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Let $0 < p < q \leq \infty$ and suppose $ E\subset \mathbb{R}^N$ with $m(E)=1$ (where $m$ is the Lebesgue measure). I am asked to find necessary and sufficient conditions for: $$ ( \int_E{|f|}^pdx)^{1/p} = (\int_E{|f|}^qdx)^{1/q}$$ I know how to prove that that LHS $\leq$ RHS, this is done with Jensen's inequality in its measure-theoretic form, considering the function $\phi(t)=t^{q/p}$, which is convex, because $p<q$. This is how it works:
$$(\int_E{|f|^pdx})^{q/p}=\phi(\int_E{|f|^p}dx) \leq \int_E{\phi(|f|^p)dx}= \int_E{|f|^qdx }$$

I was guessing that the necessary and sufficient condition required in the excercize is that $f(x) = 0$ almost everywhere (EDIT as observed in the comments, the condition is also satisfied if $f$ is such that $|f|$ is constant a.e.), but I'm stuck trying to prove it. I tried manipulating the exponents, but nothing seems to work; maybe some result on the inverse implication in Jensen's inequality could help, but I can't find any.
I hope someone can point me in the right direction, thanks in advance!

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Although I haven't tried this myself, I think a good idea would be to follow a proof of Jensen's inequality and track where the inequality emerges in the proof: then you might be able to find a condition which is necessary for equality to hold. – Clive Newstead Jan 16 '12 at 14:00
@CliveNewstead Thanks for your interest. I edited the question to include the proof I was alluding to. I know where the inequality is used in this proof, however I don't know how to use the inequality "backwards" in order to prove what the excercize is asking. – Emilio Ferrucci Jan 16 '12 at 14:10
Since $m(E)=1$, any $f$ such that $|f|$ is a.e. constant will give equality, not just a.e. zero. – Chris Eagle Jan 16 '12 at 14:11
@ChrisEagle Thanks! I hadn't noticed this. – Emilio Ferrucci Jan 16 '12 at 14:18
up vote 6 down vote accepted

I don't know how to use Jensen's inequality. But that's what I can prove using Holder's inequality:

Holder's inequality says: $$\int_E|f_1f_2|\leq\left(\int_E|f_1|^a\right)^{\frac{1}{a}}\left(\int_E|f_2|^a\right)^{\frac{1}{b}}$$ where $\displaystyle\frac{1}{a}+\frac{1}{b}=1$. Equality holds when $\alpha |f_1|^a=\beta |f_2|^b$ almost everywhere on $E$, where $\alpha$ and $\beta$ are constants.

Hence, if $1<p<q<\infty$, by Holder's inequality with $f_1=|f|^p$, $f_2=1$, $a=\frac{q}{p}$, $b=\frac{q}{q-p}$, we have $$\int_E{|f|}^p\leq \left(\int_E(|f|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}}\cdot\Big(m(E)\Big)^{1-\frac{p}{q}}=\left(\int_E|f|^q\right)^{\frac{p}{q}}.$$ Taking $p$-th root, we have $$\left(\int_E{|f|}^p\right)^{\frac{1}{p}}\leq\left(\int_E|f|^q\right)^{\frac{1}{q}}.$$

Therefore, if $\|f\|_p=\|f\|_q$, equality must hold when we apply the Holder's inequality, i.e. $$\alpha(|f|^p)^{\frac{q}{p}}=\beta\mbox{ almost everywhere on }E.$$ That is to say, $f$ is constant almost everywhere on $E$.

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Note that, since ${q\over p}>1$, you can take $0<p<q<\infty$ in your second paragraph. – David Mitra Jan 16 '12 at 14:44
Thank you very much, this is the answer I was looking for. – Emilio Ferrucci Jan 16 '12 at 14:44

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