Question concerning an application of Cauchy-Schwarz












1














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







share|cite|improve this question






















  • 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
















1














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







share|cite|improve this question






















  • 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














1












1








1


1





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






active

oldest

votes


















1














Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}






share|cite|improve this answer





















  • 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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}






share|cite|improve this answer





















  • 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
















1














Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}






share|cite|improve this answer





















  • 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














1












1








1






Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}






share|cite|improve this answer












Hint:
begin{align}
int^e_1 frac{dx}{x} = 1.
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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