Showing convergence in ${ L }^{ p }(dmu )$
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I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.
Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.
So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do
real-analysis lp-spaces
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
$endgroup$
add a comment |
$begingroup$
I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.
Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.
So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do
real-analysis lp-spaces
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
$endgroup$
add a comment |
$begingroup$
I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.
Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.
So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do
real-analysis lp-spaces
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
$endgroup$
I'm having struggles to prove the following statement because im confused about the notation and im unsure about my approch.
Let $X$ be an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
Let ${ f }_{ i }epsilon { L }^{ p }(dmu )$ with ${ f }_{ i }rightarrow fquad inquad { L }^{ p }(dmu )$. Define ${ ({ f }_{ i }) }_{ m }:=max(min({ f }_{ i },m),-m)$ show that ${ { f }_{ i } }_{ m }rightarrow f$ in ${ L }^{ p }(dmu )$ for ${ i }_{ m }rightarrow infty $.
So we want to show that ${ left| { { f }_{ i } }_{ m }-f right| }_{ p }=0$ if we choose any convergent subsequence that converges to f we get
${ left| { { f }_{ i } }_{ m }-f right| }_{ p }={ left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k }+{ { f }_{ i } }_{ k }-f right| }_{ p }le { left| { { { f }_{ i } }_{ m }-{ f }_{ i } }_{ k } right| }_{ p }+{ left| { { f }_{ i } }_{ k }-f right| }_{ p }$ by Minkowski-inequality where the first term goes to zero since it's a Cauchy sequence and the right term also goes to zero since ${ { f }_{ i } }_{ k }rightarrow f$. So the statement would follow?
But what if we have $m<{ f }_{ i }$ than the maximum is m and we can't do the above estimation anymore. Does someone have a tipp on what to do
real-analysis lp-spaces
real-analysis lp-spaces
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
asked Nov 30 '18 at 9:31
MasterPIMasterPI
1998
1998
add a comment |
add a comment |
1 Answer
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$begingroup$
There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
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$begingroup$
There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
add a comment |
$begingroup$
There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
add a comment |
$begingroup$
There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
There are some errors in the statement. You don't start with $(f_i) in L^{p}$; you start with $fin L^{p}$ and define $f_i$'s in terms of them. Also limit is taken as $m to infty$. You have written $i_m to infty$ but there is no $i_m$ here. The statement follows immediately from DCT and the fact that $|f_i| leq |f|$. [You can use the inequality $(a+b)^{p} leq 2^{p} (a^{p}+b^{p})$ for $a,b geq 0$ which gives $|f_i-f|^{p} leq 2^{p+1} |f|$.].
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
edited Nov 30 '18 at 10:10
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Nov 30 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
52.6k32055
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