Crystal Wheeler

2022-07-13

I have some problems, help:

Let $(X,{\mathcal{A}}_{\mathcal{1}},\nu )$ be a probability space. Let $A:{L}_{1}\to {L}_{1}$ be a contraction such that $A:{L}_{\mathrm{\infty}}\to {L}_{\mathrm{\infty}}$ is also a contraction. Suppose $A$ preserves positivity.

In a proof of why $A:{L}_{p}\to {L}_{p}$ is a contraction for every $1\le p<\mathrm{\infty}$, my lecture notes state that for all $f\ge 0$, we have

$\Vert Af{\Vert}_{p}=\underset{g\ge 0,\Vert g{\Vert}_{q}=1}{sup}{\int}_{X}(Af\cdot g),$

where $\frac{1}{p}+\frac{1}{q}=1$.

Why does the above equality hold? The strangest part to me is the fact that the LHS has a power of $\frac{1}{p}$ outside of the integral ${\int}_{X}|Af{|}^{p}$, but in the RHS the integral is not being raised to any power. I've tried to raise both sides to $p$ but I could not do much with this. If it helps, this came up in the study of conditional expectations.

Let $(X,{\mathcal{A}}_{\mathcal{1}},\nu )$ be a probability space. Let $A:{L}_{1}\to {L}_{1}$ be a contraction such that $A:{L}_{\mathrm{\infty}}\to {L}_{\mathrm{\infty}}$ is also a contraction. Suppose $A$ preserves positivity.

In a proof of why $A:{L}_{p}\to {L}_{p}$ is a contraction for every $1\le p<\mathrm{\infty}$, my lecture notes state that for all $f\ge 0$, we have

$\Vert Af{\Vert}_{p}=\underset{g\ge 0,\Vert g{\Vert}_{q}=1}{sup}{\int}_{X}(Af\cdot g),$

where $\frac{1}{p}+\frac{1}{q}=1$.

Why does the above equality hold? The strangest part to me is the fact that the LHS has a power of $\frac{1}{p}$ outside of the integral ${\int}_{X}|Af{|}^{p}$, but in the RHS the integral is not being raised to any power. I've tried to raise both sides to $p$ but I could not do much with this. If it helps, this came up in the study of conditional expectations.

sniokd

Beginner2022-07-14Added 22 answers

We have by Holder's that $|{\int}_{X}(Af\cdot g)|\le \Vert Af{\Vert}_{p}\Vert g{\Vert}_{q}=\Vert Af{\Vert}_{p}$ for all $g\in {L}^{q}$ with $\Vert g{\Vert}_{q}=1$. The question then becomes whether or not $g$ can be chosen to obtain equality.

Note that $1/p+1/q=1$ implies that $1+p/q=p$ and $1+q/p=q$. Hence, choose $g=\frac{(Af{)}^{p/q}}{\Vert Af{\Vert}_{p}^{p/q}}$, which is nonnegative because $f$ is and $A$ preserves nonnegativity. Moreover, $\Vert g{\Vert}_{q}=1$.

But then $Af\cdot g=\frac{(Af{)}^{1+p/q}}{\Vert Af{\Vert}_{p}^{p/q}}=\frac{(Af{)}^{p}}{\Vert Af{\Vert}_{p}^{p/q}}$. Integrating, shows that

${\int}_{X}Af\cdot g=\frac{\Vert Af{\Vert}_{p}^{p}}{\Vert Af{\Vert}_{p}^{p/q}}=\Vert Af{\Vert}_{p}.$

Note that $1/p+1/q=1$ implies that $1+p/q=p$ and $1+q/p=q$. Hence, choose $g=\frac{(Af{)}^{p/q}}{\Vert Af{\Vert}_{p}^{p/q}}$, which is nonnegative because $f$ is and $A$ preserves nonnegativity. Moreover, $\Vert g{\Vert}_{q}=1$.

But then $Af\cdot g=\frac{(Af{)}^{1+p/q}}{\Vert Af{\Vert}_{p}^{p/q}}=\frac{(Af{)}^{p}}{\Vert Af{\Vert}_{p}^{p/q}}$. Integrating, shows that

${\int}_{X}Af\cdot g=\frac{\Vert Af{\Vert}_{p}^{p}}{\Vert Af{\Vert}_{p}^{p/q}}=\Vert Af{\Vert}_{p}.$

Which expression has both 8 and n as factors???

One number is 2 more than 3 times another. Their sum is 22. Find the numbers

8, 14

5, 17

2, 20

4, 18

10, 12Perform the indicated operation and simplify the result. Leave your answer in factored form

$\left[\frac{(4x-8)}{(-3x)}\right].\left[\frac{12}{(12-6x)}\right]$ An ordered pair set is referred to as a ___?

Please, can u convert 3.16 (6 repeating) to fraction.

Write an algebraic expression for the statement '6 less than the quotient of x divided by 3 equals 2'.

A) $6-\frac{x}{3}=2$

B) $\frac{x}{3}-6=2$

C) 3x−6=2

D) $\frac{3}{x}-6=2$Find: $2.48\xf74$.

Multiplication $999\times 999$ equals.

Solve: (128÷32)÷(−4)=

A) -1

B) 2

C) -4

D) -3What is $0.78888.....$ converted into a fraction? $\left(0.7\overline{8}\right)$

The mixed fraction representation of 7/3 is...

How to write the algebraic expression given: the quotient of 5 plus d and 12 minus w?

Express 200+30+5+4100+71000 as a decimal number and find its hundredths digit.

A)235.47,7

B)235.047,4

C)235.47,4

D)234.057,7Find four equivalent fractions of the given fraction:$\frac{6}{12}$

How to find the greatest common factor of $80{x}^{3},30y{x}^{2}$?