Proving $int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$
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I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.
Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.
calculus integration
edited Nov 30 '18 at 6:24
user1551
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
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add a comment |
$begingroup$
I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.
Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.
calculus integration
edited Nov 30 '18 at 6:24
user1551
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
$endgroup$
$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43
add a comment |
$begingroup$
I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.
Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.
calculus integration
edited Nov 30 '18 at 6:24
user1551
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
$endgroup$
I was supposed to show that $$langle f,g rangle = int_{-1}^1 frac{f(x)g(x)}{sqrt{(1-x^2)}} dx$$ is an inner product. The question basically said to prove symmetry, bilinearity, and that $langle f,f rangle gt 0$ where $f ne 0$.
Now, I have already shown symmetry and bilinearity. However, I am stuck with the last part. I would have to prove that $$int_{-1}^1 frac{(f(x))^2}{sqrt{(1-x^2)}} dx gt 0$$ but I have no clue how to do that. Could someone explain where should I start? Thanks.
calculus integration
calculus integration
edited Nov 30 '18 at 6:24
user1551
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
edited Nov 30 '18 at 6:24
user1551
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
edited Nov 30 '18 at 6:24
user1551
72.3k566127
edited Nov 30 '18 at 6:24
user1551
72.3k566127
edited Nov 30 '18 at 6:24
user1551
72.3k566127
72.3k566127
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
asked Nov 30 '18 at 4:39
dmsj djsldmsj djsl
35517
35517
$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43
add a comment |
$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43
$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43
$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43
add a comment |
3 Answers
3
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You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
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add a comment |
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Hint: The integral of a positive function is positive.
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
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2
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I am curious why there is a downvote?
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– Tianlalu
Nov 30 '18 at 4:57
add a comment |
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Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,
$int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,
a contradiction.
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
$endgroup$
add a comment |
$begingroup$
You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
$endgroup$
add a comment |
$begingroup$
You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
$endgroup$
You're overthinking! If we assume the integral exists, then $f^2(x)geq0$ and $sqrt{1-x^2}geq0$ for all $xin[-1,1]$. Since we assume that $f$ is not identically $0$, then this integral is of a positive function and the integral is positive.
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
answered Dec 1 '18 at 3:04
YiFanYiFan
2,7531422
2,7531422
add a comment |
add a comment |
$begingroup$
Hint: The integral of a positive function is positive.
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
$endgroup$
2
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
add a comment |
$begingroup$
Hint: The integral of a positive function is positive.
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
$endgroup$
2
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
add a comment |
$begingroup$
Hint: The integral of a positive function is positive.
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
$endgroup$
Hint: The integral of a positive function is positive.
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
answered Nov 30 '18 at 4:41
Jimmy R.Jimmy R.
33k42157
33k42157
2
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
add a comment |
2
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
2
2
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
$begingroup$
I am curious why there is a downvote?
$endgroup$
– Tianlalu
Nov 30 '18 at 4:57
add a comment |
$begingroup$
Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,
$int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,
a contradiction.
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
$endgroup$
add a comment |
$begingroup$
Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,
$int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,
a contradiction.
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
$endgroup$
add a comment |
$begingroup$
Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,
$int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,
a contradiction.
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
$endgroup$
Assume that $g(x) > 0$ almost everywhere and $E$ is a set of positive measure, that is, $m(E) > 0$. Suppose, by way of contradiction, that $int_E g(x) dx leq 0$. Let $E_n = {x in E: g(x) > 1/n }$. Then $E = cup_{k=1}^infty E_k$ and since $g(x) > 0$ almost everywhere, there exists an $n$ such that $m(E_n)>0$. Hence,
$int_E g(x) dx geq int_{E_n} g(x) dx > frac{1}{n} m(E_n) > 0 $ ,
a contradiction.
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
edited Dec 2 '18 at 22:52
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
answered Nov 30 '18 at 4:53
Mustafa SaidMustafa Said
2,9411913
2,9411913
add a comment |
add a comment |
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$begingroup$
Note that the integrand is positive. When it exists, the Riemann integral is positive definite.
$endgroup$
– Dzoooks
Nov 30 '18 at 4:43