$L^1$ between the sections of two functions
$begingroup$
Let $f_1:mathbb{R}^2 to (0, infty)$ and $f_2:mathbb{R}^2 to (0, infty)$. Then consider the $L^1$ distance
$$
Vert f_1 -f_2 Vert_1=int_{mathbb{R}^2}|f_1(x,y)-f_2(x,y)|dxdy.
$$
Now, fix any $yin mathbb{R}$, and consider the $L^1$ distance between the sections $f_1(cdot,y)$ and $f_2(cdot,y)$, i.e.
$$
Vert f_1 -f_2 Vert_{1,y}=int_{mathbb{R}}|f_1(x,y)-f_2(x,y)|dx.
$$
It should hold that
$$
Vert f_1 -f_2 Vert_{1}=int_{mathbb{R}}Vert f_1 -f_2 Vert_{1,y}dy.
$$
Consider now sequences of functions $f_{n,1}$ and $f_{n,2}$; under regularity conditions that allow to exchange limit and integral, we should have that $$sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$$
entails that also $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$. My question is the following: could you think of any condition on the sequences $(f_{n,1})$ and $(f_{n,2})$ such that a sort of converse implication holds true: i.e. $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$ entails that
$$
Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0, quad forall y in mathbb{R}, quad (n to infty)
$$
or even that $sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$ (for the latter case, we could maybe restrict to the case of functions supported on the unit simplex or on the unit square). I'm particularly interested in the case in which $(f_{n,1})$ and $(f_{n,2})$ are sequences of probability densities.
real-analysis functional-analysis metric-spaces density-function
asked Jan 24 at 11:16
Jack LondonJack London
33318
$endgroup$
add a comment |
$begingroup$
Let $f_1:mathbb{R}^2 to (0, infty)$ and $f_2:mathbb{R}^2 to (0, infty)$. Then consider the $L^1$ distance
$$
Vert f_1 -f_2 Vert_1=int_{mathbb{R}^2}|f_1(x,y)-f_2(x,y)|dxdy.
$$
Now, fix any $yin mathbb{R}$, and consider the $L^1$ distance between the sections $f_1(cdot,y)$ and $f_2(cdot,y)$, i.e.
$$
Vert f_1 -f_2 Vert_{1,y}=int_{mathbb{R}}|f_1(x,y)-f_2(x,y)|dx.
$$
It should hold that
$$
Vert f_1 -f_2 Vert_{1}=int_{mathbb{R}}Vert f_1 -f_2 Vert_{1,y}dy.
$$
Consider now sequences of functions $f_{n,1}$ and $f_{n,2}$; under regularity conditions that allow to exchange limit and integral, we should have that $$sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$$
entails that also $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$. My question is the following: could you think of any condition on the sequences $(f_{n,1})$ and $(f_{n,2})$ such that a sort of converse implication holds true: i.e. $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$ entails that
$$
Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0, quad forall y in mathbb{R}, quad (n to infty)
$$
or even that $sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$ (for the latter case, we could maybe restrict to the case of functions supported on the unit simplex or on the unit square). I'm particularly interested in the case in which $(f_{n,1})$ and $(f_{n,2})$ are sequences of probability densities.
real-analysis functional-analysis metric-spaces density-function
asked Jan 24 at 11:16
Jack LondonJack London
33318
$endgroup$
add a comment |
$begingroup$
Let $f_1:mathbb{R}^2 to (0, infty)$ and $f_2:mathbb{R}^2 to (0, infty)$. Then consider the $L^1$ distance
$$
Vert f_1 -f_2 Vert_1=int_{mathbb{R}^2}|f_1(x,y)-f_2(x,y)|dxdy.
$$
Now, fix any $yin mathbb{R}$, and consider the $L^1$ distance between the sections $f_1(cdot,y)$ and $f_2(cdot,y)$, i.e.
$$
Vert f_1 -f_2 Vert_{1,y}=int_{mathbb{R}}|f_1(x,y)-f_2(x,y)|dx.
$$
It should hold that
$$
Vert f_1 -f_2 Vert_{1}=int_{mathbb{R}}Vert f_1 -f_2 Vert_{1,y}dy.
$$
Consider now sequences of functions $f_{n,1}$ and $f_{n,2}$; under regularity conditions that allow to exchange limit and integral, we should have that $$sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$$
entails that also $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$. My question is the following: could you think of any condition on the sequences $(f_{n,1})$ and $(f_{n,2})$ such that a sort of converse implication holds true: i.e. $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$ entails that
$$
Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0, quad forall y in mathbb{R}, quad (n to infty)
$$
or even that $sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$ (for the latter case, we could maybe restrict to the case of functions supported on the unit simplex or on the unit square). I'm particularly interested in the case in which $(f_{n,1})$ and $(f_{n,2})$ are sequences of probability densities.
real-analysis functional-analysis metric-spaces density-function
asked Jan 24 at 11:16
Jack LondonJack London
33318
$endgroup$
Let $f_1:mathbb{R}^2 to (0, infty)$ and $f_2:mathbb{R}^2 to (0, infty)$. Then consider the $L^1$ distance
$$
Vert f_1 -f_2 Vert_1=int_{mathbb{R}^2}|f_1(x,y)-f_2(x,y)|dxdy.
$$
Now, fix any $yin mathbb{R}$, and consider the $L^1$ distance between the sections $f_1(cdot,y)$ and $f_2(cdot,y)$, i.e.
$$
Vert f_1 -f_2 Vert_{1,y}=int_{mathbb{R}}|f_1(x,y)-f_2(x,y)|dx.
$$
It should hold that
$$
Vert f_1 -f_2 Vert_{1}=int_{mathbb{R}}Vert f_1 -f_2 Vert_{1,y}dy.
$$
Consider now sequences of functions $f_{n,1}$ and $f_{n,2}$; under regularity conditions that allow to exchange limit and integral, we should have that $$sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$$
entails that also $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$. My question is the following: could you think of any condition on the sequences $(f_{n,1})$ and $(f_{n,2})$ such that a sort of converse implication holds true: i.e. $Vert f_{n,1} -f_{n,2} Vert_{1}to 0$ entails that
$$
Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0, quad forall y in mathbb{R}, quad (n to infty)
$$
or even that $sup_{y in mathbb{R}}Vert f_{n,1} -f_{n,2} Vert_{1,y}to 0 quad (nto infty)$ (for the latter case, we could maybe restrict to the case of functions supported on the unit simplex or on the unit square). I'm particularly interested in the case in which $(f_{n,1})$ and $(f_{n,2})$ are sequences of probability densities.
real-analysis functional-analysis metric-spaces density-function
real-analysis functional-analysis metric-spaces density-function
asked Jan 24 at 11:16
Jack LondonJack London
33318
asked Jan 24 at 11:16
Jack LondonJack London
33318
edited Jan 25 at 9:25
Jack London
asked Jan 24 at 11:16
Jack LondonJack London
33318
asked Jan 24 at 11:16
Jack LondonJack London
33318
asked Jan 24 at 11:16
Jack LondonJack London
33318
33318
add a comment |
add a comment |
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