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#2: Post edited by user avatar Derek Elkins‭ · 2021-12-31T02:13:44Z (over 2 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 user avatar whybecause‭ · 2021-12-31T01:18:40Z (over 2 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?