Birkhoff average of $x mapsto x+1$ in $mathbb R$ with $L^p$ observable
$begingroup$
Let $f in mathcal L^p(mathbb R, lambda)$, where $lambda$ is Lebesgue measure and $p in (1,infty)$. And let $T : mathbb R to mathbb R$ be the map $T(x) = x+1$. I want to show:
$$
frac 1 n sum_{k=0}^{n-1} f circ T^k xrightarrow{L^p} 0 quad textrm{as } n to infty.
$$
This exercise is coming from a probability theory textbook, so I want to avoid advanced tools from ergodic theory. I can show the $n$th Birkhoff average is in $mathcal L^p$, but I'm having trouble proving anything about this sequence of $mathcal L^p$ functions (Cauchy, a.e. convergent, etc.). Any suggestions?
real-analysis probability-theory dynamical-systems ergodic-theory
asked Dec 27 '18 at 0:10
D FordD Ford
593213
$endgroup$
add a comment |
$begingroup$
Let $f in mathcal L^p(mathbb R, lambda)$, where $lambda$ is Lebesgue measure and $p in (1,infty)$. And let $T : mathbb R to mathbb R$ be the map $T(x) = x+1$. I want to show:
$$
frac 1 n sum_{k=0}^{n-1} f circ T^k xrightarrow{L^p} 0 quad textrm{as } n to infty.
$$
This exercise is coming from a probability theory textbook, so I want to avoid advanced tools from ergodic theory. I can show the $n$th Birkhoff average is in $mathcal L^p$, but I'm having trouble proving anything about this sequence of $mathcal L^p$ functions (Cauchy, a.e. convergent, etc.). Any suggestions?
real-analysis probability-theory dynamical-systems ergodic-theory
asked Dec 27 '18 at 0:10
D FordD Ford
593213
$endgroup$
add a comment |
$begingroup$
Let $f in mathcal L^p(mathbb R, lambda)$, where $lambda$ is Lebesgue measure and $p in (1,infty)$. And let $T : mathbb R to mathbb R$ be the map $T(x) = x+1$. I want to show:
$$
frac 1 n sum_{k=0}^{n-1} f circ T^k xrightarrow{L^p} 0 quad textrm{as } n to infty.
$$
This exercise is coming from a probability theory textbook, so I want to avoid advanced tools from ergodic theory. I can show the $n$th Birkhoff average is in $mathcal L^p$, but I'm having trouble proving anything about this sequence of $mathcal L^p$ functions (Cauchy, a.e. convergent, etc.). Any suggestions?
real-analysis probability-theory dynamical-systems ergodic-theory
asked Dec 27 '18 at 0:10
D FordD Ford
593213
$endgroup$
Let $f in mathcal L^p(mathbb R, lambda)$, where $lambda$ is Lebesgue measure and $p in (1,infty)$. And let $T : mathbb R to mathbb R$ be the map $T(x) = x+1$. I want to show:
$$
frac 1 n sum_{k=0}^{n-1} f circ T^k xrightarrow{L^p} 0 quad textrm{as } n to infty.
$$
This exercise is coming from a probability theory textbook, so I want to avoid advanced tools from ergodic theory. I can show the $n$th Birkhoff average is in $mathcal L^p$, but I'm having trouble proving anything about this sequence of $mathcal L^p$ functions (Cauchy, a.e. convergent, etc.). Any suggestions?
real-analysis probability-theory dynamical-systems ergodic-theory
real-analysis probability-theory dynamical-systems ergodic-theory
asked Dec 27 '18 at 0:10
D FordD Ford
593213
asked Dec 27 '18 at 0:10
D FordD Ford
593213
asked Dec 27 '18 at 0:10
D FordD Ford
593213
asked Dec 27 '18 at 0:10
D FordD Ford
593213
asked Dec 27 '18 at 0:10
D FordD Ford
593213
593213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First note the result is obvious if $f$ is compactly supported. Then, since compactly supported functions are dense, we are done if we can get a bound of the form $||frac{1}{N}sum_{n le N} f(cdot+n)||_p le C ||f||_p$ with $C$ independent of $f$ and $N$. But we can get $C=1$ by triangle inequality.
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
$endgroup$
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First note the result is obvious if $f$ is compactly supported. Then, since compactly supported functions are dense, we are done if we can get a bound of the form $||frac{1}{N}sum_{n le N} f(cdot+n)||_p le C ||f||_p$ with $C$ independent of $f$ and $N$. But we can get $C=1$ by triangle inequality.
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
$endgroup$
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
|
show 2 more comments
$begingroup$
First note the result is obvious if $f$ is compactly supported. Then, since compactly supported functions are dense, we are done if we can get a bound of the form $||frac{1}{N}sum_{n le N} f(cdot+n)||_p le C ||f||_p$ with $C$ independent of $f$ and $N$. But we can get $C=1$ by triangle inequality.
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
$endgroup$
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
|
show 2 more comments
$begingroup$
First note the result is obvious if $f$ is compactly supported. Then, since compactly supported functions are dense, we are done if we can get a bound of the form $||frac{1}{N}sum_{n le N} f(cdot+n)||_p le C ||f||_p$ with $C$ independent of $f$ and $N$. But we can get $C=1$ by triangle inequality.
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
$endgroup$
First note the result is obvious if $f$ is compactly supported. Then, since compactly supported functions are dense, we are done if we can get a bound of the form $||frac{1}{N}sum_{n le N} f(cdot+n)||_p le C ||f||_p$ with $C$ independent of $f$ and $N$. But we can get $C=1$ by triangle inequality.
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
edited Dec 27 '18 at 0:32
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
answered Dec 27 '18 at 0:18
mathworker21mathworker21
8,9421928
8,9421928
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
|
show 2 more comments
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
Thanks for the tip! I might try using the simple function argument without using Minkowski's inequality, only because that hasn't appeared yet in the textbook I'm referencing. Fingers crossed.
$endgroup$
– D Ford
Dec 27 '18 at 0:29
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord do you mean proving the result is true for $f$ simple?
$endgroup$
– mathworker21
Dec 27 '18 at 0:31
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
@DFord if so, I edited my answer to make life easier for you
$endgroup$
– mathworker21
Dec 27 '18 at 0:32
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
Well, the argument via simple functions or compactly supported ones still uses the $L^p$ triangle inequality, a.k.a. Minkowski's inequality. Since this theorem isn't proven in this textbook until the following section, I'd like to avoid it if possible.
$endgroup$
– D Ford
Dec 27 '18 at 4:10
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
$begingroup$
@DFord the textbook hasn't proven that $||cdot||_p$ is a norm?? also, you should be precise with language. you should say "the argument showing that the result for simple functions implies the result for all $L^p$ functions" rather than just saying "the argument via simple functions"
$endgroup$
– mathworker21
Dec 27 '18 at 4:23
|
show 2 more comments
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