174 CHAPTER 6. FIXED POINT THEOREMS

Proof: From the Ekeland variational principle with λ = 1, there exists xε such thatφ (xε) ≤ φ (x0) ≤ infx∈X φ (x)+ ε and for all x, φ (xε) < φ (x)+ ε ∥x− xε∥. Then lettingx = xε + hv where ∥v∥ = 1,φ (xε +hv)−φ (xε) > −ε |h|. Let h < 0. Then divide by it toobtain φ(xε+hv)−φ(xε )

h < ε . Passing to a limit as h→ 0 yields φ′ (x)(v) ≤ ε . Now v was

arbitrary with norm 1 and so sup∥v∥=1 |φ ′ (xε)(v)|= ∥φ ′ (xε)∥ ≤ ε ■There is another very interesting application of the Ekeland variational principle [22].

Theorem 6.6.5 Let X be a real Banach space and φ : X → R be Gateaux differen-tiable, bounded from below, and lower semicontinuous. Also suppose there exists a,c > 0such that

a∥x∥− c≤ φ (x) for all x ∈ X

Then {φ ′ (x) : x ∈ X} is dense in the ball of X ′ centered at 0 with radius a. Here φ′ (x) ∈ X ′

and is determined by

φ′ (x)(v)≡ lim

h→0

φ (x+hv)−φ (x)h

Proof: Let x∗ ∈ X ′,∥x∗∥ ≤ a. Let ψ (x) = φ (x)−x∗ (x) . This is lower semicontinuous.It is also bounded from below because

ψ (x)≥ φ (x)−a∥x∥ ≥ (a∥x∥− c)−a∥x∥=−c

It is also clearly Gateaux differentiable and lower semicontinuous because the piece addedin is actually continuous. It is clear that the Gateaux derivative is just φ

′ (x)− x∗. ByTheorem 6.6.4, there exists xε such that ∥φ ′ (xε)− x∗∥ ≤ ε ■

Thus this theorem says that if φ (x) ≥ a∥x∥− c where φ has the nice properties of thetheorem, it follows that φ

′ (x) is dense in B(0,a) in the dual space X ′. It follows that if forevery a, there exists c such that φ (x)≥ a∥x∥−c for all x ∈ X then {φ ′ (x) : x ∈ X} is densein X ′. This proves the following lemma.

Lemma 6.6.6 Let X be a real Banach space and φ : X → R be Gateaux differentiable,bounded from below, and lower semicontinuous. Suppose for all a > 0 there exists a c > 0such that φ (x)≥ a∥x∥− c for all x. Then {φ ′ (x) : x ∈ X} is dense in X ′.

If the above holds, then φ(x)∥x∥ ≥ a− c

∥x∥ and so, since a is arbitrary, it must be the casethat

lim∥x∥→∞

φ (x)∥x∥

= ∞. (6.14)

In fact, this is sufficient to conclude that for each a > 0 there is c > 0 such that φ (x) ≥a∥x∥− c. If not, there would exist a > 0 such that φ (xn)< a∥xn∥−n. Let −L be a lowerbound for φ (x). Then −L+n≤ a∥xn∥ and so ∥xn∥→ ∞. Now it follows that

a≥ φ (xn)

∥xn∥+

n∥xn∥

≥ φ (xn)

∥xn∥(6.15)

which is a contradiction to 6.14. This proves the following interesting density theorem.

Theorem 6.6.7 Let X be a real Banach space and φ : X → R be Gateaux differen-tiable, bounded from below, and lower semicontinuous. Also suppose the coercivity condi-tion

lim∥x∥→∞

φ (x)∥x∥

= ∞