Mixing
Question concerning an application of Cauchy-Schwarz
Specifically, the question is as follows:
Prove that for every integrable real-valued $f:mathbb{R}rightarrowmathbb{R}$,
$$left(int_1^ef(x)dxright)^2leqint_1^ex(f(x))^2dx.$$
I'm really just looking for assistance getting started on this problem. It is homework, so I request hints, not a full solution. For reference, this was asked in the context of linear algebra, not real analysis.
I am able to show that
$$langle f,grangle=int_1^ef(t)g(t)dt$$
is an inner product on $L^2$. We have developed the Cauchy-Schwarz inequality, but not Schwarz's inequality, so this question is presumably solvable simply using
$$|langle u,vrangle|leq|u||v|$$
(where $u$ and $v$ are elements of some vector space). So, setting $g=1$, we have
$$left(int_1^ef(x)dxright)^2=|langle f,grangle|^2leq(|f||1|)^2=(e-1)int_1^e(f(x))^2dx.$$
But I don't see how to draw the desired conclusion from here (so maybe I'm not approaching it right??).
real-analysis linear-algebra norm inner-product-space cauchy-schwarz-inequality
asked Nov 20 '18 at 3:55
Atsina
791116
|
show 1 more comment
Specifically, the question is as follows:
Prove that for every integrable real-valued $f:mathbb{R}rightarrowmathbb{R}$,
$$left(int_1^ef(x)dxright)^2leqint_1^ex(f(x))^2dx.$$
I'm really just looking for assistance getting started on this problem. It is homework, so I request hints, not a full solution. For reference, this was asked in the context of linear algebra, not real analysis.
I am able to show that
$$langle f,grangle=int_1^ef(t)g(t)dt$$
is an inner product on $L^2$. We have developed the Cauchy-Schwarz inequality, but not Schwarz's inequality, so this question is presumably solvable simply using
$$|langle u,vrangle|leq|u||v|$$
(where $u$ and $v$ are elements of some vector space). So, setting $g=1$, we have
$$left(int_1^ef(x)dxright)^2=|langle f,grangle|^2leq(|f||1|)^2=(e-1)int_1^e(f(x))^2dx.$$
But I don't see how to draw the desired conclusion from here (so maybe I'm not approaching it right??).
real-analysis linear-algebra norm inner-product-space cauchy-schwarz-inequality
asked Nov 20 '18 at 3:55
Atsina
791116
What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27
|
show 1 more comment
Specifically, the question is as follows:
Prove that for every integrable real-valued $f:mathbb{R}rightarrowmathbb{R}$,
$$left(int_1^ef(x)dxright)^2leqint_1^ex(f(x))^2dx.$$
I'm really just looking for assistance getting started on this problem. It is homework, so I request hints, not a full solution. For reference, this was asked in the context of linear algebra, not real analysis.
I am able to show that
$$langle f,grangle=int_1^ef(t)g(t)dt$$
is an inner product on $L^2$. We have developed the Cauchy-Schwarz inequality, but not Schwarz's inequality, so this question is presumably solvable simply using
$$|langle u,vrangle|leq|u||v|$$
(where $u$ and $v$ are elements of some vector space). So, setting $g=1$, we have
$$left(int_1^ef(x)dxright)^2=|langle f,grangle|^2leq(|f||1|)^2=(e-1)int_1^e(f(x))^2dx.$$
But I don't see how to draw the desired conclusion from here (so maybe I'm not approaching it right??).
real-analysis linear-algebra norm inner-product-space cauchy-schwarz-inequality
asked Nov 20 '18 at 3:55
Atsina
791116
Specifically, the question is as follows:
Prove that for every integrable real-valued $f:mathbb{R}rightarrowmathbb{R}$,
$$left(int_1^ef(x)dxright)^2leqint_1^ex(f(x))^2dx.$$
I'm really just looking for assistance getting started on this problem. It is homework, so I request hints, not a full solution. For reference, this was asked in the context of linear algebra, not real analysis.
I am able to show that
$$langle f,grangle=int_1^ef(t)g(t)dt$$
is an inner product on $L^2$. We have developed the Cauchy-Schwarz inequality, but not Schwarz's inequality, so this question is presumably solvable simply using
$$|langle u,vrangle|leq|u||v|$$
(where $u$ and $v$ are elements of some vector space). So, setting $g=1$, we have
$$left(int_1^ef(x)dxright)^2=|langle f,grangle|^2leq(|f||1|)^2=(e-1)int_1^e(f(x))^2dx.$$
But I don't see how to draw the desired conclusion from here (so maybe I'm not approaching it right??).
real-analysis linear-algebra norm inner-product-space cauchy-schwarz-inequality
real-analysis linear-algebra norm inner-product-space cauchy-schwarz-inequality
asked Nov 20 '18 at 3:55
Atsina
791116
asked Nov 20 '18 at 3:55
Atsina
791116
asked Nov 20 '18 at 3:55
Atsina
791116
asked Nov 20 '18 at 3:55
Atsina
791116
asked Nov 20 '18 at 3:55
Atsina
791116
791116
What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27
|
show 1 more comment
What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27
What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27
|
show 1 more comment
1 Answer
1
active
oldest
votes
Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
add a comment |
Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
add a comment |
Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
answered Nov 20 '18 at 4:14
Jacky Chong
17.7k21128
17.7k21128
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
add a comment |
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
Ah. This is a better formulation of the idea presented in the comments. Thank you.
– Atsina
Nov 20 '18 at 4:25
2
2
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@Atsina Basically the same thing.
– xbh
Nov 20 '18 at 4:26
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
@xbh Indeed. It's just (slightly) easier to work with.
– Atsina
Nov 20 '18 at 4:33
add a comment |
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What happens if $u = f(x)/sqrt x$ and $g = sqrt x$?
– JavaMan
Nov 20 '18 at 3:57
@JavaMan You mean $v=sqrt{x}$?
– Atsina
Nov 20 '18 at 3:59
@JavaMan Something like $left(int_1^efright)^2=left|int_1^efright|^2=|langle u,vrangle|^2leq(|u||v|)^2=left(sqrt{int_1^efrac{f^2}{x}}sqrt{int_1^ex}right)^2=int_1^efrac{f^2}{x}int_1^exleqint_1^exf^2$
– Atsina
Nov 20 '18 at 4:18
Oops. I meant $u = f(x) sqrt x$ and $v =1/ sqrt x$
– JavaMan
Nov 20 '18 at 4:26
I don't think your last inequality in your most recent comment is valid...
– JavaMan
Nov 20 '18 at 4:27