Differentiability of $L^p$ normAbsolute continuity of the kernelIncreasing rearrangement and Hardy-Littlewood inequalityMeasures agreeing on a $sigma$-field generated by some class of setsTricky detail in the proof of Haar's theoremStochastic integral density of simple functions no1What's the relationship between a measure space and a metric space?Can one always construct an equivalent measure?Uniform Lipshitz continuity implies Continuous DifferentiabilityDifferentiability of $L^p$-valued functionsSufficient conditions for zeros of a function to have measure zero
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Differentiability of $L^p$ norm
Absolute continuity of the kernelIncreasing rearrangement and Hardy-Littlewood inequalityMeasures agreeing on a $sigma$-field generated by some class of setsTricky detail in the proof of Haar's theoremStochastic integral density of simple functions no1What's the relationship between a measure space and a metric space?Can one always construct an equivalent measure?Uniform Lipshitz continuity implies Continuous DifferentiabilityDifferentiability of $L^p$-valued functionsSufficient conditions for zeros of a function to have measure zero
$begingroup$
Let $2 < p < infty$ and $f, g in L^p(Omega, mathcalF, nu)$ be fixed elements of norm $1$, where $(Omega, mathcalF, nu)$ is any measure space. Consider the function $F$ defined on $mathbbR$ by
$$F(t) = int_Omega|f+t,g|^p dnu.$$
Is $F$ twice continuously differentiable on $mathbbR$, i.e. is $F$ of class $C^2$? Theorem 2.6 in Analysis by Lieb and Loss asserts that $F$ is differentiable at $t=0$, but I wanted to know if this holds for all $t$ and if any higher derivatives exist and are continuous. I know that $x mapsto |x|^p$ is $n = leftlfloorprightrfloor - 1$ times differentiable, but does not have a continuous $n$th derivative, so my guess is that the same holds here; $F$ is also $n$-times differentiable or $n-1$-times continuously differentiable. Do any additional assumptions on the measure space allow us to infer any more differentiability class properties of $F$?
measure-theory derivatives
$endgroup$
add a comment |
$begingroup$
Let $2 < p < infty$ and $f, g in L^p(Omega, mathcalF, nu)$ be fixed elements of norm $1$, where $(Omega, mathcalF, nu)$ is any measure space. Consider the function $F$ defined on $mathbbR$ by
$$F(t) = int_Omega|f+t,g|^p dnu.$$
Is $F$ twice continuously differentiable on $mathbbR$, i.e. is $F$ of class $C^2$? Theorem 2.6 in Analysis by Lieb and Loss asserts that $F$ is differentiable at $t=0$, but I wanted to know if this holds for all $t$ and if any higher derivatives exist and are continuous. I know that $x mapsto |x|^p$ is $n = leftlfloorprightrfloor - 1$ times differentiable, but does not have a continuous $n$th derivative, so my guess is that the same holds here; $F$ is also $n$-times differentiable or $n-1$-times continuously differentiable. Do any additional assumptions on the measure space allow us to infer any more differentiability class properties of $F$?
measure-theory derivatives
$endgroup$
$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00
add a comment |
$begingroup$
Let $2 < p < infty$ and $f, g in L^p(Omega, mathcalF, nu)$ be fixed elements of norm $1$, where $(Omega, mathcalF, nu)$ is any measure space. Consider the function $F$ defined on $mathbbR$ by
$$F(t) = int_Omega|f+t,g|^p dnu.$$
Is $F$ twice continuously differentiable on $mathbbR$, i.e. is $F$ of class $C^2$? Theorem 2.6 in Analysis by Lieb and Loss asserts that $F$ is differentiable at $t=0$, but I wanted to know if this holds for all $t$ and if any higher derivatives exist and are continuous. I know that $x mapsto |x|^p$ is $n = leftlfloorprightrfloor - 1$ times differentiable, but does not have a continuous $n$th derivative, so my guess is that the same holds here; $F$ is also $n$-times differentiable or $n-1$-times continuously differentiable. Do any additional assumptions on the measure space allow us to infer any more differentiability class properties of $F$?
measure-theory derivatives
$endgroup$
Let $2 < p < infty$ and $f, g in L^p(Omega, mathcalF, nu)$ be fixed elements of norm $1$, where $(Omega, mathcalF, nu)$ is any measure space. Consider the function $F$ defined on $mathbbR$ by
$$F(t) = int_Omega|f+t,g|^p dnu.$$
Is $F$ twice continuously differentiable on $mathbbR$, i.e. is $F$ of class $C^2$? Theorem 2.6 in Analysis by Lieb and Loss asserts that $F$ is differentiable at $t=0$, but I wanted to know if this holds for all $t$ and if any higher derivatives exist and are continuous. I know that $x mapsto |x|^p$ is $n = leftlfloorprightrfloor - 1$ times differentiable, but does not have a continuous $n$th derivative, so my guess is that the same holds here; $F$ is also $n$-times differentiable or $n-1$-times continuously differentiable. Do any additional assumptions on the measure space allow us to infer any more differentiability class properties of $F$?
measure-theory derivatives
measure-theory derivatives
asked Mar 13 at 19:17
talfredtalfred
1007
1007
$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00
add a comment |
$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00
$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00
add a comment |
0
active
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$begingroup$
You need to make sure that integral that represents $F'''$ etc. remains defined.
$endgroup$
– Umberto P.
Mar 13 at 19:31
$begingroup$
@UmbertoP. is that sufficient?
$endgroup$
– talfred
Mar 13 at 20:00