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Show that $left| f right|_r$ is continuous function in $r$.
Clash Royale CLAN TAG#URR8PPP
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I am trying to prove a statement below.
If $fin L^1(0,infty)cap L^2(0,infty)$ then
$left| f right|_r$ is continuous function in $r$ on $[1,2]$.
I just proved that $fin L^r(0,infty)$ $forall rin (1,2)$ but don't know how to prove the continuity...
real-analysis analysis lp-spaces
asked Aug 6 at 23:17
Lev Ban
46716
add a comment |Â
up vote
0
down vote
favorite
I am trying to prove a statement below.
If $fin L^1(0,infty)cap L^2(0,infty)$ then
$left| f right|_r$ is continuous function in $r$ on $[1,2]$.
I just proved that $fin L^r(0,infty)$ $forall rin (1,2)$ but don't know how to prove the continuity...
real-analysis analysis lp-spaces
asked Aug 6 at 23:17
Lev Ban
46716
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to prove a statement below.
If $fin L^1(0,infty)cap L^2(0,infty)$ then
$left| f right|_r$ is continuous function in $r$ on $[1,2]$.
I just proved that $fin L^r(0,infty)$ $forall rin (1,2)$ but don't know how to prove the continuity...
real-analysis analysis lp-spaces
asked Aug 6 at 23:17
Lev Ban
46716
I am trying to prove a statement below.
If $fin L^1(0,infty)cap L^2(0,infty)$ then
$left| f right|_r$ is continuous function in $r$ on $[1,2]$.
I just proved that $fin L^r(0,infty)$ $forall rin (1,2)$ but don't know how to prove the continuity...
real-analysis analysis lp-spaces
asked Aug 6 at 23:17
Lev Ban
46716
asked Aug 6 at 23:17
Lev Ban
46716
asked Aug 6 at 23:17
Lev Ban
46716
asked Aug 6 at 23:17
Lev Ban
46716
46716
add a comment |Â
add a comment |Â
1 Answer
1
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1
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Let $f_n$ be a sequence of simple functions such that $f_n to f$ almost everywhere and $|f_n| leq |f|$ almost everywhere for each $n$. Then $f_n to f$ in $L^1$ as well as $L^2$. Claim: $|f_n - f|_rto 0$ uniformly for $1 leq r leq 2$. Let $g_n=f_n-f$. Then $$int |g_n|^r =int_ |g_n|^r +int_g_n |g_n|^r leq int |g_n|+int |g_n|^2 to 0.$$ This proves the claim. It is trivial to check that $|h|_r$ is a continuous function of $r$ when $h$ is simple. By uniform convergence the same is true for the given $f$.
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $f_n$ be a sequence of simple functions such that $f_n to f$ almost everywhere and $|f_n| leq |f|$ almost everywhere for each $n$. Then $f_n to f$ in $L^1$ as well as $L^2$. Claim: $|f_n - f|_rto 0$ uniformly for $1 leq r leq 2$. Let $g_n=f_n-f$. Then $$int |g_n|^r =int_ |g_n|^r +int_g_n |g_n|^r leq int |g_n|+int |g_n|^2 to 0.$$ This proves the claim. It is trivial to check that $|h|_r$ is a continuous function of $r$ when $h$ is simple. By uniform convergence the same is true for the given $f$.
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
add a comment |Â
up vote
1
down vote
accepted
Let $f_n$ be a sequence of simple functions such that $f_n to f$ almost everywhere and $|f_n| leq |f|$ almost everywhere for each $n$. Then $f_n to f$ in $L^1$ as well as $L^2$. Claim: $|f_n - f|_rto 0$ uniformly for $1 leq r leq 2$. Let $g_n=f_n-f$. Then $$int |g_n|^r =int_ |g_n|^r +int_g_n |g_n|^r leq int |g_n|+int |g_n|^2 to 0.$$ This proves the claim. It is trivial to check that $|h|_r$ is a continuous function of $r$ when $h$ is simple. By uniform convergence the same is true for the given $f$.
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f_n$ be a sequence of simple functions such that $f_n to f$ almost everywhere and $|f_n| leq |f|$ almost everywhere for each $n$. Then $f_n to f$ in $L^1$ as well as $L^2$. Claim: $|f_n - f|_rto 0$ uniformly for $1 leq r leq 2$. Let $g_n=f_n-f$. Then $$int |g_n|^r =int_ |g_n|^r +int_g_n |g_n|^r leq int |g_n|+int |g_n|^2 to 0.$$ This proves the claim. It is trivial to check that $|h|_r$ is a continuous function of $r$ when $h$ is simple. By uniform convergence the same is true for the given $f$.
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
Let $f_n$ be a sequence of simple functions such that $f_n to f$ almost everywhere and $|f_n| leq |f|$ almost everywhere for each $n$. Then $f_n to f$ in $L^1$ as well as $L^2$. Claim: $|f_n - f|_rto 0$ uniformly for $1 leq r leq 2$. Let $g_n=f_n-f$. Then $$int |g_n|^r =int_ |g_n|^r +int_g_n |g_n|^r leq int |g_n|+int |g_n|^2 to 0.$$ This proves the claim. It is trivial to check that $|h|_r$ is a continuous function of $r$ when $h$ is simple. By uniform convergence the same is true for the given $f$.
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
answered Aug 7 at 0:05
Kavi Rama Murthy
21.2k2830
21.2k2830
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
add a comment |Â
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
I cannot see if it is true when $h$ is simple and the measure is not finite. Could you clarify more?
– Lev Ban
Aug 7 at 1:45
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan You are dealing with a simple function that belongs to $L^1cap L^2$. So it is of the type $sum_k=1^n c_k I_E_k$ where each $E_k$ has finite measure.
– Kavi Rama Murthy
Aug 7 at 5:28
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
@LevBan $(sum_k=1^N |a_k|^rm(E_k))^1/r$ is certainly a continuous function of $r$ if $m(E_k) <infty$ for all $k$, right?
– Kavi Rama Murthy
Aug 7 at 6:21
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
Thank you! I think I got that:)
– Lev Ban
Aug 7 at 11:24
add a comment |Â
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