$int|f|^alpha|g|^{1-alpha}dmule(int|f|dmu)^alpha(int|g|dmu)^{1-alpha}$
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1
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$f,g$ on $(Omega, mathcal{A}, mu)$ and $0<alpha<1$.
Then
$$
int|f|^alpha |g|^{1-alpha} dmu
le left(int|f|dmuright)^alpha
left(int|g|dmuright)^{1-alpha}
$$
This seems like some case of Holder's inequality but I haven't found anything about it, is this inequality correct? Does anyone have a link to a proof of this? I would appreciate it! Thanks!
probability measure-theory inequality lebesgue-integral
edited Nov 15 at 19:37
gt6989b
31.9k22351
asked Nov 15 at 19:34
user610431
627
add a comment |
up vote
1
down vote
favorite
$f,g$ on $(Omega, mathcal{A}, mu)$ and $0<alpha<1$.
Then
$$
int|f|^alpha |g|^{1-alpha} dmu
le left(int|f|dmuright)^alpha
left(int|g|dmuright)^{1-alpha}
$$
This seems like some case of Holder's inequality but I haven't found anything about it, is this inequality correct? Does anyone have a link to a proof of this? I would appreciate it! Thanks!
probability measure-theory inequality lebesgue-integral
edited Nov 15 at 19:37
gt6989b
31.9k22351
asked Nov 15 at 19:34
user610431
627
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$f,g$ on $(Omega, mathcal{A}, mu)$ and $0<alpha<1$.
Then
$$
int|f|^alpha |g|^{1-alpha} dmu
le left(int|f|dmuright)^alpha
left(int|g|dmuright)^{1-alpha}
$$
This seems like some case of Holder's inequality but I haven't found anything about it, is this inequality correct? Does anyone have a link to a proof of this? I would appreciate it! Thanks!
probability measure-theory inequality lebesgue-integral
edited Nov 15 at 19:37
gt6989b
31.9k22351
asked Nov 15 at 19:34
user610431
627
$f,g$ on $(Omega, mathcal{A}, mu)$ and $0<alpha<1$.
Then
$$
int|f|^alpha |g|^{1-alpha} dmu
le left(int|f|dmuright)^alpha
left(int|g|dmuright)^{1-alpha}
$$
This seems like some case of Holder's inequality but I haven't found anything about it, is this inequality correct? Does anyone have a link to a proof of this? I would appreciate it! Thanks!
probability measure-theory inequality lebesgue-integral
probability measure-theory inequality lebesgue-integral
edited Nov 15 at 19:37
gt6989b
31.9k22351
asked Nov 15 at 19:34
user610431
627
edited Nov 15 at 19:37
gt6989b
31.9k22351
asked Nov 15 at 19:34
user610431
627
edited Nov 15 at 19:37
gt6989b
31.9k22351
edited Nov 15 at 19:37
gt6989b
31.9k22351
edited Nov 15 at 19:37
gt6989b
31.9k22351
31.9k22351
asked Nov 15 at 19:34
user610431
627
asked Nov 15 at 19:34
user610431
627
asked Nov 15 at 19:34
user610431
627
627
add a comment |
add a comment |
1 Answer
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1
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The numbers $p = dfrac 1alpha$ and $q = dfrac 1{1-alpha}$ are Holder conjugates, so that $$int |f|^alpha |g|^{1-alpha} le left( int |f|^{alpha frac 1alpha} right)^alpha left(int |g|^{(1-alpha) frac{1}{1-alpha}} right)^{1 - alpha} = left( int |f| right)^alpha left( int |g| right)^{1-alpha}.$$
answered Nov 15 at 19:41
Umberto P.
37.9k13063
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The numbers $p = dfrac 1alpha$ and $q = dfrac 1{1-alpha}$ are Holder conjugates, so that $$int |f|^alpha |g|^{1-alpha} le left( int |f|^{alpha frac 1alpha} right)^alpha left(int |g|^{(1-alpha) frac{1}{1-alpha}} right)^{1 - alpha} = left( int |f| right)^alpha left( int |g| right)^{1-alpha}.$$
answered Nov 15 at 19:41
Umberto P.
37.9k13063
add a comment |
up vote
1
down vote
The numbers $p = dfrac 1alpha$ and $q = dfrac 1{1-alpha}$ are Holder conjugates, so that $$int |f|^alpha |g|^{1-alpha} le left( int |f|^{alpha frac 1alpha} right)^alpha left(int |g|^{(1-alpha) frac{1}{1-alpha}} right)^{1 - alpha} = left( int |f| right)^alpha left( int |g| right)^{1-alpha}.$$
answered Nov 15 at 19:41
Umberto P.
37.9k13063
add a comment |
up vote
1
down vote
up vote
1
down vote
The numbers $p = dfrac 1alpha$ and $q = dfrac 1{1-alpha}$ are Holder conjugates, so that $$int |f|^alpha |g|^{1-alpha} le left( int |f|^{alpha frac 1alpha} right)^alpha left(int |g|^{(1-alpha) frac{1}{1-alpha}} right)^{1 - alpha} = left( int |f| right)^alpha left( int |g| right)^{1-alpha}.$$
answered Nov 15 at 19:41
Umberto P.
37.9k13063
The numbers $p = dfrac 1alpha$ and $q = dfrac 1{1-alpha}$ are Holder conjugates, so that $$int |f|^alpha |g|^{1-alpha} le left( int |f|^{alpha frac 1alpha} right)^alpha left(int |g|^{(1-alpha) frac{1}{1-alpha}} right)^{1 - alpha} = left( int |f| right)^alpha left( int |g| right)^{1-alpha}.$$
answered Nov 15 at 19:41
Umberto P.
37.9k13063
answered Nov 15 at 19:41
Umberto P.
37.9k13063
answered Nov 15 at 19:41
Umberto P.
37.9k13063
answered Nov 15 at 19:41
Umberto P.
37.9k13063
37.9k13063
add a comment |
add a comment |
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