Contradiction Observation to Cauchy Schwarz inequality for inequalty












-1












$begingroup$



$f:[0,1]to mathbb R$ , be continuous function then prove that
$$int_0^1f^2(x)dxgeq biggl(int_0^1|f(x)| biggr) ^2$$




I tried this for $x^2$



For that above is true



But I checked following proof Which is complete opposite to above.
Proving the Cauchy-Schwarz integral inequality in a different way



Please help me to find that where is I am making wrong ?



Any help will be appreciated






real-analysis analysis cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not? Just let $g=1$ for your result.
    $endgroup$
    – Zachary Selk
    Dec 6 '18 at 15:19
















-1












$begingroup$



$f:[0,1]to mathbb R$ , be continuous function then prove that
$$int_0^1f^2(x)dxgeq biggl(int_0^1|f(x)| biggr) ^2$$




I tried this for $x^2$



For that above is true



But I checked following proof Which is complete opposite to above.
Proving the Cauchy-Schwarz integral inequality in a different way



Please help me to find that where is I am making wrong ?



Any help will be appreciated






real-analysis analysis cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not? Just let $g=1$ for your result.
    $endgroup$
    – Zachary Selk
    Dec 6 '18 at 15:19














-1












-1








-1





$begingroup$



$f:[0,1]to mathbb R$ , be continuous function then prove that
$$int_0^1f^2(x)dxgeq biggl(int_0^1|f(x)| biggr) ^2$$




I tried this for $x^2$



For that above is true



But I checked following proof Which is complete opposite to above.
Proving the Cauchy-Schwarz integral inequality in a different way



Please help me to find that where is I am making wrong ?



Any help will be appreciated






real-analysis analysis cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$





$f:[0,1]to mathbb R$ , be continuous function then prove that
$$int_0^1f^2(x)dxgeq biggl(int_0^1|f(x)| biggr) ^2$$




I tried this for $x^2$



For that above is true



But I checked following proof Which is complete opposite to above.
Proving the Cauchy-Schwarz integral inequality in a different way



Please help me to find that where is I am making wrong ?



Any help will be appreciated






real-analysis analysis cauchy-schwarz-inequality




real-analysis analysis cauchy-schwarz-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 15:37









José Carlos Santos

157k22126227




157k22126227










asked Dec 6 '18 at 15:15









MathLoverMathLover

49710




49710












  • $begingroup$
    It's not? Just let $g=1$ for your result.
    $endgroup$
    – Zachary Selk
    Dec 6 '18 at 15:19


















  • $begingroup$
    It's not? Just let $g=1$ for your result.
    $endgroup$
    – Zachary Selk
    Dec 6 '18 at 15:19
















$begingroup$
It's not? Just let $g=1$ for your result.
$endgroup$
– Zachary Selk
Dec 6 '18 at 15:19




$begingroup$
It's not? Just let $g=1$ for your result.
$endgroup$
– Zachary Selk
Dec 6 '18 at 15:19










1 Answer
1






active

oldest

votes


















1












$begingroup$

The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,gin C^1bigl([0,1]bigr)$, then$$left(int_0^1f(x)g(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)left(int_0^1g^2(x),mathrm dxright).$$If you choose $g=f$, you get the trivial inequality$$left(int_0^1f^2(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)^2$$or$$int_0^1f^2(x),mathrm dxleqslantint_0^1f^2(x),mathrm dx.$$This in no way contradicts what you are supposed to prove.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ohh I am applying wrong result. Thanks A lot .You are always helping me.
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:37










  • $begingroup$
    ohh Sorry Sir I forget....
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:40










  • $begingroup$
    Sir ,Please can you give some hint so that I can prove above theorem
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:43










  • $begingroup$
    Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 15:45










  • $begingroup$
    Thanks a lots Sir...
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:47











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,gin C^1bigl([0,1]bigr)$, then$$left(int_0^1f(x)g(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)left(int_0^1g^2(x),mathrm dxright).$$If you choose $g=f$, you get the trivial inequality$$left(int_0^1f^2(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)^2$$or$$int_0^1f^2(x),mathrm dxleqslantint_0^1f^2(x),mathrm dx.$$This in no way contradicts what you are supposed to prove.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ohh I am applying wrong result. Thanks A lot .You are always helping me.
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:37










  • $begingroup$
    ohh Sorry Sir I forget....
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:40










  • $begingroup$
    Sir ,Please can you give some hint so that I can prove above theorem
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:43










  • $begingroup$
    Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 15:45










  • $begingroup$
    Thanks a lots Sir...
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:47
















1












$begingroup$

The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,gin C^1bigl([0,1]bigr)$, then$$left(int_0^1f(x)g(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)left(int_0^1g^2(x),mathrm dxright).$$If you choose $g=f$, you get the trivial inequality$$left(int_0^1f^2(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)^2$$or$$int_0^1f^2(x),mathrm dxleqslantint_0^1f^2(x),mathrm dx.$$This in no way contradicts what you are supposed to prove.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ohh I am applying wrong result. Thanks A lot .You are always helping me.
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:37










  • $begingroup$
    ohh Sorry Sir I forget....
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:40










  • $begingroup$
    Sir ,Please can you give some hint so that I can prove above theorem
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:43










  • $begingroup$
    Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 15:45










  • $begingroup$
    Thanks a lots Sir...
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:47














1












1








1





$begingroup$

The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,gin C^1bigl([0,1]bigr)$, then$$left(int_0^1f(x)g(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)left(int_0^1g^2(x),mathrm dxright).$$If you choose $g=f$, you get the trivial inequality$$left(int_0^1f^2(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)^2$$or$$int_0^1f^2(x),mathrm dxleqslantint_0^1f^2(x),mathrm dx.$$This in no way contradicts what you are supposed to prove.






share|cite|improve this answer









$endgroup$



The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,gin C^1bigl([0,1]bigr)$, then$$left(int_0^1f(x)g(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)left(int_0^1g^2(x),mathrm dxright).$$If you choose $g=f$, you get the trivial inequality$$left(int_0^1f^2(x),mathrm dxright)^2leqslantleft(int_0^1f^2(x),mathrm dxright)^2$$or$$int_0^1f^2(x),mathrm dxleqslantint_0^1f^2(x),mathrm dx.$$This in no way contradicts what you are supposed to prove.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 15:34









José Carlos SantosJosé Carlos Santos

157k22126227




157k22126227












  • $begingroup$
    ohh I am applying wrong result. Thanks A lot .You are always helping me.
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:37










  • $begingroup$
    ohh Sorry Sir I forget....
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:40










  • $begingroup$
    Sir ,Please can you give some hint so that I can prove above theorem
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:43










  • $begingroup$
    Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 15:45










  • $begingroup$
    Thanks a lots Sir...
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:47


















  • $begingroup$
    ohh I am applying wrong result. Thanks A lot .You are always helping me.
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:37










  • $begingroup$
    ohh Sorry Sir I forget....
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:40










  • $begingroup$
    Sir ,Please can you give some hint so that I can prove above theorem
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:43










  • $begingroup$
    Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 15:45










  • $begingroup$
    Thanks a lots Sir...
    $endgroup$
    – MathLover
    Dec 6 '18 at 15:47
















$begingroup$
ohh I am applying wrong result. Thanks A lot .You are always helping me.
$endgroup$
– MathLover
Dec 6 '18 at 15:37




$begingroup$
ohh I am applying wrong result. Thanks A lot .You are always helping me.
$endgroup$
– MathLover
Dec 6 '18 at 15:37












$begingroup$
ohh Sorry Sir I forget....
$endgroup$
– MathLover
Dec 6 '18 at 15:40




$begingroup$
ohh Sorry Sir I forget....
$endgroup$
– MathLover
Dec 6 '18 at 15:40












$begingroup$
Sir ,Please can you give some hint so that I can prove above theorem
$endgroup$
– MathLover
Dec 6 '18 at 15:43




$begingroup$
Sir ,Please can you give some hint so that I can prove above theorem
$endgroup$
– MathLover
Dec 6 '18 at 15:43












$begingroup$
Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 15:45




$begingroup$
Apply the Cauchy-Schwarz inequality to the functions $lvert frvert$ and $1$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 15:45












$begingroup$
Thanks a lots Sir...
$endgroup$
– MathLover
Dec 6 '18 at 15:47




$begingroup$
Thanks a lots Sir...
$endgroup$
– MathLover
Dec 6 '18 at 15:47


















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