Compact sets in Banach spaces and Hilbert spaces

• The characterization of compact set in C(K) is the Arzel`a-Ascoli theorem. Here, K is a com-pact set in a metric space. It basically says that any bounded and equi-continuous sequence in C(K) has convergent subsequence. One of its application is to prove existence theorem for ODEs ˙x = f (t, x), where f is only continuous in x and L¹in t.

• Characterization of compact set in L²(Ω) is the Rellich theorem. It states that any bounded set in H¹(Ω) is pre-compact in L²(Ω). The condition on Ω is bounded and ∂Ω is Lipschitz continuous.

Here, we recall that a set is pre-compact if its closure is compact.

6.2.1 Compact set in C(K)

We will show in this section that a family of functions F ⊂ C[a, b] is pre-compact if it is bounded in C¹[a, b]. The key part is the boundedness of f⁰for all f ∈ F . This implies that there exists an L such that |f (x) − f (y)| ≤ L|x − y| for all f ∈ F . This condition can be weaken to the following equi-continuity.

6.2. COMPACT SETS IN BANACH SPACES AND HILBERT SPACES 117 Definition 6.4. Let X, Y be metric spaces. A family of function F ⊂ C(X, Y ) is said to be equicontinuous at x if for any  > 0 there exists a δ > 0 such that for any f ∈ F , we have d(f (y), f (x)) <  whenever d(x, y) < δ. The family F is said to be equi-continuous in X if it is equicontinuous at everyx ∈ X. It is called uniformly equi-continuous if the above δ can be chosen independent ofx.

Theorem 6.7. Let K be a compact metric space. If F ⊂ C(K) is equi-continuous, then it is uniformly equi-continuous.

The proof is a simple modification of Theorem 6.2.

Theorem 6.8 (Arzel`a-Ascoli). Let K be a compact metric space. A subset F ⊂ C(K) is pre-compact if and only if it is bounded and equi-continuous.

Proof. (⇒)

1. We show that if F is unbounded, then it cannot be pre-compact. Since F is unbounded, we can find f_n ∈ F such that kf_n+1k_∞ ≥ kf_nk_∞ + 1. Then kf_nk_∞ → ∞. If {f_n} has a convergent subsequence {fnk}, then n_k → ∞ and kf_nkk_∞ is bounded. This is a contradiction.

2. We show that if F is precompact, then F is equicontinuous. For any  > 0,

F ⊂ [

f ∈F

B_/3(f ).

From compactness of F , there exist {f1, ..., fn} such that F ⊂

n

[

i=1

B_/3(f_i).

Each fi is uniformly continuous on K, thus there exist δi > 0 such that

|f_i(x) − fi(y)| < /3 whenever d(x, y) < δi. We choose

δ = min

1≤i≤nδ_i.

Then for every f ∈ F , there exists fisuch that kf − fik_∞< /3. For any d(x, y) < δ,

|f (x) − f (y)| ≤ |f (x) − f_i(x)| + |fi(x) − fi(y)| + |fi(y) − f (y)| < .

Thus, F is equicontinuous.

(⇐) We show that if F is bounded and equicontinuous, then it is pre-compact, i.e. any sequence {f_n} in F has convergent subsequence.

1. First, from K being a compact metric space, then it is separable. Thus, there exist countable {x_i}_i∈N which is dense in K.

2. From boundedness of {fn(x1)} in R, there exists a subsequence from {fn}, called {f1,n}, such that {f1,n(x₁)} converges. Repeating this process, we can choose subsequence {f_2,n} from {f1,n} such that {f2,n(x2)} converges, and so on. Eventually, we obtain nested sub-sequence {f_k,n} such that {f_k,n} ⊂ {f_k−1,n} and f_k,n(x_k) converges. In fact, f_k,n(xi) converges for all xiwith i ≤ k.

3. Let gk = fk,k. {gk} is a subsequence of {fn}. Then for any i ≤ k, {gk} is a subsequence of {f_i,n}. Thus, {g_k(x_i)} converges.

4. Since F is equicontinuous, there exists a δ > 0 such that

|g_k(x) − g_k(y)| < 

3 whenever d(x, y) < δ.

5. Since {xi} is dense in K, we have

K ⊂

∞

[

i=1

B_δ(x_i).

From compactness of K, there exists a finite subset, call it {x1, ..., xp} such that

K ⊂

p

[

i=1

B_δ(xi).

6. Since {g_k(x_i)} are Cauchy, there exists an N such that for any i = 1, ..., p and for any n, m ≥ N , we have

|g_n(x_i) − g_m(x_i)| < /3.

7. Now, for any x ∈ K, there exists x_isuch that x ∈ B_δ(x_i). We have for any n, m ≥ N

|g_n(x) − g_m(x)| ≤ |g_n(x) − g_n(x_i)| + |g_n(x_i) − g_m(x_i)| + |g_m(x_i) − g_m(x)| < .

Thus, {g_k} converges uniformly in K.

6.2.2 Compact sets in Hilbert spaces

Roughly speaking, the compact set in a Hilbert space is almost like a bounded set in a finite dimen-sional subspace. We have the following abstract theorem, which says that the “tail” of a compact set has to be equally small.

6.2. COMPACT SETS IN BANACH SPACES AND HILBERT SPACES 119 Theorem 6.9. Let H be a separable Hilbert space. Suppose D ⊂ H is a bounded set. If in addition, there is an orthonormal basis{e_n} of H such that for any  > 0, there exists an N such that

∞

X

n=N +1

|(x, e_n)|² <  for all x ∈ D, (6.1) thenD is precompact in H.

Proof. Let (xn) be a sequence in D. By the boundedness of D, the sequence (xn) is bounded, say by 1. Let Vn = he1, · · · , eni and Pnbe the orthogonal projection onto Vn. We shall use diagonal process to construct a convergent subsequence of (x_n).

By assumption, given  = 1/k, there exists an Nk, which satisfies Nk≥ N_k−1and

∞

X

Nk+1

|(x, e_n)|² ≤ 1

k for all x ∈ D.

The sequence (PN1(x_n)) is bounded in the finite dimensional space V_N1, by Hein-Borel theorem, it has a convergent subsequence (PN1x1,n). In fact, we select (x1,n) such that

kP_N1(x_1,n− x_1,m)k² ≤ 1

n for all n < m.

Next, we consider P_N2(x_1,n) in V_N2. We can choose convergent subsequence (P_N2x_2,n), n ≥ 2 such that

kP_N2(x2,n− x_2,m)k² ≤ 1

n for all n < m.

Continuing this process, we construct (x_k,n)_n≥k, which is a subsequence of (x_k−1,n)_n≥k−1, such that

kP_Nk(xk,n− x_k,m)k² ≤ 1

n for all n < m.

Now, we claim x_k,kis a Cauchy sequence. For any large k, l with l > k, we have x_l,l ∈ {x_k,n|n ≥ k} and

kx_k,k− x_l,lk² = kP_Nk(x_k,k− x_l,l)k²+ k(I − P_Nk)(x_k,k− x_l,l)k²

≤ kP_Nk(x_k,k− x_l,l)k²+ 2 k

≤ 3

k.

This shows (xk,k) is a Cauchy sequence in H. Hence D is precompact.

The Sobolev space H^s(T), s > 0, is defined to be {u ∈ L²(T)|X

n

(1 + |n|²)^s|(u, e_n)|² < ∞}

Here, en= ^√1

2πe^inx. The Sobolev norm of u is defined by kuk²_s=X

n

(1 + |n|²)^s|(u, e_n)|²

Theorem 6.10. A bounded set D in H^s(T) with s > 0 is a precompact set in L²(T).

Proof. For, if u ∈ D, then

∞

X

N +1

|(u, e_n)|²≤ 1 1 + N^2s

∞

X

N +1

(1 + |n|²)^s|(u, e_n)|²≤ 1

1 + N^2skuk²_s Thus, D has uniformly small tails. Hence it is precompact in L²(T).

Remarks

1. Consider the generalization of Aszel`a-Ascoli theorem to continuous functions in the whole space R^d. We can compactize Rⁿ by Rⁿ∗ := Rⁿ ∪ {∞}. The topology is generated by the union of the neighborhoods of ∞ and the topology of Rⁿ. The topological space Rⁿ∗ is compact. Indeed, for any open cover {Uα|α ∈ A} = Rⁿ∗, there exists Uβ which is open and covers ∞. This means that there exists an M > 0 such that the outer open ball BM(0)^c :=

{x| |x| > M } ⊂ U_β. On the other hand, {U_α|α ∈ A} is also an open cover of the compact set B_M(0). Thus, there exists a finite subcover α₁, ..., α_n. These together with U_βcovers Rⁿ∗. Alternatively, one can identify Rⁿ∗ with Sⁿ in Rⁿ⁺¹ by the polar stereographic transform:

x ∈ Rⁿ∗ 7→ (x⁰, z⁰) ∈ Sⁿwith

|x⁰|

|x| = 1 − z⁰

1 , |x⁰|²+ z⁰²= 1.

These give

z⁰ = |x|²− 1

|x|²+ 1, x⁰ = (1 − |z⁰|)x.

The polar stereographic projection induces a natural metric on Rⁿ∗.

With this, then we can define C(Rⁿ∗) and the equi-continuity for F ⊂ C(Rⁿ∗) as usual. In fact, the equi-continuity at ∞ is: for any  > 0, there exists an M > 0 such that for any f ∈ F and for any x ∈ Rⁿwith |x| > M , we have

|f (x) − f (∞)| < .

2. Consider L²(R). Let α, β > 0, M > 0. Define D = {u ∈ L²(R) |

Z

(1 + |x|²)^α|u(x)|²dx + Z

(1 + |ξ|²)^β|ˆu(ξ)|²dξ ≤ M } Show that D is a precompact set in L²(R). Hint: use Haar basis.

3. In quantum mechanics, one considers the space H := H¹(R³) ∩ L²_V(R³), where V : R³ → R⁺satisfying

V (x) → ∞ as |x| → ∞.

6.3. WEAK CONVERGENCE 121