Post History
#2: Post edited
by
Derek Elkins
·
2021-12-31T02:13:44Z (almost 3 years ago)
Fix braces.
$\int_{E_n} |g|^q = \left| \int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g| \right|$
I am trying to understand why the following equation is true. Here $E$ is a measurable set and all functions are defined and measurable on it. $1<p,q,<\infty$ such that $\frac 1 p+\frac 1 q=1$ and $g\in L^q(E)$. $E_n= \{ x \in E:|g|\le n \}$ (for some reason the LaTeX isn't rendering the set curly braces here). And there exists a number $M$ such that for every $f\in L^p(E)$, we have $\left|\int_Efg ight|\le M- ||f||_p$.
- $$\int_{E_n} |g|^q = \left|\int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g|\right|$$
- My main problem with this is that it seems like the powers are off by one. I would think that on $E_n$,
- $$|g|^q = \chi_{E_n}\cdot |g|\cdot |g|^{q-1} = \chi_{E_n}\cdot\text{sgn}(g)\cdot g \cdot |g|^{q-1}$$
- But even if we correct the extra factor of $|g|$ then I don't understand how we can write this as an equality. If we're using the "integral triangle inequalty" or whatever $\left|\int_Ef\right|\le\int_E|f|$ is called, then shouldn't the equality actually be an inequality? Is there some reason why in this particular setting we can actually have equality?
- I am trying to understand why the following equation is true. Here $E$ is a measurable set and all functions are defined and measurable on it. $1<p,q,<\infty$ such that $\frac 1 p+\frac 1 q=1$ and $g\in L^q(E)$. $E_n= \\{ x \in E:|g|\le n \\}$. And there exists a number $M$ such that for every $f\in L^p(E)$, we have $\left|\int_Efg ight|\le M
- ||f||_p$.
- $$\int_{E_n} |g|^q = \left|\int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g|\right|$$
- My main problem with this is that it seems like the powers are off by one. I would think that on $E_n$,
- $$|g|^q = \chi_{E_n}\cdot |g|\cdot |g|^{q-1} = \chi_{E_n}\cdot\text{sgn}(g)\cdot g \cdot |g|^{q-1}$$
- But even if we correct the extra factor of $|g|$ then I don't understand how we can write this as an equality. If we're using the "integral triangle inequalty" or whatever $\left|\int_Ef\right|\le\int_E|f|$ is called, then shouldn't the equality actually be an inequality? Is there some reason why in this particular setting we can actually have equality?
#1: Initial revision
by
whybecause
·
2021-12-31T01:18:40Z (almost 3 years ago)
$\int_{E_n} |g|^q = \left| \int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g| \right|$
I am trying to understand why the following equation is true. Here $E$ is a measurable set and all functions are defined and measurable on it. $1<p,q,<\infty$ such that $\frac 1 p+\frac 1 q=1$ and $g\in L^q(E)$. $E_n= \{ x \in E:|g|\le n \}$ (for some reason the LaTeX isn't rendering the set curly braces here). And there exists a number $M$ such that for every $f\in L^p(E)$, we have $\left|\int_Efg\right|\le M
||f||_p$.
$$\int_{E_n} |g|^q = \left|\int_E \chi_{E_n}\cdot \text{sgn}(g)\cdot g \cdot |g|^{q-1}\cdot |g|\right|$$
My main problem with this is that it seems like the powers are off by one. I would think that on $E_n$,
$$|g|^q = \chi_{E_n}\cdot |g|\cdot |g|^{q-1} = \chi_{E_n}\cdot\text{sgn}(g)\cdot g \cdot |g|^{q-1}$$
But even if we correct the extra factor of $|g|$ then I don't understand how we can write this as an equality. If we're using the "integral triangle inequalty" or whatever $\left|\int_Ef\right|\le\int_E|f|$ is called, then shouldn't the equality actually be an inequality? Is there some reason why in this particular setting we can actually have equality?