4
$\begingroup$

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. 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!

  • 0
    @ChrisEagle Thanks! I hadn't noticed this.2012-01-16

1 Answers 1

6

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, 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 $E$.

  • 0
    Thank you very much, this is the answer I was looking for.2012-01-16