Finite-time bounds for Banach-valued derivatives

Given: Countable Choice, a finite interval with Lebesgue measure, a real or complex Banach value space, strongly measurable derivative representatives, and the stated uniform norm bound.

Goal: Convert the derivative hypothesis to the exact Bochner $L^1$ bound without changing its value space.

The derivative is the Bochner distributional identity, whose norm is the norm of its representative (Distributional time derivatives with values in a Banach space).

The essential supremum is an almost-everywhere bound (The essential supremum is attained as the least essential bound, The essential supremum of a measurable function with respect to a measure).

The Bochner norm convention and finite-exponent conversion are supplied by Bochner spaces and bounded inclusion of value spaces; the conversion is its last assertion, obtained by Holder's inequality for integrals, including the endpoint cases.

A Banach space has its specified real or complex scalar field and complete norm metric (Banach space). Strong measurability is almost-everywhere norm approximation by measurable simple functions (Strongly measurable Banach-valued function).

Under Countable Choice, the one-dimensional open-box case of A box in ℝⁿ with parameters aᵢ≤ bᵢ is Lebesgue measurable of measure ∏_i<n(bᵢ-aᵢ), whichever of its faces are included, with endpoints $0$ and $T$, gives ${\lambda }_1((0,T))=T$.

Countable Choice (The Axiom of Countable Choice (AC_ω)) is assumed to meet the hypotheses of the derivative convention in [F1], the Bochner-space supplier in [F3], and the interval-measure supplier in [F5]. No necessity claim for the scalar norm estimates is asserted.

The endpoint $(\mathrm{\infty },1)$ case of Holder's inequality for integrals, including the endpoint cases gives ${\int }_I\mathrm{\mid }ab\mathrm{\mid }dt\le \mathrm{\parallel }a{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I)}\mathrm{\parallel }b{\mathrm{\parallel }}_{L^1(I)}$ for measurable real-valued $a\in L^{\mathrm{\infty }}(I)$ and $b\in L^1(I)$, where $I=(0,T)$.

On $I=(0,T)$, the constant function $1={\chi }_I$ has simple integral ${\lambda }_1(I)$ by The integral of a nonnegative simple function. Its nonnegative Lebesgue integral is the same by The nonnegative integral agrees with the simple integral on simple functions. Thus [F5] gives $\mathrm{\parallel }1{\mathrm{\parallel }}_{L^1(I)}=T$.