24.1. MOUNTAIN PASS THEOREM IN HILBERT SPACE 813

Proof: First consider whether G(x), the set of pseudogradients of φ at x is nonemptyfor φ

′ (x) ̸= 0. From the definition of the operator norm, there exists u such that ∥u∥X = 1and ⟨φ ′ (x) ,u⟩ ≥ δ ∥φ ′ (x)∥X ′ where δ ∈ (0,1). Then let v = ru∥φ ′ (x)∥X ′ where r ∈ (1,2).⟨

φ′ (x) ,v

⟩=⟨φ′ (x) ,ru

∥∥φ′ (x)

∥∥⟩= r⟨φ′ (x) ,u

⟩∥∥φ′ (x)

∥∥≥ rδ∥∥φ′ (x)

∥∥2

Then choose r,δ such that rδ > 1 and r < 2. Then if these were chosen this way in theabove reasoning, it follows that

∥v∥< 2∥∥φ′ (x)

∥∥ and⟨φ′ (x) ,v

⟩>∥∥φ′ (x)

∥∥2.

That φ′ (x) ̸= 0 is needed to insure that the above strict inequalities hold.

Thus, letting Y be the metric space consisting of the regular points of φ ,the continuityof φ

′ implies that the above inequalities persist for all y close enough to x. Thus there isan open set U containing x such that v satisfies the above inequalities for x replaced witharbitrary y ∈U . Thus

v ∈ ∩y∈U G(y)

Since it is clear that each G(y) is convex, Lemma 24.1.6 implies the existence of a locallyLipschitz selection from G. That is x→V (x) is locally Lipschitz and V (x) ∈ G(x) for allregular x.

It will be important to consider y′ = f (y) where f is locally Lipschitz and y is just ina Banach space. This is more complicated than in Hilbert space because of the lack of aconvenient projection map.

Theorem 24.1.8 Let f : U → X be locally Lipschitz where X is a Banach space and U isan open set. Then there exists a unique local solution to the IVP

y′ = f (y) , y(0) = y0 ∈U

Proof: Let B be a closed ball of radius R centered at y0 such that f has Lipschitzconstant K on B. Then

y1 (t) = y0 +∫ t

0f (y0)ds

and if yn (t) has been obtained,

yn+1 (t) = y0 +∫ t

0f (yn (s))ds (24.1.3)

Now t < T where T is so small that ∥ f (y0)∥TeKT < R.Claim: ∥yn (t)− yn−1 (t)∥ ≤ ∥ f (y0)∥ tnKn−1 1

(n−1)! .

Proof of claim: First

∥y1 (t)− y0∥ ≤∫ t

0∥ f (y0)∥ds≤ ∥ f (y0)∥ t

Now suppose it is so for n. Then

∥yn+1 (t)− yn (t)∥ ≤∫ t

0∥ f (yn (s))− f (yn−1 (s))∥ds