18.6. A VARIATIONAL PRINCIPLE OF EKELAND 509

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 suchthat

φ (yλ )≥ φ (z)+ε

λd (z,yλ )

so thatφ (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.

18.6.1 Cariste Fixed Point TheoremAs mentioned in [55], the above result can be used to prove a fixed point theorem calledthe Cariste fixed point theorem.

Theorem 18.6.3 Let φ be lower semicontinuous, proper, and bounded below on a completemetric space X and let F : X →P (X) be set valued such that F (x) ̸= /0 for all x. Alsosuppose 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) (18.6.28)

for all y ̸= x0.