508 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

Thus you would have φ (yλ )− ε

λd (yλ ,x) ≤ φ (x) for all x which is seen to be what is

wanted.Proof: Let x1 = x0 and define

S1 ≡{

z ∈ X : φ (z)≤ φ (x1)−ε

λd (z,x1)

}Then S1 contains x1 so it is nonempty. It is also clear that S1 is a closed set. This followsfrom the lower semicontinuity of φ . Let x2 be a point of S1, possibly different than x1 andlet

S2 ≡{

z ∈ X : φ (z)≤ φ (x2)−ε

λd (z,x2)

}Continue in this way. Now let there be a sequence of points {xk} such that xk ∈ Sk−1 anddefine Sk by

Sk ≡{

z ∈ X : φ (z)≤ φ (xk)−ε

λd (z,xk)

}where xk is some point of Sk−1. Then xk is a point of Sk. Will this yield a nested sequenceof nonempty closed sets? Yes, it appears that it would because if z ∈ Sk then

φ (z) ≤∈Sk−1

φ (xk)−ε

λd (z,xk)≤

(φ (xk−1)−

ε

λd (xk−1,xk)

)− ε

λd (z,xk)

≤ φ (xk−1)−ε

λd (z,xk−1)

showing that z has what it takes to be in Sk−1. Thus we would obtain a sequence of nested,nonempty, closed sets according to this scheme.

Now here is how to choose the xk ∈ Sk−1. Let

φ (xk)< infx∈Sk−1

φ (x)+12k

Then for z ∈ Sn+1 ⊆ Sn,

φ (z)≤ φ (xn+1)−ε

λd (z,xn+1)

and so

ε

λd (z,xn+1) ≤ φ (xn+1)−φ (z)≤ inf

x∈Snφ (x)+

12n+1 −φ (z)

≤ φ (z)+1

2n+1 −φ (z) =1

2n+1

Thus every z ∈ Sn+1 is within 12n+1 of the single point xn+1 and so the diameter of Sn

converges to 0 as n→ ∞. By completeness of X , there exists a unique yλ ∈ ∩nSn. Then itfollows in particular that for x0 = x1 as above,

φ (yλ )≤ φ (x0)−ε

λd (yλ ,x0)≤ φ (x0)

which verifies the first of the above conclusions.