Bourbaki

Exercise 10, § 6

Integration, IV, p. 136

Bourbaki marks this one as harder than the rest.

a) In a Banach space FF, let a\mathbf{a} and b\mathbf{b} be two vectors such that ∣a∣=∣b∣=1|\mathbf{a}| = |\mathbf{b}| = 1. Show that for every number tt such that 0≤t≤10 \leq t \leq 1, and for every pp such that 1≤p<+∞1 \leq p < +\infty, (1)

∣a−tb∣p≤2p∣a−tpb∣ |\mathbf{a} - t\mathbf{b}|^p \leq 2^p |\mathbf{a} - t^p\mathbf{b}|
(2)
∣a−tpb∣≤3p∣a−tb∣ |\mathbf{a} - t^p\mathbf{b}| \leq 3^p |\mathbf{a} - t\mathbf{b}|
(express a−tpb\mathbf{a} - t^p\mathbf{b} as a linear combination of a−tb\mathbf{a} - t\mathbf{b} and a−b\mathbf{a} - \mathbf{b} and note that for 0≤p≤10 \leq p \leq 1, one has ∣a−pb∣≥1−p|\mathbf{a} - p\mathbf{b}| \geq 1 - p and ∣a−b∣≤2∣a−pb∣|\mathbf{a} - \mathbf{b}| \leq 2|\mathbf{a} - p\mathbf{b}|). From this, deduce that if y\mathbf{y} and z\mathbf{z} are any two vectors in FF, then (3)
∣y−z∣p≤2p∣∣y∣p−1⋅y−∣z∣p−1⋅z∣ |\mathbf{y} - \mathbf{z}|^p \leq 2^p \left| |\mathbf{y}|^{p-1} \cdot \mathbf{y} - |\mathbf{z}|^{p-1} \cdot \mathbf{z} \right|
(4)
∣∣y∣p−1⋅y−∣z∣p−1⋅z∣≤3p∣y−z∣(∣y∣+∣z∣)p−1. \left| |\mathbf{y}|^{p-1} \cdot \mathbf{y} - |\mathbf{z}|^{p-1} \cdot \mathbf{z} \right| \leq 3^p |\mathbf{y} - \mathbf{z}|(|\mathbf{y}| + |\mathbf{z}|)^{p-1}.

b) Show that the mapping f↦∣f∣(1/p)−1⋅ff \mapsto |f|^{(1/p)-1} \cdot f is a uniformly continuous bijective mapping of LF1\mathscr{L}_F^1 onto LFp\mathscr{L}_F^p (make use of the inequality (3)).

c) Show that the mapping f↦∣f∣p−1⋅ff \mapsto |f|^{p-1} \cdot f of LFp\mathscr{L}_F^p onto LF1\mathscr{L}_F^1 is uniformly continuous on every bounded subset of the space LFp\mathscr{L}_F^p (make use of the inequality (4) and Hölder’s inequality). Deduce from this that the topological spaces LF1\mathscr{L}_F^1 and LFp\mathscr{L}_F^p are homeomorphic.

Solution

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Transcribed from the printed volume. Not translated.