Does $f^2inmathcal{L}^1(mathbb{R})$ imply $finmathcal{L}^1(mathbb{R})$? [closed]
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Suppose $f^2inmathcal{L}^1(mathbb{R})$.
i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.
Is it true that $finmathcal{L}^1(mathbb{R})$?
real-analysis integration lebesgue-integral lebesgue-measure lp-spaces
asked Nov 21 at 1:16
usmndj
484
closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41
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If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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Suppose $f^2inmathcal{L}^1(mathbb{R})$.
i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.
Is it true that $finmathcal{L}^1(mathbb{R})$?
real-analysis integration lebesgue-integral lebesgue-measure lp-spaces
asked Nov 21 at 1:16
usmndj
484
closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18
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up vote
0
down vote
favorite
Suppose $f^2inmathcal{L}^1(mathbb{R})$.
i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.
Is it true that $finmathcal{L}^1(mathbb{R})$?
real-analysis integration lebesgue-integral lebesgue-measure lp-spaces
asked Nov 21 at 1:16
usmndj
484
Suppose $f^2inmathcal{L}^1(mathbb{R})$.
i.e. $int |f^2|~dlambda=int f^2~dlambda<infty$ where $lambda$ is the Lebesgue measure.
Is it true that $finmathcal{L}^1(mathbb{R})$?
real-analysis integration lebesgue-integral lebesgue-measure lp-spaces
real-analysis integration lebesgue-integral lebesgue-measure lp-spaces
asked Nov 21 at 1:16
usmndj
484
asked Nov 21 at 1:16
usmndj
484
asked Nov 21 at 1:16
usmndj
484
asked Nov 21 at 1:16
usmndj
484
asked Nov 21 at 1:16
usmndj
484
484
closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom Nov 21 at 8:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Cesareo, Lord Shark the Unknown, Jean-Claude Arbaut, Tom-Tom
If this question can be reworded to fit the rules in the help center, please edit the question.
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18
add a comment |
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18
add a comment |
2 Answers
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Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.
This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$
Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.
answered Nov 21 at 1:31
user3482749
2,086414
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No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.
answered Nov 21 at 1:30
Saucy O'Path
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.
This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$
Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.
answered Nov 21 at 1:31
user3482749
2,086414
add a comment |
up vote
2
down vote
Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.
This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$
Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.
answered Nov 21 at 1:31
user3482749
2,086414
add a comment |
up vote
2
down vote
up vote
2
down vote
Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.
This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$
Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.
answered Nov 21 at 1:31
user3482749
2,086414
Clearly $f^2$ is in $mathcal{L}^1(mathbb{R})$ if and only if $f$ is in $mathcal{L}^2(mathbb{R})$, so your question is equivalent to asking whether $mathcal{L}^2(mathbb{R})$ is a subset of $mathcal{L}^1(mathbb{R})$.
This is false, with a particular example being $$f(x) = left{array{0&xin[-1,1]\frac{1}{|x|}&text{otherwise}}right.$$
Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.
answered Nov 21 at 1:31
user3482749
2,086414
answered Nov 21 at 1:31
user3482749
2,086414
answered Nov 21 at 1:31
user3482749
2,086414
answered Nov 21 at 1:31
user3482749
2,086414
2,086414
add a comment |
add a comment |
up vote
1
down vote
No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
add a comment |
up vote
1
down vote
No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
add a comment |
up vote
1
down vote
up vote
1
down vote
No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
No, $L^2(Bbb R)nsubseteq L^1(Bbb R)$. It would be the case if you were considering a space of finite measure, though.
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
answered Nov 21 at 1:30
Saucy O'Path
5,7341626
5,7341626
add a comment |
add a comment |
What's the motivation for this question? Can you add your own thoughts on the question?
– amWhy
Nov 21 at 1:18