— Let p and q be two real numbers such that 1≤p<+∞, 1≤q<+∞. If a function f belongs to LFp then the function ∣f∣(p/q)−1⋅f belongs to LFq, and conversely.
By hypothesis, there exists a sequence (fn) of continuous functions with compact support such that ∑n=1∞Np(fn)<+∞ and f(x)=∑n=1∞fn(x) almost everywhere (No. 4, Th. 3). Set
gn=∣f1+f2+⋯+fn∣(p/q)−1⋅(f1+f2+⋯+fn);
the function gn is continuous with compact support; on the other hand,
∣gn∣q=∣f1+f2+⋯+fn∣p≤(n=1∑∞∣fn∣)p=hq,
where the numerical function h⩾0 (finite or not) satisfies the inequality
(Nq(h))q=(Np(n=1∑∞∣fn∣))p⩽(n=1∑∞Np(fn))p<+∞
by the countable convexity theorem. Moreover, gn(x) tends almost everywhere to g(x)=∣f(x)∣(p/q)−1⋅f(x), therefore Lebesgue’s theorem shows that g∈LFq. The converse is immediate, since f=∣g∣(q/p)−1⋅g.
It can be shown that the mapping f↦∣f∣qp−1⋅f is a homeomorphism of LFp onto LFq (§ 6, Exer. 10).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.