6.6. EKELAND’S VARIATIONAL PRINCIPLE 171

1. φ (yλ )≤ φ (x0)

2. d (yλ ,x0)≤ λ

3. φ (yλ )− ε

λd (x,yλ )< φ (x) for all x ̸= yλ

To motivate the proof, see the following picture which illustrates the first two steps.The Siwill be sets in X but are denoted symbolically by labeling them in X× (−∞,∞].

x1

x2

S1 S1

Then the end result of this iteration would be a picture like the following.

yλ

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 con-

tains x1 so it is nonempty. It is also clear that S1 is a closed set. This follows from the lowersemicontinuity of φ . Suppose

Sk ≡{

z ∈ X : φ (z)≤ φ (xk)−ε

λd (z,xk)

}where xk ∈ Sk−1. Pick xk+1 ∈ Sk and define Sk+1 similarly. 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