How to show that $L^p$ norm is monotone increasing?

Question

I am trying to solve the following (very standard) exercise:

Let $(X,\mathcal M,\mu)$ be a measure space and $f\in L^r\cap L^\infty$ for some $1\leqslant r<\infty$. Then $f\in L^p$ for $1\leqslant p < r$ and $$\lim_{p\to\infty} \|f\|_p = \|f\|_\infty.$$

I have worked through the proof here: Limit of $L^p$ norm and find it satisfactory, however I am trying to take a different approach. I'd like to show that $$r\leqslant p \implies \|f\|_r\leqslant \|f\|_p$$ and then use monotone convergence to prove the result. I am stuck on how to prove this inequality though.

No, I am not assuming that $\mu$ is a probability measure. Actually, that is a good point - I have read that the reverse inequality holds in $\ell^p$ spaces (sequence spaces). — Math1000, Aug 14 '15 at 13:51
Read this (and possibly the references): https://en.wikipedia.org/wiki/Lp_space#Embeddings — Siminore, Aug 14 '15 at 13:53

score 10 · Accepted Answer · answered Aug 15 '15 at 10:03

10

In general, the $p$-norms of a function aren't increasing. If there is a measurable set $E$ with $1 < \mu(E) < +\infty$, then

$$\lVert \chi_E\rVert_p = \mu(E)^{\frac{1}{p}}$$

is decreasing. Generally, there are $f$ for which $\lVert f\rVert_p$ is (eventually) decreasing and $f$ for which $\lVert f\rVert_p$ is (eventually) increasing.

If $\mu$ is a probability measure however, then Jensen's inequality shows that for $r \leqslant p <\infty$ we have

$$\int_X \lvert f\rvert^p\,d\mu \geqslant \Biggl(\int_X \lvert f\rvert^r\,d\mu\Biggr)^{\frac{p}{r}},$$

and thus $\lVert f\rVert_r \leqslant \lVert f\rVert_p$ for all $f$.

answered Aug 15 '15 at 10:03

Daniel Fischer

206,697

2

How can we show that $\phi(x)=x^{\frac{p}{r}}$ is convex? – Mog Oct 30 '18 at 09:39
2

@Mog A twice differentiable function is convex if and only if its second derivative is positive. – WeakestTopology Feb 05 '19 at 16:38

How to show that $L^p$ norm is monotone increasing?

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