Notes for Functional Analysis

Wang Zuoqin (typed by Xiyu Zhai)

September 18, 2015

1 Lecture 6

1.1 Topology Deﬁned by Seminorms Let’s start with the example we mentioned at the beginning of lecture 4. 0 X = C0 (R) = {compactly supported continuous functions on R} (1) Then as we have seen, ∞ [ X = Xn, n=1 where 0 Xn = C0 ([−n, n]) = {compactly supported continuous functions whose support lies in (−n, n)} (2) 0 The topology of Xn, as a closed subspace of C ([−n, n]), is determined by open sets ( )

0 B(f, r) = g ∈ C0 ([−n, n]) sup |f(x) − g(x)| < r . x∈[−n,n] In particular, if we let pn(f) = sup |f(x)|, x∈[−n,n] then a local base for this topology is the collection of sets of the form 1 f pn(f) < . k

We would like to assign a topology to X, so that as a subspace of X, the spaces Xn has the same topology mentioned above. The easiest way is to let ( ) 1 0 Un,k = f ∈ C0 (R) pn(f) = sup |f(x)| < (3) x∈[−n,n] k and take F to be the topology generated by these Un,k’s.

1 Remark 1.1. i) As a map, 0 pn : C0 (R) → R, 0 0 is NOT a norm on C0 (R), (although it is a norm on C ([−n, n])), since one can easily 0 ﬁnd a function 0 6= f ∈ C0 (R) such that pn(f) = 0. However, it is easy to see that

• pn ≥ 0,

• pn(f + g) ≤ pn(f) + pn(g),

• pn(αf) = |α|pn(f). ii) With this topology, (X, F) is a locally convex topological vector space.

iii) There are many other interesting spaces of this type.

Deﬁnition 1.2. A semi-norm on a vector space X is a function p : X → [0, +∞) s.t.

i) p(x + y) ≤ p(x) + p(y) (subadditivity)

ii) p(αx) = |α|p(x) (positive homogeneity)

A family of seminorms {pλ} on X is called separating if for any x 6= 0, there exists pλ in the family such that pλ(x) 6= 0. Example 1.1. The family of functions

pn(f) = sup |f(x)| x∈[−n,n]

0 is a countable family of separating semi-norms on C0 (R). Our main theorem in this lecture is

Theorem 1.3. Let P = {pλ} be a separating family of seminorms on a vector space X. For each pλ ∈ P and each k ∈ N we let 1 Uλ,k = x ∈ X pλ(x) < . (4) k

Let B be the collection of all ﬁnite intersection of sets of the form Uλ.k. Let F be the translation-invariant topology on X that has B as a local base at 0. Then:

1. (X, F) is a topological vector space.

2. B is a convex balanced local base.

2 3. Each pλ ∈ P is continuous.

4. A set E ⊂ X is bounded iﬀ every pλ in P is bounded on E. Proof. 1. (X, F) is a topological vector space. a) F is Hausdorﬀ. By translation invariance, it is enough to separate 0 with x 6= 0. We choose pλ ∈ P such that pλ(x) 6= 0. We denote pλ(x) by . Take k large enough such that 1 < . k 2 Then Uλ,k is a neighborhood of 0, x + Uλ,k is a neighborhood of x, and

Uλ,k ∩ (x + Uλ,k) = ∅.

So F is Hausdorﬀ. b) The vector addition is continuous.

Let U be any open neighborhood of x + y. By deﬁnition of F, one can ﬁnd pλ1 , ··· , pλl in P and k1, ··· , kl in N such that

x + y + (Uλ1,k1 ∩ · · · ∩ Uλl,kl ) ⊂ U. (5) Now let

U1 = x + Uλ1,2k1 ∩ · · · ∩ Uλl,2kl which is an open neighborhood of x, and let

U2 = y + Uλ1,2k1 ∩ · · · ∩ Uλl,2kl , which is an open neighborhood of y, then

U1 + U2 ⊂ U.

c)The scalar multiplication is continuous. Let U be a neighborhood of αx, so as above,

αx + Uλ1,k1 ∩ · · · ∩ Uλl,kl ⊂ U for some pλ1 , ··· , pλl ∈ P and k1, ··· , kl ∈ N. Case 1: α = 0. Choose

A > max{pλ1 (x), ··· , pλl (x)}, then for 0 < δ < min( 1 , ··· , 1 , 1), we have 2Ak1 2Akl

(−δ, δ) · (x + Uλ1,2k1 , ··· ,Uλl,2kl ) ⊂ U.

3 [Check this!] Case 2: α 6= 0. Choose 1 1 A > max{pλ1 (x), ··· , pλl (x), , ··· , }, 3k1|α| 3kl|α| and choose δ < min( 1 , ··· , 1 , 1). Then 3k1A 3knA

(α − δ, α + δ) · (x + Uλ1,[3k1|α|]+1 ∩ · · · ∩ Uλl,[3kl|α|]+1) ⊂ U.

2. By deﬁnition B is a local base for F.

To prove Uλ1,k1 ∩· · ·∩Uλl,kl is convex and balanced, it is enough to prove Uλ,k is convex and balanced. It is balanced by positive homogeneity. It is convex since for any x, y ∈ Uλ,k and any 0 ≤ α ≤ 1, 1 1 1 p (αx + (1 − α)y) ≤ αp (x) + (1 − α)p (y) < α · + (1 − α) = . λ λ λ k k k

3. By deﬁnition each pλ is continuous at 0. The continuity of pλ at x follows from 1 1 pλ(x + Uλ, 1 ) ⊂ pλ(x) − , pλ(x) + . k k k

(This can be proved using the triangle inequality) 4. Suppose E is bounded, and pλ ∈ P. Then there exists t > 0 such that

E ⊂ tUλ,1,

1 i.e., t E ⊂ Uλ,1. So pλ(x) < t, ∀x ∈ E.

Conversely suppose each pλ is bounded on E. Then for any neighborhood

U ⊃ Uλ1,k1 ∩ · · · ∩ Uλl,kl of 0, one pick t1, ··· , tl such that

pλi (x) < ti for all x ∈ E. Then for t > max(t1k1, ··· , tlkl), we have

E ⊂ t(Uλ1,k1 ∩ · · · ∩ Uλl,kl ).

4 1.2 Fr´echet Spaces 0 In many applications, as in the case of C0 (R), the topology is deﬁned by a countable sequence of seminorms. As a result, the local base B contains only countable elements, and thus F is metrizable. In fact, we can explicitly write down a compatible translation- invariant metric in this case.

Proposition 1.4. If the topology on X is deﬁned by a separating sequence {pn} of seminorms, then for any sequence of positive numbers {ci} that tends to 0, c p (x − y) d(x, y) = max i i (6) i 1 + pi(x − y) is a compatible translation-invariant metric on X.

Proof. Exercise.

Deﬁnition 1.5. A Frech´etspace is a topological vector space whose topology is induced by a countable family of seminorms, and the compatible translation-invariant metric is complete.

Remark 1.6. If d1 and d2 are both compatible translation-invariant metric for a topological vector space X, then d1 is complete iff d2 is complete. So the definition above is indepen- dent of the choice of compatible translation-invariant metrics. [However, we have seen in problem set 2 part 2, problem 1, that one may find two compatible metrics, not necessary translation-invariant, so that one is complete while the other is not complete.] Example 1.2. X = C(R) is Fréchet. In fact, for

pn(f) = max |f(x)|, −n≤x≤n it is easy to see that pn form a countable separating family of seminorms. Since p1 ≤ p2 ≤ p3 ≤ · · · , the sets 1 Vn = f ∈ C(R) pn(f) < (7) n form a convex local base for the topology on X, which is compatible with

1 p (f − g) d(f, g) = max n . (8) n n 1 + pn(f − g)

This metric is complete: if {fn} is a Cauchy sequence in (X, d), then for any n,

pn(fi − fj) → 0

5 as i, j → ∞. It follows that fi converges to a function f(n) uniformly on [−n, n]. Obviously

f(n) = f(m) on [−n, n] if n < m. So we get a function f on R. Since

pn(fi − f) → 0 as i → ∞ for each n, we see that d(fi, f) → 0. 0 Remark 1.7. C0 (R) is NOT Fréchet since it is NOT complete. In general, for any open set n Ω ⊂ R , the space C(Ω) is Fréchet. The proof is the same as above: one just need to find a sequence of compact sets K1 ⊂ K2 ⊂ K3 ⊂ · · · so that ∪Kn = Ω.