Is the following set dense in $L^2$?
up vote
13
down vote
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Lately I was talking to a friend of mine and we came up with the following question
Denote by $mathcal{P}$ the set of all real valued polynomial functions. Is the set
$$ { p(x) e^{- alpha vert x vert} : pin mathcal{P}, alpha in mathbb{R}_{>0} } $$
dense in $L^2(mathbb{R} ,mathbb{R})?$
My feeling is that it should be true. I was thinking about using the Stone-Weierstrass, however, I cannot control the $L^2$-norm of the polynomial function outside the compact set. Clearly it is finite (the exponential decay beats the polynomial growth), but it is not clear to me whether one can choose it to be small.
functional-analysis hilbert-spaces
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
add a comment |
up vote
13
down vote
favorite
Lately I was talking to a friend of mine and we came up with the following question
Denote by $mathcal{P}$ the set of all real valued polynomial functions. Is the set
$$ { p(x) e^{- alpha vert x vert} : pin mathcal{P}, alpha in mathbb{R}_{>0} } $$
dense in $L^2(mathbb{R} ,mathbb{R})?$
My feeling is that it should be true. I was thinking about using the Stone-Weierstrass, however, I cannot control the $L^2$-norm of the polynomial function outside the compact set. Clearly it is finite (the exponential decay beats the polynomial growth), but it is not clear to me whether one can choose it to be small.
functional-analysis hilbert-spaces
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
Lately I was talking to a friend of mine and we came up with the following question
Denote by $mathcal{P}$ the set of all real valued polynomial functions. Is the set
$$ { p(x) e^{- alpha vert x vert} : pin mathcal{P}, alpha in mathbb{R}_{>0} } $$
dense in $L^2(mathbb{R} ,mathbb{R})?$
My feeling is that it should be true. I was thinking about using the Stone-Weierstrass, however, I cannot control the $L^2$-norm of the polynomial function outside the compact set. Clearly it is finite (the exponential decay beats the polynomial growth), but it is not clear to me whether one can choose it to be small.
functional-analysis hilbert-spaces
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
Lately I was talking to a friend of mine and we came up with the following question
Denote by $mathcal{P}$ the set of all real valued polynomial functions. Is the set
$$ { p(x) e^{- alpha vert x vert} : pin mathcal{P}, alpha in mathbb{R}_{>0} } $$
dense in $L^2(mathbb{R} ,mathbb{R})?$
My feeling is that it should be true. I was thinking about using the Stone-Weierstrass, however, I cannot control the $L^2$-norm of the polynomial function outside the compact set. Clearly it is finite (the exponential decay beats the polynomial growth), but it is not clear to me whether one can choose it to be small.
functional-analysis hilbert-spaces
functional-analysis hilbert-spaces
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 1 at 22:20
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 1 at 22:20
Severin Schraven
5,6381832
asked Nov 1 at 22:20
Severin Schraven
5,6381832
asked Nov 1 at 22:20
Severin Schraven
5,6381832
5,6381832
I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44
add a comment |
I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44
I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44
add a comment |
1 Answer
1
active
oldest
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up vote
9
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accepted
Yes, the answer is affirmative. It is equivalent to
${p(x)e^{-|x|} : pinmathcal{P}}$ is dense in $L_2(Bbb{R},Bbb{R})$.
In our context of Hilbert spaces, we need to show that
$$L_0 = left{fin L_2(Bbb{R}) : (forall ninBbb{Z}_{geq 0}) int_{Bbb{R}}x^n e^{-|x|}f(x),dx=0right}$$
consists of $fequiv 0$ only. Here and below, all functions are complex-valued.
Let $Lambda={lambdainBbb{C} : |Relambda|<1}$; for $fin L_2(Bbb{R})$, the function
$$B_f(lambda)=int_{Bbb{R}}e^{-|x|+lambda x}f(x),dx$$
is analytic in $Lambda$ (differentiation is admissible under the integral sign). Further, for any $ninBbb{Z}_{geq 0}$ we have $B_f^{(n)}(0)=displaystyleint_{Bbb{R}}x^n e^{-|x|}f(x),dx$ and, therefore, $fin L_0$ if and only if $B_fequiv 0$.
Now let $fin L_0$ and $gin L_1(Bbb{R})$ (fixme... much less is enough). Then
$$0=int_{Bbb{R}}g(lambda)B_f(ilambda),dlambda=int_{Bbb{R}}e^{-|x|}hat{g}(x)f(x),dx,quadhat{g}(x)=int_{Bbb{R}}e^{ilambda x}g(lambda),dlambda.$$
Thus, $e^{-|x|}f(x)$ is orthogonal to ${hat{g} : gin L_1(Bbb{R})}$. This space is dense in $L_2(Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from
$$hat{g}_0(x) = max{0,1-|x|} impliedby g_0(lambda)=frac{2}{pi}left(frac{sinlambda/2}{lambda}right)^2$$
using linear combinations and shifts; $hat{g}_1(x)=hat{g}(x+a)impliedby g_1(lambda)=e^{ilambda a}g(lambda)$.
(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)
answered Nov 5 at 8:07
metamorphy
3,0071520
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Yes, the answer is affirmative. It is equivalent to
${p(x)e^{-|x|} : pinmathcal{P}}$ is dense in $L_2(Bbb{R},Bbb{R})$.
In our context of Hilbert spaces, we need to show that
$$L_0 = left{fin L_2(Bbb{R}) : (forall ninBbb{Z}_{geq 0}) int_{Bbb{R}}x^n e^{-|x|}f(x),dx=0right}$$
consists of $fequiv 0$ only. Here and below, all functions are complex-valued.
Let $Lambda={lambdainBbb{C} : |Relambda|<1}$; for $fin L_2(Bbb{R})$, the function
$$B_f(lambda)=int_{Bbb{R}}e^{-|x|+lambda x}f(x),dx$$
is analytic in $Lambda$ (differentiation is admissible under the integral sign). Further, for any $ninBbb{Z}_{geq 0}$ we have $B_f^{(n)}(0)=displaystyleint_{Bbb{R}}x^n e^{-|x|}f(x),dx$ and, therefore, $fin L_0$ if and only if $B_fequiv 0$.
Now let $fin L_0$ and $gin L_1(Bbb{R})$ (fixme... much less is enough). Then
$$0=int_{Bbb{R}}g(lambda)B_f(ilambda),dlambda=int_{Bbb{R}}e^{-|x|}hat{g}(x)f(x),dx,quadhat{g}(x)=int_{Bbb{R}}e^{ilambda x}g(lambda),dlambda.$$
Thus, $e^{-|x|}f(x)$ is orthogonal to ${hat{g} : gin L_1(Bbb{R})}$. This space is dense in $L_2(Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from
$$hat{g}_0(x) = max{0,1-|x|} impliedby g_0(lambda)=frac{2}{pi}left(frac{sinlambda/2}{lambda}right)^2$$
using linear combinations and shifts; $hat{g}_1(x)=hat{g}(x+a)impliedby g_1(lambda)=e^{ilambda a}g(lambda)$.
(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)
answered Nov 5 at 8:07
metamorphy
3,0071520
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
|
show 2 more comments
up vote
9
down vote
accepted
Yes, the answer is affirmative. It is equivalent to
${p(x)e^{-|x|} : pinmathcal{P}}$ is dense in $L_2(Bbb{R},Bbb{R})$.
In our context of Hilbert spaces, we need to show that
$$L_0 = left{fin L_2(Bbb{R}) : (forall ninBbb{Z}_{geq 0}) int_{Bbb{R}}x^n e^{-|x|}f(x),dx=0right}$$
consists of $fequiv 0$ only. Here and below, all functions are complex-valued.
Let $Lambda={lambdainBbb{C} : |Relambda|<1}$; for $fin L_2(Bbb{R})$, the function
$$B_f(lambda)=int_{Bbb{R}}e^{-|x|+lambda x}f(x),dx$$
is analytic in $Lambda$ (differentiation is admissible under the integral sign). Further, for any $ninBbb{Z}_{geq 0}$ we have $B_f^{(n)}(0)=displaystyleint_{Bbb{R}}x^n e^{-|x|}f(x),dx$ and, therefore, $fin L_0$ if and only if $B_fequiv 0$.
Now let $fin L_0$ and $gin L_1(Bbb{R})$ (fixme... much less is enough). Then
$$0=int_{Bbb{R}}g(lambda)B_f(ilambda),dlambda=int_{Bbb{R}}e^{-|x|}hat{g}(x)f(x),dx,quadhat{g}(x)=int_{Bbb{R}}e^{ilambda x}g(lambda),dlambda.$$
Thus, $e^{-|x|}f(x)$ is orthogonal to ${hat{g} : gin L_1(Bbb{R})}$. This space is dense in $L_2(Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from
$$hat{g}_0(x) = max{0,1-|x|} impliedby g_0(lambda)=frac{2}{pi}left(frac{sinlambda/2}{lambda}right)^2$$
using linear combinations and shifts; $hat{g}_1(x)=hat{g}(x+a)impliedby g_1(lambda)=e^{ilambda a}g(lambda)$.
(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)
answered Nov 5 at 8:07
metamorphy
3,0071520
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
|
show 2 more comments
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Yes, the answer is affirmative. It is equivalent to
${p(x)e^{-|x|} : pinmathcal{P}}$ is dense in $L_2(Bbb{R},Bbb{R})$.
In our context of Hilbert spaces, we need to show that
$$L_0 = left{fin L_2(Bbb{R}) : (forall ninBbb{Z}_{geq 0}) int_{Bbb{R}}x^n e^{-|x|}f(x),dx=0right}$$
consists of $fequiv 0$ only. Here and below, all functions are complex-valued.
Let $Lambda={lambdainBbb{C} : |Relambda|<1}$; for $fin L_2(Bbb{R})$, the function
$$B_f(lambda)=int_{Bbb{R}}e^{-|x|+lambda x}f(x),dx$$
is analytic in $Lambda$ (differentiation is admissible under the integral sign). Further, for any $ninBbb{Z}_{geq 0}$ we have $B_f^{(n)}(0)=displaystyleint_{Bbb{R}}x^n e^{-|x|}f(x),dx$ and, therefore, $fin L_0$ if and only if $B_fequiv 0$.
Now let $fin L_0$ and $gin L_1(Bbb{R})$ (fixme... much less is enough). Then
$$0=int_{Bbb{R}}g(lambda)B_f(ilambda),dlambda=int_{Bbb{R}}e^{-|x|}hat{g}(x)f(x),dx,quadhat{g}(x)=int_{Bbb{R}}e^{ilambda x}g(lambda),dlambda.$$
Thus, $e^{-|x|}f(x)$ is orthogonal to ${hat{g} : gin L_1(Bbb{R})}$. This space is dense in $L_2(Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from
$$hat{g}_0(x) = max{0,1-|x|} impliedby g_0(lambda)=frac{2}{pi}left(frac{sinlambda/2}{lambda}right)^2$$
using linear combinations and shifts; $hat{g}_1(x)=hat{g}(x+a)impliedby g_1(lambda)=e^{ilambda a}g(lambda)$.
(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)
answered Nov 5 at 8:07
metamorphy
3,0071520
Yes, the answer is affirmative. It is equivalent to
${p(x)e^{-|x|} : pinmathcal{P}}$ is dense in $L_2(Bbb{R},Bbb{R})$.
In our context of Hilbert spaces, we need to show that
$$L_0 = left{fin L_2(Bbb{R}) : (forall ninBbb{Z}_{geq 0}) int_{Bbb{R}}x^n e^{-|x|}f(x),dx=0right}$$
consists of $fequiv 0$ only. Here and below, all functions are complex-valued.
Let $Lambda={lambdainBbb{C} : |Relambda|<1}$; for $fin L_2(Bbb{R})$, the function
$$B_f(lambda)=int_{Bbb{R}}e^{-|x|+lambda x}f(x),dx$$
is analytic in $Lambda$ (differentiation is admissible under the integral sign). Further, for any $ninBbb{Z}_{geq 0}$ we have $B_f^{(n)}(0)=displaystyleint_{Bbb{R}}x^n e^{-|x|}f(x),dx$ and, therefore, $fin L_0$ if and only if $B_fequiv 0$.
Now let $fin L_0$ and $gin L_1(Bbb{R})$ (fixme... much less is enough). Then
$$0=int_{Bbb{R}}g(lambda)B_f(ilambda),dlambda=int_{Bbb{R}}e^{-|x|}hat{g}(x)f(x),dx,quadhat{g}(x)=int_{Bbb{R}}e^{ilambda x}g(lambda),dlambda.$$
Thus, $e^{-|x|}f(x)$ is orthogonal to ${hat{g} : gin L_1(Bbb{R})}$. This space is dense in $L_2(Bbb{R})$ because, e.g., it contains all continuous piecewise linear finite functions obtained from
$$hat{g}_0(x) = max{0,1-|x|} impliedby g_0(lambda)=frac{2}{pi}left(frac{sinlambda/2}{lambda}right)^2$$
using linear combinations and shifts; $hat{g}_1(x)=hat{g}(x+a)impliedby g_1(lambda)=e^{ilambda a}g(lambda)$.
(I'm sure I've duplicated some known facts about integral transforms. Thus, it might be good to replace some parts of the above with references to these...)
answered Nov 5 at 8:07
metamorphy
3,0071520
edited Nov 5 at 21:51
answered Nov 5 at 8:07
metamorphy
3,0071520
answered Nov 5 at 8:07
metamorphy
3,0071520
answered Nov 5 at 8:07
metamorphy
3,0071520
3,0071520
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
|
show 2 more comments
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
1
1
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
Great answer. Thanks!
– Severin Schraven
Nov 5 at 13:53
2
2
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
I don't think it's stronger. The set of polynomials is invariant under $xmapsto frac{x}{alpha}$.
– mathworker21
Nov 5 at 18:58
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@mathworker21: +1, edited.
– metamorphy
Nov 5 at 21:56
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy very nice solution! I just read through it. Can you give me a reference for where you have seen such arguments? (please forgive me, but I am under the assumption that this solution is not completely original). Also, a remark: it is enough to consider $g$ Schwarz, and then density is obvious because the fourier transform is a bijection from Schwarz functions to Schwarz functions.
– mathworker21
Nov 15 at 15:10
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
@metamorphy also, one more question. it seems you used fourier inversion to conclude that $widehat{B_f(icdot)}(x) = e^{-|x|}f(x)$. To apply fourier inversion, don't you need to know that $B_f(icdot) in L^1$? How would you show this?
– mathworker21
Nov 15 at 16:18
|
show 2 more comments
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I'd guess that the answer is "no", with $1_{[0,1]}$ not being able to be approximated arbitrarily well by elements of your set... but I'm having trouble proving it. The reason why I think the answer is no is that if you have some polynomial $p$ approximating a given $f$, you must take $alpha$ to be really large to ensure quick enough decay, but then you need to make $p$ larger so that $pe^{-alpha |x|}$ is close enough to $f$, but then you need to make $alpha$ larger to ensure quick enough decay... etc. These are just some obscure thoughts
– mathworker21
Nov 4 at 22:13
@mathworker21. Given $epsilon>0$ let $f(x) = 1_{[0,1]}(x) , e^{x}$ and apply Stone-Weierstrass to get a polynomial $p$ with $|f-p|_{infty} < epsilon.$ Then $|1_{[0,1]} - e^{-x}p|_infty = |e^{-x}(f-p)|_infty < |f-p|_infty < epsilon$ since $|e^{-x}|_infty = 1$ on $[0,1]$.
– md2perpe
Nov 4 at 22:36
@md2perpe but you're doing $L^infty$ over $[0,1]$ not over $mathbb{R}$. My comment is saying that if you want to get a good $L^infty$ bound over $[0,1]$, you won't have the right decay in $mathbb{R}$. And conversely if you do have the right decay in $mathbb{R}$, you won't have an $L^infty$ bound on $[0,1]$.
– mathworker21
Nov 4 at 23:11
@mathworker21. Ah, you're right.
– md2perpe
Nov 5 at 6:44