Mixing
Fourier transform $mathcal F colon (mathcal S(mathbb R^d), lVert cdot rVert_1) to L^1(mathbb R^d)$ unbounded?
$begingroup$
I want to prove or disprove that the Fourier transform $mathcal F colon (mathcal S(mathbb R^d), lVert cdot rVert_1) to L^1(mathbb R^d)$ is unbounded, where $lVertcdot rVert_1$ denotes the $L^1(mathbb R^d)$-norm.
Having thought about this for a moment, I believe it is indeed unbounded. So I tried to find a sequence of Schwartz functions $(f_n)_{nin mathbb N} subseteq mathcal S(mathbb R^d)$ with $forall n: lVert f_n rVert_1 = 1$ and $$lVert mathcal F f_n rVert to +infty.$$
Of course I first thought about Gaussians but couldn't quite find a suitable sequence. Any help appreciated!
real-analysis functional-analysis analysis fourier-analysis mathematical-physics
asked Jan 23 at 19:18
bavor42bavor42
320110
$endgroup$
add a comment |
$begingroup$
I want to prove or disprove that the Fourier transform $mathcal F colon (mathcal S(mathbb R^d), lVert cdot rVert_1) to L^1(mathbb R^d)$ is unbounded, where $lVertcdot rVert_1$ denotes the $L^1(mathbb R^d)$-norm.
Having thought about this for a moment, I believe it is indeed unbounded. So I tried to find a sequence of Schwartz functions $(f_n)_{nin mathbb N} subseteq mathcal S(mathbb R^d)$ with $forall n: lVert f_n rVert_1 = 1$ and $$lVert mathcal F f_n rVert to +infty.$$
Of course I first thought about Gaussians but couldn't quite find a suitable sequence. Any help appreciated!
real-analysis functional-analysis analysis fourier-analysis mathematical-physics
asked Jan 23 at 19:18
bavor42bavor42
320110
$endgroup$
1
$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11
add a comment |
$begingroup$
I want to prove or disprove that the Fourier transform $mathcal F colon (mathcal S(mathbb R^d), lVert cdot rVert_1) to L^1(mathbb R^d)$ is unbounded, where $lVertcdot rVert_1$ denotes the $L^1(mathbb R^d)$-norm.
Having thought about this for a moment, I believe it is indeed unbounded. So I tried to find a sequence of Schwartz functions $(f_n)_{nin mathbb N} subseteq mathcal S(mathbb R^d)$ with $forall n: lVert f_n rVert_1 = 1$ and $$lVert mathcal F f_n rVert to +infty.$$
Of course I first thought about Gaussians but couldn't quite find a suitable sequence. Any help appreciated!
real-analysis functional-analysis analysis fourier-analysis mathematical-physics
asked Jan 23 at 19:18
bavor42bavor42
320110
$endgroup$
I want to prove or disprove that the Fourier transform $mathcal F colon (mathcal S(mathbb R^d), lVert cdot rVert_1) to L^1(mathbb R^d)$ is unbounded, where $lVertcdot rVert_1$ denotes the $L^1(mathbb R^d)$-norm.
Having thought about this for a moment, I believe it is indeed unbounded. So I tried to find a sequence of Schwartz functions $(f_n)_{nin mathbb N} subseteq mathcal S(mathbb R^d)$ with $forall n: lVert f_n rVert_1 = 1$ and $$lVert mathcal F f_n rVert to +infty.$$
Of course I first thought about Gaussians but couldn't quite find a suitable sequence. Any help appreciated!
real-analysis functional-analysis analysis fourier-analysis mathematical-physics
real-analysis functional-analysis analysis fourier-analysis mathematical-physics
asked Jan 23 at 19:18
bavor42bavor42
320110
asked Jan 23 at 19:18
bavor42bavor42
320110
asked Jan 23 at 19:18
bavor42bavor42
320110
asked Jan 23 at 19:18
bavor42bavor42
320110
asked Jan 23 at 19:18
bavor42bavor42
320110
320110
1
$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11
add a comment |
1
$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11
1
1
$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11
$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're right to consider Gaussians!
If you define the Fourier transform of a Schwarz function $f$ to be
$$mathcal F f(xi)=int_{mathbb R^d}f(t)e^{-2ipilangle xi, xrangle}dx$$
then consider the family of Gaussian functions parametrized by $sigma >0$
$$f_sigma(t)= frac 1 {(2pi)^{frac d 2}sigma^d }e^{-frac {|x|^2}{2sigma^2}}$$
The corresponding Fourier transforms are
$$mathcal F f_sigma(xi)=e^{-2pisigma^2|xi|^2}$$
Now $$|f_sigma|_1=mathcal F f_sigma(0) =1$$ while $$|mathcal F f_sigma|_1=frac {C} {sigma^d}$$ for some constant $C$.
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
$endgroup$
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
|
show 2 more comments
$begingroup$
Begin with $f=mathbb1 _{[-1/2,1/2]}in L^1(mathbb R)$. This satisfies
$$ mathcal F f(xi) = frac{sin(pi xi)}{pi xi} notin L^1(mathbb R)$$
Now take any $f_n in mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $|f_n|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence
$$mathcal F f_n(xi) = int_{mathbb R} f_n(x)e^{-2pi i xxi} dx to mathcal F f(xi) $$
By Fatou's lemma,
$$ infty = |mathcal Ff |_{L^1} le liminf_{ntoinfty}|mathcal Ff_n |_{L^1}$$
For dimensions $d>1$, one can use $f = mathbb 1_{[-1/2,1/2]^d} $.
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're right to consider Gaussians!
If you define the Fourier transform of a Schwarz function $f$ to be
$$mathcal F f(xi)=int_{mathbb R^d}f(t)e^{-2ipilangle xi, xrangle}dx$$
then consider the family of Gaussian functions parametrized by $sigma >0$
$$f_sigma(t)= frac 1 {(2pi)^{frac d 2}sigma^d }e^{-frac {|x|^2}{2sigma^2}}$$
The corresponding Fourier transforms are
$$mathcal F f_sigma(xi)=e^{-2pisigma^2|xi|^2}$$
Now $$|f_sigma|_1=mathcal F f_sigma(0) =1$$ while $$|mathcal F f_sigma|_1=frac {C} {sigma^d}$$ for some constant $C$.
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
$endgroup$
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
|
show 2 more comments
$begingroup$
You're right to consider Gaussians!
If you define the Fourier transform of a Schwarz function $f$ to be
$$mathcal F f(xi)=int_{mathbb R^d}f(t)e^{-2ipilangle xi, xrangle}dx$$
then consider the family of Gaussian functions parametrized by $sigma >0$
$$f_sigma(t)= frac 1 {(2pi)^{frac d 2}sigma^d }e^{-frac {|x|^2}{2sigma^2}}$$
The corresponding Fourier transforms are
$$mathcal F f_sigma(xi)=e^{-2pisigma^2|xi|^2}$$
Now $$|f_sigma|_1=mathcal F f_sigma(0) =1$$ while $$|mathcal F f_sigma|_1=frac {C} {sigma^d}$$ for some constant $C$.
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
$endgroup$
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
|
show 2 more comments
$begingroup$
You're right to consider Gaussians!
If you define the Fourier transform of a Schwarz function $f$ to be
$$mathcal F f(xi)=int_{mathbb R^d}f(t)e^{-2ipilangle xi, xrangle}dx$$
then consider the family of Gaussian functions parametrized by $sigma >0$
$$f_sigma(t)= frac 1 {(2pi)^{frac d 2}sigma^d }e^{-frac {|x|^2}{2sigma^2}}$$
The corresponding Fourier transforms are
$$mathcal F f_sigma(xi)=e^{-2pisigma^2|xi|^2}$$
Now $$|f_sigma|_1=mathcal F f_sigma(0) =1$$ while $$|mathcal F f_sigma|_1=frac {C} {sigma^d}$$ for some constant $C$.
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
$endgroup$
You're right to consider Gaussians!
If you define the Fourier transform of a Schwarz function $f$ to be
$$mathcal F f(xi)=int_{mathbb R^d}f(t)e^{-2ipilangle xi, xrangle}dx$$
then consider the family of Gaussian functions parametrized by $sigma >0$
$$f_sigma(t)= frac 1 {(2pi)^{frac d 2}sigma^d }e^{-frac {|x|^2}{2sigma^2}}$$
The corresponding Fourier transforms are
$$mathcal F f_sigma(xi)=e^{-2pisigma^2|xi|^2}$$
Now $$|f_sigma|_1=mathcal F f_sigma(0) =1$$ while $$|mathcal F f_sigma|_1=frac {C} {sigma^d}$$ for some constant $C$.
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
edited Jan 23 at 21:32
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
answered Jan 23 at 19:40
Stefan LafonStefan Lafon
2,92519
2,92519
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
|
show 2 more comments
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
And this blows up as $sigma to 0$... this was exactly what I thought but somehow wasn't able to write down. Thank you very much!
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
Jan 23 at 19:53
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Minor typo, you meant to write $f_sigma(x)$ not $t$. Secondly I believe there should be a $d$th power of $sigma$ in $f_sigma$ if you intended $mathcal F f_sigma$ to have this formula. (Conclusion is of course right)
$endgroup$
– Calvin Khor
Jan 23 at 19:54
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
$begingroup$
Note that you changed your Fourier transform definition to $$e^{-2ipilangle xi, xrangle} int_{mathbb R^d}f$$
$endgroup$
– Calvin Khor
Jan 23 at 20:09
1
1
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
$begingroup$
Arghhh, need coffee...
$endgroup$
– Stefan Lafon
Jan 23 at 21:33
|
show 2 more comments
$begingroup$
Begin with $f=mathbb1 _{[-1/2,1/2]}in L^1(mathbb R)$. This satisfies
$$ mathcal F f(xi) = frac{sin(pi xi)}{pi xi} notin L^1(mathbb R)$$
Now take any $f_n in mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $|f_n|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence
$$mathcal F f_n(xi) = int_{mathbb R} f_n(x)e^{-2pi i xxi} dx to mathcal F f(xi) $$
By Fatou's lemma,
$$ infty = |mathcal Ff |_{L^1} le liminf_{ntoinfty}|mathcal Ff_n |_{L^1}$$
For dimensions $d>1$, one can use $f = mathbb 1_{[-1/2,1/2]^d} $.
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
add a comment |
$begingroup$
Begin with $f=mathbb1 _{[-1/2,1/2]}in L^1(mathbb R)$. This satisfies
$$ mathcal F f(xi) = frac{sin(pi xi)}{pi xi} notin L^1(mathbb R)$$
Now take any $f_n in mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $|f_n|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence
$$mathcal F f_n(xi) = int_{mathbb R} f_n(x)e^{-2pi i xxi} dx to mathcal F f(xi) $$
By Fatou's lemma,
$$ infty = |mathcal Ff |_{L^1} le liminf_{ntoinfty}|mathcal Ff_n |_{L^1}$$
For dimensions $d>1$, one can use $f = mathbb 1_{[-1/2,1/2]^d} $.
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
add a comment |
$begingroup$
Begin with $f=mathbb1 _{[-1/2,1/2]}in L^1(mathbb R)$. This satisfies
$$ mathcal F f(xi) = frac{sin(pi xi)}{pi xi} notin L^1(mathbb R)$$
Now take any $f_n in mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $|f_n|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence
$$mathcal F f_n(xi) = int_{mathbb R} f_n(x)e^{-2pi i xxi} dx to mathcal F f(xi) $$
By Fatou's lemma,
$$ infty = |mathcal Ff |_{L^1} le liminf_{ntoinfty}|mathcal Ff_n |_{L^1}$$
For dimensions $d>1$, one can use $f = mathbb 1_{[-1/2,1/2]^d} $.
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
Begin with $f=mathbb1 _{[-1/2,1/2]}in L^1(mathbb R)$. This satisfies
$$ mathcal F f(xi) = frac{sin(pi xi)}{pi xi} notin L^1(mathbb R)$$
Now take any $f_n in mathcal S $ converging to $f$ in $L^1$ and almost everywhere. We have $|f_n|_{L^1} < 2$ eventually. By dominated convergence, we have the pointwise convergence
$$mathcal F f_n(xi) = int_{mathbb R} f_n(x)e^{-2pi i xxi} dx to mathcal F f(xi) $$
By Fatou's lemma,
$$ infty = |mathcal Ff |_{L^1} le liminf_{ntoinfty}|mathcal Ff_n |_{L^1}$$
For dimensions $d>1$, one can use $f = mathbb 1_{[-1/2,1/2]^d} $.
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
answered Jan 23 at 19:40
Calvin KhorCalvin Khor
12.4k21439
12.4k21439
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
add a comment |
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
Thanks for this alternative approach! :)
$endgroup$
– bavor42
Jan 23 at 19:51
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 You're welcome :)
$endgroup$
– Calvin Khor
Jan 23 at 19:56
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
@bavor42 I believe the other approach is more or less the same idea, the gaussians (after the typo I pointed out is fixed) forms an approximation to the identity, and $mathcal F delta = 1 notin L^1$. Then the (explicit) approximation sequence was used to find that the $L^1$ norm blew up
$endgroup$
– Calvin Khor
Jan 23 at 19:59
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
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– bavor42
Jan 23 at 20:06
$begingroup$
Yeah that makes sense. Still, 2 nice ways to look at it ;)
$endgroup$
– bavor42
Jan 23 at 20:06
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$begingroup$
Some uncertainty principle thing going on here...
$endgroup$
– copper.hat
Jan 23 at 20:11