Calculus of Real and Complex Variables

7.2 Compactness in C

(K, ℝn)

Let K be a nonempty compact set in ℝ^m and consider all the continuous functions defined on this set having values in ℝⁿ. It is desired to give conditions which will show that a subset of C

is compact. First is an important observation about compact sets.

Proof: The last claim is obvious. Indeed, if B

≡, then consider D_1∕k where < r. Then the given ball must contain a point of D_1∕k since its center is within 1∕k of some point of D_k. Now consider the first claime about the ε net. Pick x₁ ∈ K. If B ⊇ K, stop. You have your ε net. Otherwise pick x₂B. If K ⊆ B ∪B, stop. You have found your ε net. Continue this way. Eventually, the process must stop since otherwise, you would have an infinite sequence of points with not limit point because they are all ε apart. This contradicts the compactness of K. ■

Recall Lemma 7.1.4. If you consider C

it is automatically equal to BC because of the extreme value theorem applied to x → for x ∈ K. Therefore, the space C is complete with respect to the norm defined there.

The following is the Arzela Ascoli theorem . Actually, the converse is also true but I will only give the direction of most use in applications.

Proof: Let

_j=1^∞ be a sequence of functions in A. Let D be a countable dense subset of K. Say D ≡_k=1^∞. Then _j=1^∞ is a bounded set of points in ℝⁿ. By the Heine Borel theorem, there is a subsequence, denoted by _j=1^∞ which converges. Now apply what was just done with to and feature d₂ instead of d₁. Thus _j=1^∞ is a subsequence of which converges at d₂. This new subsequence still converges at d₁ thanks to Theorem 2.2.18. Continue this way. Thus we get the following

f(1,1) f(2,1) f(3,1) ⋅⋅⋅ converges at d1 f(1,2) f(2,2) f(3,2) ⋅⋅⋅ converges at d1,d2 f f f ⋅⋅⋅ converges at d ,d ,d (1.,3) (2.,3) (3,.3) . 1 2 3 .. .. .. .. f(1,l) f(2,l) f(3,l) ⋅⋅⋅ converges at dj,j ≤ l . . . . .. .. .. ..

Each subsequence

_j=1^∞ is a subsequence of the one above it _j=1^∞ and converges at d_j for j ≤ r. Consider the subsequence _r=1^∞ the diagonal subsequence. Then _r=j^∞ is a subsequence of _i=1^∞ and so _r=j^∞ converges for each i ≤ j. Since j is arbitrary, this shows that _r=1^∞converges at every point of D as r →∞.

From now on, denote this subsequence of the original sequence as

_k=1^∞. It has the property that it converges at every point of D.

Claim: 

_k=1^∞ converges at every x ∈ K.

Proof of claim: Let ε > 0 be given. Let δ go with ε∕4 in the definition of equicontinuity. Then pick d ∈ D such that

< δ. Then

There exists N such that  if k,l ≥ N, then < . Thus, if k,l ≥ N,

ε ε ε |gk(x)− gl(x)| < 4 + 3 + 4 < ε

which shows that, since ε is arbitrary,

_k=1^∞ is a Cauchy sequence and so it converges to some g. This shows the claim.

It remains to verify that g is continuous. This is implied if we can show uniform convergence of g_k and thus is obtained from the following claim which is a general result about equicontinuous functions.

Claim: If

_k=1^∞ is equicontinuous and converges pointwise to g on a compact set K, then it converges uniformly on K.

Proof of claim: Let ε > 0 be given and let δ go with ε∕4 in the definition of equicontinuous. By compactness, there are finitely many points of K

such that K ⊆∪_i=1^sB. There exists N_i such that if k,l ≥ N_i, then < . Thus if N ≥ max, then for all x_i, < if k ≥ N. Then for k,l ≥ N, and x arbitrary, let x ∈ B. Then

Thus for k,l ≥ N

3ε ∥gl − gk∥∞ <-- < ε 4

This shows

is a Cauchy sequence in C which is complete. Thus there exists g ∈ C such that _∞→ 0. ■

+ class=”left” align=”middle”(K, ℝⁿ)7.3.  THE L^P SPACES