in H_n for every n, it follows that m(S) ≤ m(T_n) = |λ_n| for every n, hence m(S) = 0. This implies again that S = 0.

The fact that S = 0 means that F is contained in the kernel of T. Conversely, if x is in ker(T), then by self-adjointness, x is orthogonal to every eigenvector e_n with non-zero eigenvalue. It follows that F = ker(T), and span{e_n} is the orthogonal complement of the kernel of T. One can complete the diagonalization of T by selecting an orthonormal basis of the kernel.