Kenneth Kuttler

12.3. HAHN BANACH THEOREM 309

5.) From 4.) and algebra,

0 = limk→∞

∣∣∣∣∣x− k

∑n=1

(x,dn)dn

∣∣∣∣∣2

= limk→∞

(|x|2 +

∑n=1|(x,dn)|2−2

∑n=1|(x,dn)|2

)

= limk→∞

(|x|2−

∑n=1|(x,dn)|2

)■

12.3 Hahn Banach TheoremThe closed graph, open mapping, and uniform boundedness theorems are the three majortopological theorems in functional analysis. The other major theorem is the Hahn-Banachtheorem which has nothing to do with topology.

12.3.1 Gauge Functions and Hahn Banach Theorem

Definition 12.3.1 Let X be a real vector space ρ : X→R is called a gauge functionif

ρ(x+ y)≤ ρ(x)+ρ(y), ρ(ax) = aρ(x) if a≥ 0. (12.9)

Suppose M is a subspace of X and z /∈ M. Suppose also that f is a linear real-valuedfunction having the property that f (x) ≤ ρ(x) for all x ∈ M. Consider the problem ofextending f to M⊕Rz such that if F is the extended function, F(y) ≤ ρ(y) for all y ∈M⊕Rz and F is linear. Since F is to be linear, it suffices to determine how to define F(z).Letting a > 0, it is required to define F (z) such that the following hold for all x,y ∈M.

f (x)︷︸︸︷F (x)+aF (z) = F(x+az)≤ ρ(x+az),

f (y)︷︸︸︷F (y)−aF (z) = F(y−az)≤ ρ(y−az). (12.10)

Therefore, multiplying by a−1 12.9 implies that what is needed is to choose F (z) such thatfor all x,y ∈M,

f (x)+F(z)≤ ρ(x+ z), f (y)−ρ(y− z)≤ F(z)

and that if F (z) can be chosen in this way, this will satisfy 12.10 for all x,y and the problemof extending f will be solved. Hence it is necessary to choose F(z) such that for all x,y∈M

f (y)−ρ(y− z)≤ F(z)≤ ρ(x+ z)− f (x). (12.11)

Is there any such number between f (y)−ρ(y− z) and ρ(x+ z)− f (x) for every pair x,y ∈M? This is where f (x)≤ ρ(x) on M and that f is linear is used. For x,y ∈M,

ρ(x+ z)− f (x)− [ f (y)−ρ(y− z)]

= ρ(x+ z)+ρ(y− z)− ( f (x)+ f (y))≥ ρ(x+ y)− f (x+ y)≥ 0.

Therefore there exists a number between the following two numbers

sup{ f (y)−ρ(y− z) : y ∈M} , inf{ρ(x+ z)− f (x) : x ∈M} .

Choose F(z) to satisfy 12.11. This has proved the following lemma.

12.3. HAHN BANACH THEOREM 3095.) From 4.) and algebra,0 = tlmk—- 00klim Gi y (sah 7kx= y? (x, dn) dnn=1 n=1 n=12 k k= him (I+ Y |(x,dn)?-2¥. (sn?)k-00 nel12.3. Hahn Banach TheoremThe closed graph, open mapping, and uniform boundedness theorems are the three majortopological theorems in functional analysis. The other major theorem is the Hahn-Banachtheorem which has nothing to do with topology.12.3.1 Gauge Functions and Hahn Banach TheoremDefinition 12.3.1 Lex be a real vector space p :X — Ris called a gauge functionifp(xty) <p) +ply), plax) = ap(x) ifa > 0. (12.9)Suppose M is a subspace of X and z ¢ M. Suppose also that f is a linear real-valuedfunction having the property that f(x) < p(x) for all x € M. Consider the problem ofextending f to M@Rz such that if F is the extended function, F(y) < p(y) for all y €M ®Rzand F is linear. Since F is to be linear, it suffices to determine how to define F(z).Letting a > 0, it is required to define F (z) such that the following hold for all x,y € M.J (x)“——sF (x) +-aF (z) = F(x+az) < p(x+az),f—y)=}y) —aF (z) = F(y—az) < p(y—az). (12.10)Therefore, multiplying by a~! 12.9 implies that what is needed is to choose F (z) such thatfor all x,y € M,is)F(x) + F(z) S p(x+z), fy) -py—2) S Fle)and that if F (z) can be chosen in this way, this will satisfy 12.10 for all x,y and the problemof extending f will be solved. Hence it is necessary to choose F(z) such that for all x,y € Mf(y) —p(y-z) S F(z) S$ p(x+z) — f(). (12.11)Is there any such number between f(y) — p(y—z) and p(x+z) — f(x) for every pair x,y €M? This is where f(x) < p(x) on M and that f is linear is used. For x,y € M,p(x+z) — f(x) -[f) -—p(v—-2)]=p(x+z)+ply—z)— (f(x) + f(y) 2 p(wt+y) — f(xty) 2 0.Therefore there exists a number between the following two numberssup {f(y) —p(y—z):y EM}, inf{p(x+z)— f(x) :x eM}.Choose F(z) to satisfy 12.11. This has proved the following lemma.