Prove that $||f||_p=sup|int fg dmu|$ The 2019 Stack Overflow Developer Survey Results Are InAn exercise in Rudin's RCAProve that lim sup $a_n$ $leq$ lim sup $b_n$.If $big|int gvarphi , dmubig|le Mintlvertvarphirvert,dmu,$ for all simple functions $varphi,,$ then $lvert grvert le M,$ a. e.Show that $sup f_n = sup g_n$ almost everywhere.Real Analysis, Folland Problem 6.2.17 Dual of $L^p$Does functions in $L^p(mathbbR)$ vanish outside of a set of finite measure?Fatou's Lemma Proof MisunderstandingProve that $sup(S)leqsup(T)$ and $inf(T)leqinf(S)$Show that $sup(S)=inf(T)$.Given $mu$ is semifinite and $mu(E)=infty$, prove that if $C>0$, then $exists$ measurable set $A subseteq E$ such that $C < mu(A) < infty.$
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Prove that $||f||_p=sup|int fg dmu|$
The 2019 Stack Overflow Developer Survey Results Are InAn exercise in Rudin's RCAProve that lim sup $a_n$ $leq$ lim sup $b_n$.If $big|int gvarphi , dmubig|le Mintlvertvarphirvert,dmu,$ for all simple functions $varphi,,$ then $lvert grvert le M,$ a. e.Show that $sup f_n = sup g_n$ almost everywhere.Real Analysis, Folland Problem 6.2.17 Dual of $L^p$Does functions in $L^p(mathbbR)$ vanish outside of a set of finite measure?Fatou's Lemma Proof MisunderstandingProve that $sup(S)leqsup(T)$ and $inf(T)leqinf(S)$Show that $sup(S)=inf(T)$.Given $mu$ is semifinite and $mu(E)=infty$, prove that if $C>0$, then $exists$ measurable set $A subseteq E$ such that $C < mu(A) < infty.$
$begingroup$
Let $mu$ be a $sigma$-finite measure and $1leq pleqinfty$,show that $||f||_p=sup|int fg dmu|$ where the supremum is taken over all bounded measurable functions $g$ that vanish outside a set (depending on $g$) of finite measure, and for which $||g||_p^primeleq 1$, ($p^prime$ is exponent conjugate of $p$) and $int fgdmu$
exists.
real-analysis analysis
asked Mar 24 at 4:14
ZoeZoe
134
$endgroup$
add a comment |
$begingroup$
Let $mu$ be a $sigma$-finite measure and $1leq pleqinfty$,show that $||f||_p=sup|int fg dmu|$ where the supremum is taken over all bounded measurable functions $g$ that vanish outside a set (depending on $g$) of finite measure, and for which $||g||_p^primeleq 1$, ($p^prime$ is exponent conjugate of $p$) and $int fgdmu$
exists.
real-analysis analysis
asked Mar 24 at 4:14
ZoeZoe
134
$endgroup$
add a comment |
$begingroup$
Let $mu$ be a $sigma$-finite measure and $1leq pleqinfty$,show that $||f||_p=sup|int fg dmu|$ where the supremum is taken over all bounded measurable functions $g$ that vanish outside a set (depending on $g$) of finite measure, and for which $||g||_p^primeleq 1$, ($p^prime$ is exponent conjugate of $p$) and $int fgdmu$
exists.
real-analysis analysis
asked Mar 24 at 4:14
ZoeZoe
134
$endgroup$
Let $mu$ be a $sigma$-finite measure and $1leq pleqinfty$,show that $||f||_p=sup|int fg dmu|$ where the supremum is taken over all bounded measurable functions $g$ that vanish outside a set (depending on $g$) of finite measure, and for which $||g||_p^primeleq 1$, ($p^prime$ is exponent conjugate of $p$) and $int fgdmu$
exists.
real-analysis analysis
real-analysis analysis
asked Mar 24 at 4:14
ZoeZoe
134
asked Mar 24 at 4:14
ZoeZoe
134
asked Mar 24 at 4:14
ZoeZoe
134
asked Mar 24 at 4:14
ZoeZoe
134
asked Mar 24 at 4:14
ZoeZoe
134
134
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $p>1$ : choose sets $A_n$ of finite measure increasing to $X$. LHS $leq$ RHS is immediate form Holder's inequality. Now consider $g=frac 1 c I_A_n |f|^p-1 sgn(f) I_x:$ where $c=(int_x: I_A_n |f|^p)^1/p'$. you can see that RHS $
geq int fg$ for each $N$ and tends to LHS as $N to infty$. [$sgn(f)$ stands for sign of $f$].
I will let you handle the case $p=1$
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
For $p>1$ : choose sets $A_n$ of finite measure increasing to $X$. LHS $leq$ RHS is immediate form Holder's inequality. Now consider $g=frac 1 c I_A_n |f|^p-1 sgn(f) I_x:$ where $c=(int_x: I_A_n |f|^p)^1/p'$. you can see that RHS $
geq int fg$ for each $N$ and tends to LHS as $N to infty$. [$sgn(f)$ stands for sign of $f$].
I will let you handle the case $p=1$
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
add a comment |
$begingroup$
For $p>1$ : choose sets $A_n$ of finite measure increasing to $X$. LHS $leq$ RHS is immediate form Holder's inequality. Now consider $g=frac 1 c I_A_n |f|^p-1 sgn(f) I_x:$ where $c=(int_x: I_A_n |f|^p)^1/p'$. you can see that RHS $
geq int fg$ for each $N$ and tends to LHS as $N to infty$. [$sgn(f)$ stands for sign of $f$].
I will let you handle the case $p=1$
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
add a comment |
$begingroup$
For $p>1$ : choose sets $A_n$ of finite measure increasing to $X$. LHS $leq$ RHS is immediate form Holder's inequality. Now consider $g=frac 1 c I_A_n |f|^p-1 sgn(f) I_x:$ where $c=(int_x: I_A_n |f|^p)^1/p'$. you can see that RHS $
geq int fg$ for each $N$ and tends to LHS as $N to infty$. [$sgn(f)$ stands for sign of $f$].
I will let you handle the case $p=1$
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
$endgroup$
For $p>1$ : choose sets $A_n$ of finite measure increasing to $X$. LHS $leq$ RHS is immediate form Holder's inequality. Now consider $g=frac 1 c I_A_n |f|^p-1 sgn(f) I_x:$ where $c=(int_x: I_A_n |f|^p)^1/p'$. you can see that RHS $
geq int fg$ for each $N$ and tends to LHS as $N to infty$. [$sgn(f)$ stands for sign of $f$].
I will let you handle the case $p=1$
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
edited Mar 24 at 5:27
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
answered Mar 24 at 5:10
Kavi Rama MurthyKavi Rama Murthy
74.1k53270
74.1k53270
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
add a comment |
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
But, where uses the assumption $mu$ is $sigma$-finite?
$endgroup$
– Zoe
Mar 24 at 5:13
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
This approach still holds when the case $||f||_p=infty$?
$endgroup$
– Zoe
Mar 24 at 5:21
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
@Zoe I have edited the answer. It does hold even when LHS is $infty$ and sigma finiteness is required for the argument.
$endgroup$
– Kavi Rama Murthy
Mar 24 at 5:30
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
$begingroup$
Thinks, now I think I can finish proof for the the case p=1
$endgroup$
– Zoe
Mar 24 at 5:34
add a comment |
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