Prove the inequality $int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right)...
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I want to prove the inequality
begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}
Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!
integration functional-analysis analysis
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
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add a comment |
$begingroup$
I want to prove the inequality
begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}
Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!
integration functional-analysis analysis
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
$endgroup$
5
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
1
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21
add a comment |
$begingroup$
I want to prove the inequality
begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}
Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!
integration functional-analysis analysis
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
$endgroup$
I want to prove the inequality
begin{align} int^{3}_{-3}left|f(t) right|dtleq sqrt{6} left( int^{3}_{-3}left|f(t) right|^2dt right) ^{1/2}end{align}
Actually, I don't know where to start. However, does it have anything to with the Cauchy Schwarz Inequality for Integrals? Please, I need your help!
integration functional-analysis analysis
integration functional-analysis analysis
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
asked Dec 15 '18 at 13:15
Omojola MichealOmojola Micheal
1,853324
1,853324
5
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
1
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21
add a comment |
5
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
1
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21
5
5
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
1
1
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21
add a comment |
1 Answer
1
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$begingroup$
Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
$endgroup$
Recall that the Cauchy-Schwarz says that $int|fg|dx le sqrt{int |f|^2dxint|g|^2dx}$. Consider $g=mathbf 1_Omega$, where
$$
mathbf 1_Omega(x):=begin{cases}1 &; xinOmega \
0 &;xne Omega.
end{cases}
$$
Here you have $Omega=[-3,3]$. Can you finish the proof?
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
answered Dec 16 '18 at 14:15
BigbearZzzBigbearZzz
8,69121652
8,69121652
add a comment |
add a comment |
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5
$begingroup$
Cauchy-Schwarz should work if you work with $1cdot |f(t)|$.
$endgroup$
– Michael Burr
Dec 15 '18 at 13:17
1
$begingroup$
$ sqrt{6}=left(int_{-3}^31^2 dtright)^{1/2} $
$endgroup$
– user587192
Dec 16 '18 at 14:21