Fourier transform of vanishing function in $L^2$
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Let $u$ be a function in $L^{2}(mathbb{R}^{3})$,
Question1:
Is $left | u right |_{L^{2}} =0 $ implies that $Fleft(uright)=0$?
Question2:
Is $left | u right |_{L^{2}} > 0$ implies that there exists a positive constant $c> 0$ such that $Fleft ( u right )> c$?
Note that $F(u)$ is the Fourier transform of $u$.
integration functional-analysis fourier-transform
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
asked Nov 17 at 17:56
user326064
63
add a comment |
up vote
0
down vote
favorite
Let $u$ be a function in $L^{2}(mathbb{R}^{3})$,
Question1:
Is $left | u right |_{L^{2}} =0 $ implies that $Fleft(uright)=0$?
Question2:
Is $left | u right |_{L^{2}} > 0$ implies that there exists a positive constant $c> 0$ such that $Fleft ( u right )> c$?
Note that $F(u)$ is the Fourier transform of $u$.
integration functional-analysis fourier-transform
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
asked Nov 17 at 17:56
user326064
63
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $u$ be a function in $L^{2}(mathbb{R}^{3})$,
Question1:
Is $left | u right |_{L^{2}} =0 $ implies that $Fleft(uright)=0$?
Question2:
Is $left | u right |_{L^{2}} > 0$ implies that there exists a positive constant $c> 0$ such that $Fleft ( u right )> c$?
Note that $F(u)$ is the Fourier transform of $u$.
integration functional-analysis fourier-transform
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
asked Nov 17 at 17:56
user326064
63
Let $u$ be a function in $L^{2}(mathbb{R}^{3})$,
Question1:
Is $left | u right |_{L^{2}} =0 $ implies that $Fleft(uright)=0$?
Question2:
Is $left | u right |_{L^{2}} > 0$ implies that there exists a positive constant $c> 0$ such that $Fleft ( u right )> c$?
Note that $F(u)$ is the Fourier transform of $u$.
integration functional-analysis fourier-transform
integration functional-analysis fourier-transform
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
asked Nov 17 at 17:56
user326064
63
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
asked Nov 17 at 17:56
user326064
63
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
edited Nov 18 at 4:41
Lord Shark the Unknown
98.1k958131
98.1k958131
asked Nov 17 at 17:56
user326064
63
asked Nov 17 at 17:56
user326064
63
asked Nov 17 at 17:56
user326064
63
63
add a comment |
add a comment |
2 Answers
2
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0
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The answer for question 1 is yes, because the Fourier transform $mathcal{F}:L^2to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $mathcal{F}$ to be a linear mapping, which is easy to verify).
Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $mathcal{F}(u)=0$ a.e.
The second part is false. To see it, take any $uin L^2, une0$ (and hence with strictly positive norm) and consider $mathcal{F}(u)$. If there is a constant $C$ for which $mathcal{F}(u)>C$, then $mathcal{F}(-u)=-mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.
answered Nov 17 at 18:13
Marco
1909
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
add a comment |
up vote
0
down vote
If $u neq 0$ on a subest of $mathbb{R}^{3}$ then $F(u) neq 0$ and thus $/F(u)/ neq 0$
Therefore, $left | u right |_{L^{2}} $ = $left | F(u) right |_{L^{2}} $$neq 0$.
the inverse is true.
answered Nov 18 at 14:06
user326064
63
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The answer for question 1 is yes, because the Fourier transform $mathcal{F}:L^2to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $mathcal{F}$ to be a linear mapping, which is easy to verify).
Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $mathcal{F}(u)=0$ a.e.
The second part is false. To see it, take any $uin L^2, une0$ (and hence with strictly positive norm) and consider $mathcal{F}(u)$. If there is a constant $C$ for which $mathcal{F}(u)>C$, then $mathcal{F}(-u)=-mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.
answered Nov 17 at 18:13
Marco
1909
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
add a comment |
up vote
0
down vote
The answer for question 1 is yes, because the Fourier transform $mathcal{F}:L^2to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $mathcal{F}$ to be a linear mapping, which is easy to verify).
Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $mathcal{F}(u)=0$ a.e.
The second part is false. To see it, take any $uin L^2, une0$ (and hence with strictly positive norm) and consider $mathcal{F}(u)$. If there is a constant $C$ for which $mathcal{F}(u)>C$, then $mathcal{F}(-u)=-mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.
answered Nov 17 at 18:13
Marco
1909
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
add a comment |
up vote
0
down vote
up vote
0
down vote
The answer for question 1 is yes, because the Fourier transform $mathcal{F}:L^2to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $mathcal{F}$ to be a linear mapping, which is easy to verify).
Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $mathcal{F}(u)=0$ a.e.
The second part is false. To see it, take any $uin L^2, une0$ (and hence with strictly positive norm) and consider $mathcal{F}(u)$. If there is a constant $C$ for which $mathcal{F}(u)>C$, then $mathcal{F}(-u)=-mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.
answered Nov 17 at 18:13
Marco
1909
The answer for question 1 is yes, because the Fourier transform $mathcal{F}:L^2to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $mathcal{F}$ to be a linear mapping, which is easy to verify).
Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $mathcal{F}(u)=0$ a.e.
The second part is false. To see it, take any $uin L^2, une0$ (and hence with strictly positive norm) and consider $mathcal{F}(u)$. If there is a constant $C$ for which $mathcal{F}(u)>C$, then $mathcal{F}(-u)=-mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.
answered Nov 17 at 18:13
Marco
1909
edited Nov 17 at 18:27
answered Nov 17 at 18:13
Marco
1909
answered Nov 17 at 18:13
Marco
1909
answered Nov 17 at 18:13
Marco
1909
1909
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
add a comment |
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
In question2, I take the implication $/Fleft ( u right )/$>$ c$.
– user326064
Nov 17 at 20:56
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
@user326064 Still false. Remark that $mathcal{F}(u')(xi)=2pi iximathcal{F}(u)(xi)$ and apply this rule to the Gaussian $u(x)=e^{-pi x^2}$, which is a fixed point for $mathcal{F}$. You have that $Vert uVert_2>0$ and $mathcal{F}(u)(0)=0$.
– Marco
Nov 17 at 22:10
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Using the Plancherel's theorem in $L^2$ , what is the condition on $/F(u)/$ such that assures that $left | u right |_{L^{2}} > c$?
– user326064
Nov 18 at 9:44
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Plancherel states that $mathcal{F}:L^2to L^2$ is a linear isometry, i.e. $Vertmathcal{F}(u)Vert_2=Vert uVert_2$, hence you only need to require that $$left(int_{mathbb{R}^3}|mathcal{F}(u)(xi)|^2dxiright)^{1/2}>c$$
– Marco
Nov 18 at 9:47
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
Is it possible to write this condition with only $/F(u)/$ ? This is my question.
– user326064
Nov 18 at 9:56
add a comment |
up vote
0
down vote
If $u neq 0$ on a subest of $mathbb{R}^{3}$ then $F(u) neq 0$ and thus $/F(u)/ neq 0$
Therefore, $left | u right |_{L^{2}} $ = $left | F(u) right |_{L^{2}} $$neq 0$.
the inverse is true.
answered Nov 18 at 14:06
user326064
63
add a comment |
up vote
0
down vote
If $u neq 0$ on a subest of $mathbb{R}^{3}$ then $F(u) neq 0$ and thus $/F(u)/ neq 0$
Therefore, $left | u right |_{L^{2}} $ = $left | F(u) right |_{L^{2}} $$neq 0$.
the inverse is true.
answered Nov 18 at 14:06
user326064
63
add a comment |
up vote
0
down vote
up vote
0
down vote
If $u neq 0$ on a subest of $mathbb{R}^{3}$ then $F(u) neq 0$ and thus $/F(u)/ neq 0$
Therefore, $left | u right |_{L^{2}} $ = $left | F(u) right |_{L^{2}} $$neq 0$.
the inverse is true.
answered Nov 18 at 14:06
user326064
63
If $u neq 0$ on a subest of $mathbb{R}^{3}$ then $F(u) neq 0$ and thus $/F(u)/ neq 0$
Therefore, $left | u right |_{L^{2}} $ = $left | F(u) right |_{L^{2}} $$neq 0$.
the inverse is true.
answered Nov 18 at 14:06
user326064
63
answered Nov 18 at 14:06
user326064
63
answered Nov 18 at 14:06
user326064
63
answered Nov 18 at 14:06
user326064
63
63
add a comment |
add a comment |
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