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Is $L^infty(lambda^d) cong L^infty(lambda^d-1) otimes L^infty(lambda)$?
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Let $lambda^d$ be Lebesgue measure on $[0,1]^d$. If we take products of functions in $L^infty(lambda^d-1)$ with functions in $L^infty(lambda)$, we should get functions in $L^infty(lambda^d)$. Can we get all such functions this way? Are all functions in $L^infty(lambda^d)$ included in the closed span of linear combinations of elements of $L^infty(lambda^d-1) otimes L^infty(lambda)$? That is, is
$$L^infty(lambda^d) cong L^infty(lambda^d-1) otimes L^infty(lambda)$$
Edit: (to include more information about what I have considered)
What I would like to say is that we can approximate indicator functions (and hence simple functions) in $L^infty(lambda^d)$ by simple functions in $L^infty(lambda^d-1) otimes L^infty(lambda)$, but such a statement seems to rely on being able to approximate a measurable set $A subseteq [0,1]^d$ by measurable rectangles $B_i times C_i$ (with $B_i subseteq[0,1]^d-1$ and $C_i subseteq [0,1]$). I'm not sure if such an approximation exists in general. So far, I have only seen results that guarantee pointwise convergence of such functions, not $L^infty$ convergence. $L^infty$ convergence seems harder in this case because if $lambda(A triangle bigcup_i (B_i times C_i)) > 0$, then $d(1_A,sum_i 1_B_i times C_i) = 1$.
Currently, I think that the answer should be in the negative, but I don't know how to prove that a counterexample exists.
functional-analysis
asked Jul 16 at 15:55
optionalsampling
11
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up vote
0
down vote
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Let $lambda^d$ be Lebesgue measure on $[0,1]^d$. If we take products of functions in $L^infty(lambda^d-1)$ with functions in $L^infty(lambda)$, we should get functions in $L^infty(lambda^d)$. Can we get all such functions this way? Are all functions in $L^infty(lambda^d)$ included in the closed span of linear combinations of elements of $L^infty(lambda^d-1) otimes L^infty(lambda)$? That is, is
$$L^infty(lambda^d) cong L^infty(lambda^d-1) otimes L^infty(lambda)$$
Edit: (to include more information about what I have considered)
What I would like to say is that we can approximate indicator functions (and hence simple functions) in $L^infty(lambda^d)$ by simple functions in $L^infty(lambda^d-1) otimes L^infty(lambda)$, but such a statement seems to rely on being able to approximate a measurable set $A subseteq [0,1]^d$ by measurable rectangles $B_i times C_i$ (with $B_i subseteq[0,1]^d-1$ and $C_i subseteq [0,1]$). I'm not sure if such an approximation exists in general. So far, I have only seen results that guarantee pointwise convergence of such functions, not $L^infty$ convergence. $L^infty$ convergence seems harder in this case because if $lambda(A triangle bigcup_i (B_i times C_i)) > 0$, then $d(1_A,sum_i 1_B_i times C_i) = 1$.
Currently, I think that the answer should be in the negative, but I don't know how to prove that a counterexample exists.
functional-analysis
asked Jul 16 at 15:55
optionalsampling
11
2
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $lambda^d$ be Lebesgue measure on $[0,1]^d$. If we take products of functions in $L^infty(lambda^d-1)$ with functions in $L^infty(lambda)$, we should get functions in $L^infty(lambda^d)$. Can we get all such functions this way? Are all functions in $L^infty(lambda^d)$ included in the closed span of linear combinations of elements of $L^infty(lambda^d-1) otimes L^infty(lambda)$? That is, is
$$L^infty(lambda^d) cong L^infty(lambda^d-1) otimes L^infty(lambda)$$
Edit: (to include more information about what I have considered)
What I would like to say is that we can approximate indicator functions (and hence simple functions) in $L^infty(lambda^d)$ by simple functions in $L^infty(lambda^d-1) otimes L^infty(lambda)$, but such a statement seems to rely on being able to approximate a measurable set $A subseteq [0,1]^d$ by measurable rectangles $B_i times C_i$ (with $B_i subseteq[0,1]^d-1$ and $C_i subseteq [0,1]$). I'm not sure if such an approximation exists in general. So far, I have only seen results that guarantee pointwise convergence of such functions, not $L^infty$ convergence. $L^infty$ convergence seems harder in this case because if $lambda(A triangle bigcup_i (B_i times C_i)) > 0$, then $d(1_A,sum_i 1_B_i times C_i) = 1$.
Currently, I think that the answer should be in the negative, but I don't know how to prove that a counterexample exists.
functional-analysis
asked Jul 16 at 15:55
optionalsampling
11
Let $lambda^d$ be Lebesgue measure on $[0,1]^d$. If we take products of functions in $L^infty(lambda^d-1)$ with functions in $L^infty(lambda)$, we should get functions in $L^infty(lambda^d)$. Can we get all such functions this way? Are all functions in $L^infty(lambda^d)$ included in the closed span of linear combinations of elements of $L^infty(lambda^d-1) otimes L^infty(lambda)$? That is, is
$$L^infty(lambda^d) cong L^infty(lambda^d-1) otimes L^infty(lambda)$$
Edit: (to include more information about what I have considered)
What I would like to say is that we can approximate indicator functions (and hence simple functions) in $L^infty(lambda^d)$ by simple functions in $L^infty(lambda^d-1) otimes L^infty(lambda)$, but such a statement seems to rely on being able to approximate a measurable set $A subseteq [0,1]^d$ by measurable rectangles $B_i times C_i$ (with $B_i subseteq[0,1]^d-1$ and $C_i subseteq [0,1]$). I'm not sure if such an approximation exists in general. So far, I have only seen results that guarantee pointwise convergence of such functions, not $L^infty$ convergence. $L^infty$ convergence seems harder in this case because if $lambda(A triangle bigcup_i (B_i times C_i)) > 0$, then $d(1_A,sum_i 1_B_i times C_i) = 1$.
Currently, I think that the answer should be in the negative, but I don't know how to prove that a counterexample exists.
functional-analysis
asked Jul 16 at 15:55
optionalsampling
11
edited Jul 17 at 3:36
asked Jul 16 at 15:55
optionalsampling
11
asked Jul 16 at 15:55
optionalsampling
11
asked Jul 16 at 15:55
optionalsampling
11
11
2
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54
add a comment |Â
2
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54
2
2
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54
add a comment |Â
1 Answer
1
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oldest
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up vote
1
down vote
For $d=2$, on $[0,1]^2$ it is hard to approximate the diagonal functions
$$
f(x,y) = begincases 1 & text if x>y\
0& text if xle yendcases
$$
by products of the type $g(x)h(y)$ in $L^infty([0,1]^2)$. It is not possible to do this even with infinite series of functions of this type.
Take such a series
$$
F(x,y) = sum_k=1^infty g_k(x) h_k(y)$.
$$
Then the set $ F(cdot,y): yin[0,1]$ is separable, while the same set for the diagonal function $ f(cdot,y):yin[0,1]$ is not separable.
answered Jul 16 at 20:14
daw
22k1542
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For $d=2$, on $[0,1]^2$ it is hard to approximate the diagonal functions
$$
f(x,y) = begincases 1 & text if x>y\
0& text if xle yendcases
$$
by products of the type $g(x)h(y)$ in $L^infty([0,1]^2)$. It is not possible to do this even with infinite series of functions of this type.
Take such a series
$$
F(x,y) = sum_k=1^infty g_k(x) h_k(y)$.
$$
Then the set $ F(cdot,y): yin[0,1]$ is separable, while the same set for the diagonal function $ f(cdot,y):yin[0,1]$ is not separable.
answered Jul 16 at 20:14
daw
22k1542
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
add a comment |Â
up vote
1
down vote
For $d=2$, on $[0,1]^2$ it is hard to approximate the diagonal functions
$$
f(x,y) = begincases 1 & text if x>y\
0& text if xle yendcases
$$
by products of the type $g(x)h(y)$ in $L^infty([0,1]^2)$. It is not possible to do this even with infinite series of functions of this type.
Take such a series
$$
F(x,y) = sum_k=1^infty g_k(x) h_k(y)$.
$$
Then the set $ F(cdot,y): yin[0,1]$ is separable, while the same set for the diagonal function $ f(cdot,y):yin[0,1]$ is not separable.
answered Jul 16 at 20:14
daw
22k1542
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $d=2$, on $[0,1]^2$ it is hard to approximate the diagonal functions
$$
f(x,y) = begincases 1 & text if x>y\
0& text if xle yendcases
$$
by products of the type $g(x)h(y)$ in $L^infty([0,1]^2)$. It is not possible to do this even with infinite series of functions of this type.
Take such a series
$$
F(x,y) = sum_k=1^infty g_k(x) h_k(y)$.
$$
Then the set $ F(cdot,y): yin[0,1]$ is separable, while the same set for the diagonal function $ f(cdot,y):yin[0,1]$ is not separable.
answered Jul 16 at 20:14
daw
22k1542
For $d=2$, on $[0,1]^2$ it is hard to approximate the diagonal functions
$$
f(x,y) = begincases 1 & text if x>y\
0& text if xle yendcases
$$
by products of the type $g(x)h(y)$ in $L^infty([0,1]^2)$. It is not possible to do this even with infinite series of functions of this type.
Take such a series
$$
F(x,y) = sum_k=1^infty g_k(x) h_k(y)$.
$$
Then the set $ F(cdot,y): yin[0,1]$ is separable, while the same set for the diagonal function $ f(cdot,y):yin[0,1]$ is not separable.
answered Jul 16 at 20:14
daw
22k1542
edited Jul 17 at 19:43
answered Jul 16 at 20:14
daw
22k1542
answered Jul 16 at 20:14
daw
22k1542
answered Jul 16 at 20:14
daw
22k1542
22k1542
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
add a comment |Â
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
Can these functions be approximated by linear combinations of products $g(x)h(y)$ in $L^infty([0,1]^2)$?
– optionalsampling
Jul 16 at 21:06
add a comment |Â
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2
This is hardly true, but I feel that $L^infty(lambda^d-1)otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$, if the $otimes$ is the usual algebraic tensor product. If we define $otimes$ as its completion, then the above equation becomes true, I suppose.
– lisyarus
Jul 16 at 16:13
I had accidentally included an extra $d$ on the lone $lambda$ on the right (both in the title and body of the question). What I am basically asking is how one would show that $L^infty(lambda^d-1) otimes L^infty(lambda)$ is dense in $L^infty(lambda^d)$.
– optionalsampling
Jul 16 at 18:36
Nevermind - what I said is true for $L^2$, but apparently not for $L^infty$.
– lisyarus
Jul 16 at 21:54