Let (X,μ) be a finite measure space and f:X→[0,∞] be a measurable function. Define a function φ:R+→R+ by φ(r)=(1μ(X)∫Xfrdμ)1r. Prove that φ(r)≤φ(s) if $r
Answer
We have
φ(r)=(1μ(X)∫Xfr dμ)1r=(∫Xfr 1μ(X) dμ)1r.
Now, by Hölder's inequality, in general, for measurable functions F,G and conjugate exponents p,q, that is, 1p+1q=1 where $1
ϕ(r)=(∫XFG dμ)1r⩽(∫XFsr dμ)rrs(∫XGss−r)s−rrs=(∫Xfs dμ)1s(∫X(1μ(X))ss−r dx)s−rrs=(∫Xfs dμ)1s(μ(X)sr−sμ(X))s−rrs=(1μ(X)∫Xfs dμ)1s=φ(s),
as desired.
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