Understanding reverse Hölder and Minkowski inequalities

927 Views Asked by At

This is a problem from my measure theory book:

Let $(\Omega,\mathcal{A},\mu)$ be a measure space, $p\in(0,1)$ and $q<0$ be defined by $\frac{1}{p}+\frac{1}{q}=1$. Consider non-negative $f\in\mathcal{L}^p$ and measurable $g:\Omega\to (0,\infty)$ satisfying $0<N_q(g):= (\int g^q )^{1/q}<\infty$. By an appropriate application of Hölder's inequality show that $$ N_1(fg)\geq N_p(f)N_q(g).$$ Infer that $$ N_p(f+g)\geq N_p(f)+N_p(g)$$ and find an example to show that generally equality does not prevail here.

For the first inequality I did the following: Let $p'=1/p>1$ and $q'>1$ be defined by $\frac{1}{p'}+\frac{1}{q'}=1$. Then Hölder's inequality implies that

$$\int f^p=\int (fg)^p g^{-p}\leq N_{p'}((fg)^p) N_{q'}(g^{-p}).$$ We check that $p'\cdot p=1$ and $-p\cdot q'=q$ so that we actually have

$$\int f^p\leq N_1(fg)^p N_q(g)^{-p}$$

Raising both sides to the power $1/p$ we get

$$ N_p(f)\leq N_1(fg)N_q(g)^{-1}$$

Multiplying both sides by $N_q(g)$ gives the result.

From there I was able to show the second inequality using my book's approach for proving the normal Minkowski's inequality and the inequality just proven. However I never used the fact that $f \in \mathcal{L}^p$. Is this a redundant assumption? Do you have examples of a strict inequality for both?

Thanks a lot for your help.

1

There are 1 best solutions below

0
On

I came up with following example where both inequalities are strict. Let $(\Omega,\mathcal{A},\mu)$ be a probability space. Let $p=0.5$ and $q=-1$ so that $\frac{1}{p}+\frac{1}{q}=1$. Let $A\in \mathcal{A}$ be such that $0<\mu(A)<1$. Let $f=1_A$ and $g=1$. Then

$$N_1(fg)=\mu(A)$$

$$N_p(f)=\mu(A)^2$$

$$N_q(g)=\mu(\Omega)^{-1}=1$$

so $N_1(fg)> N_p(f)N_q(g)$. Also,

$$N_p(g)=\mu(\Omega)^2=1$$

$$N_p(f+g)=[ \sqrt{2}\mu(A)+\mu(A^c)]^2=\mu(A)^2+1+2(\sqrt{2}-1)\mu(A)\mu(A^c)$$

so $N_p(f+g)> N_p(f)+N_p(g)$ holds as well.