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 &sub; R be a closed and bounded interval. If F = {&fnof;} is an infinite set of functions &fnof; : I &rarr; R which is uniformly bounded and equicontinuous, then there is a sequence &fnof;n of elements of F such that &fnof;n converges uniformly on I.

Fix an enumeration {xi}undefined of rational numbers in I. Since F is uniformly bounded, the set of points {&fnof;(x1)}&fnof;&isin;F is bounded, and hence by the Bolzano-Weierstrass theorem, there is a sequence {&fnof;n 1 } of distinct functions in F such that {&fnof;n 1 (x1)} converges. Repeating the same argument for the sequence of points {&fnof;n 1 (x2)}, there is a subsequence {&fnof;n 2 } of {&fnof;n 1 } such that {&fnof;n 2 (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 {&fnof;n k } converges at x1,...,xk. 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, &fnof;m converges at every rational point of I.

Therefore, given any &epsilon; > 0 and rational xk in I, there is an integer N = N(&epsilon;,xk) 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 &epsilon; and for every x in I, there is an open interval Ux containing x such that


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

for all &fnof; &isin; F and all s, t in I such that s, t &isin; Ux.

The collection of intervals Ux, x &isin; I, forms an open cover of I. Since I is compact, this covering admits a finite subcover U1, ..., UJ. There exists an integer K such that each open interval Uj, 1 &le; j &le; J, contains a rational xk with 1 &le; k &le; K. Finally, for any t &isin; I, there are j and k so that t and xk belong to the same interval Uj. For this choice of k,



\begin{align} & {} < \varepsilon/3 + \varepsilon/3 +\varepsilon/3 \end{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)| \\

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

Examples

 * The set F of functions &fnof; on [0, 1] that are bounded by 1 and satisfy a Hölder condition of order α, 0 < &alpha; &le; 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&thinsp; $$C^{0, \alpha}([0, 1])$$ 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 &le; &infin;, 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 Lp([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 &alpha; = 1/q and constant M = 1.


 * It follows that F is compact in C([0, 1]). This means that the correspondence g &rarr; G defines a compact linear operator T&thinsp; between the Banach spaces Lp([0, 1]) and C([0, 1]).


 * When T is a compact linear operator from a Banach space X&thinsp; to a Banach space Y, its transpose T&lowast; is compact from the (continuous) dual Y&lowast; to X&lowast;. 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&lowast; of Y&lowast; defines, by restricting to K, a set F&thinsp; of (linear) continuous functions on K that is bounded and equicontinuous. For every sequence {y&lowast;n} in B&lowast;, there is a subsequence that converges uniformly on K, and this implies that the image $$T^*(y^*_{n_k})$$&thinsp; of that subsequence is Cauchy in X&lowast;.


 * When &fnof; is holomorphic in a disk D1 = B(z0, r), with modulus bounded by M, then (for example by Cauchy's formula) its derivative &fnof;&prime; 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 {en} 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 en. Suppose that orthonormal eigenvectors e0, &hellip;, en&minus;1 of T have been found. Then span(e0, &hellip;, en&minus;1)  is invariant under T, and by self-adjointness,  the orthogonal  complement Hn of span(e0, &hellip;, en&minus;1) is an invariant subspace of  T. Let Tn denote the restriction of T to Hn.  If m(Tn) = 0, then  Tn = 0, and the construction stops. Otherwise, by the claim above, applied to Tn, there is a norm one eigenvector en of T in  Hn,  with corresponding non-zero eigenvalue  &lambda;n =  &plusmn;&thinsp;m(Tn).

At this point, a finite or infinite orthonormal sequence {en} of eigenvectors of T, with non-zero  eigenvalues, has been constructed. Let F = (span{en})&perp;; 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 em&minus;1, then  F = Hm and S =  Tm = 0 by construction.

In the infinite case, compactness of T and the weak-convergence of en to 0 imply that T&thinsp;en=  &lambda;n en &rarr; 0,  therefore &lambda;n &rarr; 0. Since F is contained in Hn for every n, it follows that m(S) &le; m(Tn) =  |&lambda;n| for every n, hence  m(S) = 0. This implies again that.

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  en with non-zero eigenvalue. It follows that, and span{en} is the  orthogonal complement of the kernel of T. One can complete the diagonalization of T by selecting an orthonormal basis of the kernel.