172 CHAPTER 6. FIXED POINT THEOREMS

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.As to the second, φ (x0)≤ infx∈X φ (x)+ ε and so, for any x,

φ (yλ )≤ φ (x0)−ε

λd (yλ ,x0)≤ φ (x)+ ε− ε

λd (yλ ,x0) ,

this being true for x = yλ . Hence ε

λd (yλ ,x0)≤ ε and so d (yλ ,x0)≤ λ .

Finally consider the third condition. If it does not hold, then there exists z ̸= yλ such thatφ (yλ ) ≥ φ (z)+ ε

λd (z,yλ ) so that φ (z) ≤ φ (yλ )− ε

λd (z,yλ ) . But then, by the definition

of yλ as being in all the Sn,φ (yλ )≤ φ (xn)− ε

λd (xn,yλ ) and so

φ (z) ≤ φ (xn)−ε

λ(d (xn,yλ )+d (z,yλ ))

≤ φ (xn)−ε

λd (xn,z)

Since n is arbitrary, this shows that z ∈ ∩nSn but there is only one element of this intersec-tion and it is yλ so z must equal yλ , a contradiction. ■

Note how if you make λ very small, you could pick ε very small such that the conelooks pretty flat. Of course, you can always consider an equivalent metric d̂ (x,y) ≡ε

λd (x,y) in all of these considerations.

6.6.1 Cariste Fixed Point TheoremAs mentioned in [22], the above result can be used to prove the Cariste fixed point theorem.

Theorem 6.6.3 Let φ be lower semicontinuous, proper, and bounded below on acomplete metric space X and let F : X →P (X) be set valued such that F (x) ̸= /0 for allx. Also suppose that for each x ∈ X , there exists y ∈ F (x) such that φ (y)≤ φ (x)−d (x,y).Then there exists x0 such that x0 ∈ F (x0).

Proof: In the above Ekeland variational principle, let ε = 1 = λ . Then there exists x0such that for all y ̸= x0

φ (x0)−d (y,x0)< φ (y) , so φ (x0)< φ (y)+d (y,x0) (6.13)

for all y ̸= x0.

x→ φ(x0)−d(x,x0)

φ(x0)