512 CHAPTER 18. TOPOLOGICAL VECTOR SPACES

Theorem 18.6.5 Let X be a Banach space and φ : X → R be Gateaux differentiable,bounded from below, and lower semicontinuous. Also suppose there exists a,c > 0 suchthat

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 18.6.4, 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 dense in X ′. This proves the following lemma.

Lemma 18.6.6 Let X be a Banach space and φ : X→R be Gateaux differentiable, boundedfrom below, and lower semicontinuous. Suppose for all a > 0 there exists a c > 0 such that

φ (x)≥ a∥x∥− c for all x

Then {φ ′ (x) : x ∈ X} is dense in X ′.

If the above holds, thenφ (x)∥x∥

≥ a− c∥x∥

and so, since a is arbitrary, it must be the case that

lim∥x∥→∞

φ (x)∥x∥

= ∞. (18.6.29)

In fact, this is sufficient. If not, there would exist a > 0 such that φ (xn) < a∥xn∥−n. Let−L be a lower bound for φ (x). Then

−L+n≤ a∥xn∥