Is the following operator a compact operator?
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1
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I have to choose whether the following operator
$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$
is a compact operator. I have tried to use the Cauchy Schwarz inequality
$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$
but how can I identify the compactness of an operator?
functional-analysis operator-theory cauchy-schwarz-inequality
asked Nov 20 at 15:26
MathCracky
445212
add a comment |
up vote
1
down vote
favorite
I have to choose whether the following operator
$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$
is a compact operator. I have tried to use the Cauchy Schwarz inequality
$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$
but how can I identify the compactness of an operator?
functional-analysis operator-theory cauchy-schwarz-inequality
asked Nov 20 at 15:26
MathCracky
445212
2
Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have to choose whether the following operator
$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$
is a compact operator. I have tried to use the Cauchy Schwarz inequality
$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$
but how can I identify the compactness of an operator?
functional-analysis operator-theory cauchy-schwarz-inequality
asked Nov 20 at 15:26
MathCracky
445212
I have to choose whether the following operator
$$fin L^2(mathbb{R})mapstoint_{mathbb{R}}f(x) e^{-x^2} dx$$
is a compact operator. I have tried to use the Cauchy Schwarz inequality
$$langle f, e^{-x^2}rangle leq |f|_{L^2} cdot underbrace{|e^{-x^2}|_{L^2}}_{leqpi} leqpicdot|f|_{L^2} $$
but how can I identify the compactness of an operator?
functional-analysis operator-theory cauchy-schwarz-inequality
functional-analysis operator-theory cauchy-schwarz-inequality
asked Nov 20 at 15:26
MathCracky
445212
asked Nov 20 at 15:26
MathCracky
445212
asked Nov 20 at 15:26
MathCracky
445212
asked Nov 20 at 15:26
MathCracky
445212
asked Nov 20 at 15:26
MathCracky
445212
445212
2
Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36
add a comment |
2
Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36
2
2
Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36
Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36
add a comment |
1 Answer
1
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votes
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3
down vote
accepted
$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.
answered Nov 20 at 15:31
Federico
3,775511
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.
answered Nov 20 at 15:31
Federico
3,775511
add a comment |
up vote
3
down vote
accepted
$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.
answered Nov 20 at 15:31
Federico
3,775511
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.
answered Nov 20 at 15:31
Federico
3,775511
$T:Xto Y$ is compact if it sends bounded sets to relatively compact sets. Your operator $T:L^2tomathbb R$ sends bounded sets to bounded sets (because of Cauchy-Schwarz), but bounded sets in $mathbb R$ are relatively compact, so the operator is compact.
answered Nov 20 at 15:31
Federico
3,775511
answered Nov 20 at 15:31
Federico
3,775511
answered Nov 20 at 15:31
Federico
3,775511
answered Nov 20 at 15:31
Federico
3,775511
3,775511
add a comment |
add a comment |
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Any continuous linear map into a finite dimensional vector space is necessarily compact.
– Omnomnomnom
Nov 20 at 15:36