By Reyna J. A.
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Example text
Proof. Let x ∈ X with x = 1. Find x∗ ∈ X ∗ such that x∗ = x∗ (x) = 1. For every u ∈ X we choose a norm one functional yu∗ in X ∗ such that yu∗ (x+u) = x+u . Then yu∗ + x∗ ≥ yu∗ (x) + x∗ (x) = x + u − yu∗ (u) + 1. Hence yu∗ + x∗ → 2 as u → 0. By the local uniform convexity of X ∗ we deduce that yu∗ − x∗ → 0 as u → 0. It follows that ∗ 0 ≤ x + u + x − u − 2 = yu∗ (x + u) + y−u (x − u) − 2 ∗ ∗ = yu∗ (x) + y−u (x) − 2 + (yu∗ − y−u )(u) ∗ ≤ yu∗ − y−u u = o( u ). 1. Since both X and X ∗ are separable there are a dense sequence (xn )∞ n=1 in the unit sphere of X and an increasing sequence of finite dimensional sub∗ spaces (Fn )∞ n=1 of X such that X∗ = Fn .
This is enough for our purposes, but let us note that we will actually use only the following consequence of uniform equicontinuity. For every ε > 0 there is δ > 0 such that whenever F ∈ F(U ), u, v ∈ U p , y ∈ Y p , and u − v < δ, then |F (u, y) − F (v, y)| < ε. 3. If ε and δ are as above and W is a subspace of X, then for every x, x ˜ ∈ W with x − x ˜ < δ, β(f, x, W, F) ≤ β(f, x ˜, W, F) + ε. Proof. Fix any 0 < c < β(f, x, W, F). By definition of β(f, x, W, F) there is a separable subspace U ⊂ W such that β(f, x, U, F ) > c for every F ∈ F(U ).
F ) + C 2 + Cy ∗ (f (x)(e) − w) > C 2 + C − 2, implying that y ∗ (f (x)(e) − w)) > − 2 (1 + Lip C · (f )), which, multiplied by b − a and for C large enough, gives the statement. So far we have considered what could be termed one-dimensional mean value estimates: although the range could be even infinite dimensional, the estimate involved only derivative in a single direction. 3). The natural formulation may then sound somewhat awkward, but we may follow the same reasoning as above. If, instead of a line segment, we imagine an n-dimensional (say C 1 ) surface in X, then for any Lipschitz f : X −→ Y and u∗1 , .
