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Mean value inequality for a differentiable Banach-valued curve

Statement

Let X be a Banach space, a<b, and let φ:[a,b]→X be continuous on [a,b] and differentiable on (a,b) in the sense of Fréchet derivative between Banach spaces. If there is C≥0 with ∥φ′(s)∥≤C for all s∈(a,b), then ∥φ(b)−φ(a)∥≤C(b−a). In particular, if φ is continuous, differentiable on (a,b) and φ′=0 there, then φ is constant.

Facts & Assumptions

Given: A real or complex Banach space X (Banach space), read as a real vector space for differentiation so that the real-variable Fréchet derivative of Fréchet derivative between Banach spaces applies (the complex case uses the underlying real structure, and complex differentiability is a special case); real numbers a<b, a curve φ:[a,b]→X continuous on [a,b] and differentiable on (a,b), and a constant C≥0 with ∥φ′(s)∥≤C for all s∈(a,b).

[F1]

At each s∈(a,b) the Fréchet derivative φ′(s):R→X is bounded linear and there is a remainder with ∥φ(s+h)−φ(s)−φ′(s)h∥/∣h∣→0 as h→0 (Fréchet derivative between Banach spaces).

[F2]

The norm function is continuous on X and satisfies the reverse triangle inequality ∣∥x∥−∥y∥∣≤∥x−y∥ (The reverse triangle inequality in a normed space).

[F3]

Vector addition and scalar multiplication on X are continuous (Vector addition and scalar multiplication are continuous in a normed space), so limits of sums and scalar multiples may be taken termwise.

Proof

technique · direct, by a maximal-interval argument on the auxiliary function $d(\tau)=\|\varphi(\tau)-\varphi(r)\|-\widetilde C(\tau-r)$
1.1F2F3

Fix C~>C and r∈(a,b), and put d(τ):=∥φ(τ)−φ(r)∥−C~(τ−r) for τ∈[r,b]. By [F2] and [F3] the function d is continuous, so S:={τ∈[r,b]:d(τ)≤0} is a nonempty closed subset of [r,b], since r∈S; being nonempty and bounded above it has a supremum τ0∈S, which is therefore its maximum.

2.1F1F2step 1.1

If τ0<b then τ0>r: if τ0=r, then [F1] at r (legitimate because r>a) gives φ(r+h)=φ(r)+φ′(r)h+o(h) with ∥φ′(r)∥≤C, so by [F2] d(r+h)≤(C−C~)h+o(h)<0 for all sufficiently small h>0, contradicting the maximality of r=τ0; hence τ0∈(r,b)⊆(a,b), where φ is differentiable.

3.1F1F2step 2.1

If τ0<b, then differentiability at τ0 gives φ(τ0+h)=φ(τ0)+φ′(τ0)h+o(h) with ∥φ′(τ0)∥≤C, so for small h>0 [F2] gives d(τ0+h)≤d(τ0)+(C−C~)h+o(h)≤(C−C~)h+o(h)<0, since d(τ0)≤0; then τ0+h∈S, contradicting the maximality of τ0. Hence τ0=b.

4.1step 3.1

At τ0=b the defining inequality of S reads d(b)≤0, that is ∥φ(b)−φ(r)∥≤C~(b−r).

5.1F2F3step 4.1

Letting r↓a along a sequence: φ(r)→φ(a) by continuity and [F3], so [F2] gives ∥φ(b)−φ(r)∥→∥φ(b)−φ(a)∥, and b−r→b−a; hence ∥φ(b)−φ(a)∥≤C~(b−a).

6.1step 5.1algebra

Since C~>C was arbitrary, ∥φ(b)−φ(a)∥≤C(b−a).

7.1step 6.1algebra∎

If in addition φ′=0 on (a,b), take C=0 in [step 6.1]; then for every t∈(a,b] the same argument applied to the restriction of φ to [a,t] gives φ(t)=φ(a), and φ(a)=φ(a), so φ is constant on [a,b].

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