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Does $int_{-infty}^infty f(x) dx < infty$ where $f ge 0$ implies $sup_{x in mathbb R} f(x)<M $?
Question. Does $,int_{-infty}^infty, f(x), dx < infty,,$ where $,f ge 0,,$ imply that $,,mathrm{ess}sup_{x in mathbb R}, f(x)<infty,? $
My attempt
Since $f ge 0$ and $int_{-infty}^infty f(x) dx < infty$ we get $lim_{x to infty} f(x)=0$ and $lim_{x to -infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $sup_{x in mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?
I am trying to show that if $f,g$ are non-negative and the integral exists then $fcdot g(x)$ exists for almost all $x$.
If not can you give a counterexample and explain the problem in my reasoning?
probability integration definite-integrals lebesgue-integral convolution
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
add a comment |
Question. Does $,int_{-infty}^infty, f(x), dx < infty,,$ where $,f ge 0,,$ imply that $,,mathrm{ess}sup_{x in mathbb R}, f(x)<infty,? $
My attempt
Since $f ge 0$ and $int_{-infty}^infty f(x) dx < infty$ we get $lim_{x to infty} f(x)=0$ and $lim_{x to -infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $sup_{x in mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?
I am trying to show that if $f,g$ are non-negative and the integral exists then $fcdot g(x)$ exists for almost all $x$.
If not can you give a counterexample and explain the problem in my reasoning?
probability integration definite-integrals lebesgue-integral convolution
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
3
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
1
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11
add a comment |
Question. Does $,int_{-infty}^infty, f(x), dx < infty,,$ where $,f ge 0,,$ imply that $,,mathrm{ess}sup_{x in mathbb R}, f(x)<infty,? $
My attempt
Since $f ge 0$ and $int_{-infty}^infty f(x) dx < infty$ we get $lim_{x to infty} f(x)=0$ and $lim_{x to -infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $sup_{x in mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?
I am trying to show that if $f,g$ are non-negative and the integral exists then $fcdot g(x)$ exists for almost all $x$.
If not can you give a counterexample and explain the problem in my reasoning?
probability integration definite-integrals lebesgue-integral convolution
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
Question. Does $,int_{-infty}^infty, f(x), dx < infty,,$ where $,f ge 0,,$ imply that $,,mathrm{ess}sup_{x in mathbb R}, f(x)<infty,? $
My attempt
Since $f ge 0$ and $int_{-infty}^infty f(x) dx < infty$ we get $lim_{x to infty} f(x)=0$ and $lim_{x to -infty} f(x)=0$ (we can show this easily by contradiction, the proof of which I skip). Since the limits exist can we say that the sequence is bounded and use this to conclude that $sup_{x in mathbb{R}} f(x) < M $ (Lebesgue almost surely). Do we need continuity for this?
I am trying to show that if $f,g$ are non-negative and the integral exists then $fcdot g(x)$ exists for almost all $x$.
If not can you give a counterexample and explain the problem in my reasoning?
probability integration definite-integrals lebesgue-integral convolution
probability integration definite-integrals lebesgue-integral convolution
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
edited Nov 22 '18 at 10:20
Yiorgos S. Smyrlis
62.7k1383163
62.7k1383163
asked Nov 21 '18 at 0:21
user3503589
1,2011721
asked Nov 21 '18 at 0:21
user3503589
1,2011721
asked Nov 21 '18 at 0:21
user3503589
1,2011721
1,2011721
3
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
1
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11
add a comment |
3
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
1
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11
3
3
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
1
1
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11
add a comment |
3 Answers
3
active
oldest
votes
No. A very simple counter-example can be constructed as follows.
Let $f:mathbb{R}rightarrowmathbb{R}$ be defined by
$$
f(x)=begin{cases}
x^{-frac{1}{2}}, & mbox{ if }xin(0,1]\
0, & mbox{ otherwise}
end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $int f<infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that ${xmid|f(x)|>M}=(0,frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
add a comment |
No.
Consider
$$
f(x)=left{
begin{array}{ccc}
cos pi x & if & |x|le 1/2,\
0 & if & |x|>1/2.
end{array}
right.
$$
Set
$$
g(x)=sum_{n=1}^infty n,fbig(n^3(x-n)big)
$$
Then
$$
g(x)ge0, qquad int_{-infty}^infty g(x) ,dx
=sum_{n=1}^inftyint_{-pi//n^3}^{pi/n^3}
ncos(n^3pi x),dx
=sum_{n=1}^infty frac{2}{pi n^2}=frac{pi}{3}$$
and
$$
sup_{xinmathbb R} g(x)=infty.
$$
Note. If $f(x)ge 0$ and $int_{-infty}^infty f(x),dx<infty$, then this does not imply that $lim_{xtoinfty} f(x)=0$.
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
add a comment |
Neither if $f$ is continue, let $rho:mathbb Rrightarrow [0, +infty[$ a continue non identically zero function such that $rho(x)neq 0$ only inside $[-1, 1]$ and
$$
int^{+infty}_{-infty}rho(x)dx = 1
$$
Then let
$$
f(x)=sum^{+infty}_{i=1}2^irholeft(i+4^ixright)
$$
$f$ is continuous and
$$
int^{+infty}_{-infty}2^irholeft(i+4^ixright)dx=int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^irholeft(i+4^ixright)dx=2^i4^{-i}int^{1}_{-1}rho(y)dyleq 2^{-i}
$$
then $int fdx<+infty$ but $f$ isn't bounded almost everywhere: let $k=max_{xinmathbb R}rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $rho$ is continue exists an open set $Usubseteqmathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $xin U$
$$
M < 2^irho(i+4^ix)Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $mathbb R$ because statement
$$
lim_{xrightarrow infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $mathbb R$ and suppose exists $epsilon>0$ and increasing sequence $x_n$ such that $x_nrightarrow +infty$ and $f(x_n)>epsilon$.
From uniform continuity exists $delta>0$ such that for every $x, yinmathbb R$ such that $lvert x-yrvert <delta$ then $lvert f(x)-f(y)rvert<frac{epsilon}{2}$. If $xin ]x_n-delta, x_n+delta[$ then
$$
f(x)geq f(x_n)-lvert f(x)-f(x_n)rvert>epsilon-frac{epsilon}{2}=frac{epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
int^{+infty}_{-infty}f(x)dxgeqsum^{+infty}_{k=0}frac{epsilon}{2}left(x_{n_k}+delta-x_{n_k}+deltaright)=sum^{+infty}_{k=0}epsilondelta=+infty
$$
absurd.
Your last assertion is true: let $f:mathbb Rrightarrow [0, +infty]$ such that integral is finite let $K={xinmathbb R : f(x)=+infty}$ then
$$
int^{+infty}_{-infty}f(x)dxgeqint_Kf(x)dx=begin{cases}
+infty & text{ if }lvert Krvert >0\
0 & text{ if }lvert K rvert =0
end{cases}
$$
so $f(x)<+infty$ for almost every $xinmathbb R$.
answered Nov 21 '18 at 0:41
P De Donato
4147
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
No. A very simple counter-example can be constructed as follows.
Let $f:mathbb{R}rightarrowmathbb{R}$ be defined by
$$
f(x)=begin{cases}
x^{-frac{1}{2}}, & mbox{ if }xin(0,1]\
0, & mbox{ otherwise}
end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $int f<infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that ${xmid|f(x)|>M}=(0,frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
add a comment |
No. A very simple counter-example can be constructed as follows.
Let $f:mathbb{R}rightarrowmathbb{R}$ be defined by
$$
f(x)=begin{cases}
x^{-frac{1}{2}}, & mbox{ if }xin(0,1]\
0, & mbox{ otherwise}
end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $int f<infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that ${xmid|f(x)|>M}=(0,frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
add a comment |
No. A very simple counter-example can be constructed as follows.
Let $f:mathbb{R}rightarrowmathbb{R}$ be defined by
$$
f(x)=begin{cases}
x^{-frac{1}{2}}, & mbox{ if }xin(0,1]\
0, & mbox{ otherwise}
end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $int f<infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that ${xmid|f(x)|>M}=(0,frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
No. A very simple counter-example can be constructed as follows.
Let $f:mathbb{R}rightarrowmathbb{R}$ be defined by
$$
f(x)=begin{cases}
x^{-frac{1}{2}}, & mbox{ if }xin(0,1]\
0, & mbox{ otherwise}
end{cases}.
$$
Clearly $f$ is a non-negative Borel function and $int f<infty$.
We go to show that $f$ is not essentially bounded. Let $M>0$ be
arbitrary. Note that ${xmid|f(x)|>M}=(0,frac{1}{M^{2}})$, which
has a positive Lebesgue measure. This shows that $|f|$ is not essentially
bounded.
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
answered Nov 21 '18 at 11:51
Danny Pak-Keung Chan
2,19038
2,19038
add a comment |
add a comment |
No.
Consider
$$
f(x)=left{
begin{array}{ccc}
cos pi x & if & |x|le 1/2,\
0 & if & |x|>1/2.
end{array}
right.
$$
Set
$$
g(x)=sum_{n=1}^infty n,fbig(n^3(x-n)big)
$$
Then
$$
g(x)ge0, qquad int_{-infty}^infty g(x) ,dx
=sum_{n=1}^inftyint_{-pi//n^3}^{pi/n^3}
ncos(n^3pi x),dx
=sum_{n=1}^infty frac{2}{pi n^2}=frac{pi}{3}$$
and
$$
sup_{xinmathbb R} g(x)=infty.
$$
Note. If $f(x)ge 0$ and $int_{-infty}^infty f(x),dx<infty$, then this does not imply that $lim_{xtoinfty} f(x)=0$.
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
add a comment |
No.
Consider
$$
f(x)=left{
begin{array}{ccc}
cos pi x & if & |x|le 1/2,\
0 & if & |x|>1/2.
end{array}
right.
$$
Set
$$
g(x)=sum_{n=1}^infty n,fbig(n^3(x-n)big)
$$
Then
$$
g(x)ge0, qquad int_{-infty}^infty g(x) ,dx
=sum_{n=1}^inftyint_{-pi//n^3}^{pi/n^3}
ncos(n^3pi x),dx
=sum_{n=1}^infty frac{2}{pi n^2}=frac{pi}{3}$$
and
$$
sup_{xinmathbb R} g(x)=infty.
$$
Note. If $f(x)ge 0$ and $int_{-infty}^infty f(x),dx<infty$, then this does not imply that $lim_{xtoinfty} f(x)=0$.
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
add a comment |
No.
Consider
$$
f(x)=left{
begin{array}{ccc}
cos pi x & if & |x|le 1/2,\
0 & if & |x|>1/2.
end{array}
right.
$$
Set
$$
g(x)=sum_{n=1}^infty n,fbig(n^3(x-n)big)
$$
Then
$$
g(x)ge0, qquad int_{-infty}^infty g(x) ,dx
=sum_{n=1}^inftyint_{-pi//n^3}^{pi/n^3}
ncos(n^3pi x),dx
=sum_{n=1}^infty frac{2}{pi n^2}=frac{pi}{3}$$
and
$$
sup_{xinmathbb R} g(x)=infty.
$$
Note. If $f(x)ge 0$ and $int_{-infty}^infty f(x),dx<infty$, then this does not imply that $lim_{xtoinfty} f(x)=0$.
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
No.
Consider
$$
f(x)=left{
begin{array}{ccc}
cos pi x & if & |x|le 1/2,\
0 & if & |x|>1/2.
end{array}
right.
$$
Set
$$
g(x)=sum_{n=1}^infty n,fbig(n^3(x-n)big)
$$
Then
$$
g(x)ge0, qquad int_{-infty}^infty g(x) ,dx
=sum_{n=1}^inftyint_{-pi//n^3}^{pi/n^3}
ncos(n^3pi x),dx
=sum_{n=1}^infty frac{2}{pi n^2}=frac{pi}{3}$$
and
$$
sup_{xinmathbb R} g(x)=infty.
$$
Note. If $f(x)ge 0$ and $int_{-infty}^infty f(x),dx<infty$, then this does not imply that $lim_{xtoinfty} f(x)=0$.
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
edited Nov 21 '18 at 12:09
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
answered Nov 21 '18 at 11:22
Yiorgos S. Smyrlis
62.7k1383163
62.7k1383163
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
add a comment |
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
how did you find the integral of $g$ please? thanks in advance
– John11
Nov 21 '18 at 12:02
add a comment |
Neither if $f$ is continue, let $rho:mathbb Rrightarrow [0, +infty[$ a continue non identically zero function such that $rho(x)neq 0$ only inside $[-1, 1]$ and
$$
int^{+infty}_{-infty}rho(x)dx = 1
$$
Then let
$$
f(x)=sum^{+infty}_{i=1}2^irholeft(i+4^ixright)
$$
$f$ is continuous and
$$
int^{+infty}_{-infty}2^irholeft(i+4^ixright)dx=int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^irholeft(i+4^ixright)dx=2^i4^{-i}int^{1}_{-1}rho(y)dyleq 2^{-i}
$$
then $int fdx<+infty$ but $f$ isn't bounded almost everywhere: let $k=max_{xinmathbb R}rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $rho$ is continue exists an open set $Usubseteqmathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $xin U$
$$
M < 2^irho(i+4^ix)Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $mathbb R$ because statement
$$
lim_{xrightarrow infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $mathbb R$ and suppose exists $epsilon>0$ and increasing sequence $x_n$ such that $x_nrightarrow +infty$ and $f(x_n)>epsilon$.
From uniform continuity exists $delta>0$ such that for every $x, yinmathbb R$ such that $lvert x-yrvert <delta$ then $lvert f(x)-f(y)rvert<frac{epsilon}{2}$. If $xin ]x_n-delta, x_n+delta[$ then
$$
f(x)geq f(x_n)-lvert f(x)-f(x_n)rvert>epsilon-frac{epsilon}{2}=frac{epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
int^{+infty}_{-infty}f(x)dxgeqsum^{+infty}_{k=0}frac{epsilon}{2}left(x_{n_k}+delta-x_{n_k}+deltaright)=sum^{+infty}_{k=0}epsilondelta=+infty
$$
absurd.
Your last assertion is true: let $f:mathbb Rrightarrow [0, +infty]$ such that integral is finite let $K={xinmathbb R : f(x)=+infty}$ then
$$
int^{+infty}_{-infty}f(x)dxgeqint_Kf(x)dx=begin{cases}
+infty & text{ if }lvert Krvert >0\
0 & text{ if }lvert K rvert =0
end{cases}
$$
so $f(x)<+infty$ for almost every $xinmathbb R$.
answered Nov 21 '18 at 0:41
P De Donato
4147
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
add a comment |
Neither if $f$ is continue, let $rho:mathbb Rrightarrow [0, +infty[$ a continue non identically zero function such that $rho(x)neq 0$ only inside $[-1, 1]$ and
$$
int^{+infty}_{-infty}rho(x)dx = 1
$$
Then let
$$
f(x)=sum^{+infty}_{i=1}2^irholeft(i+4^ixright)
$$
$f$ is continuous and
$$
int^{+infty}_{-infty}2^irholeft(i+4^ixright)dx=int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^irholeft(i+4^ixright)dx=2^i4^{-i}int^{1}_{-1}rho(y)dyleq 2^{-i}
$$
then $int fdx<+infty$ but $f$ isn't bounded almost everywhere: let $k=max_{xinmathbb R}rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $rho$ is continue exists an open set $Usubseteqmathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $xin U$
$$
M < 2^irho(i+4^ix)Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $mathbb R$ because statement
$$
lim_{xrightarrow infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $mathbb R$ and suppose exists $epsilon>0$ and increasing sequence $x_n$ such that $x_nrightarrow +infty$ and $f(x_n)>epsilon$.
From uniform continuity exists $delta>0$ such that for every $x, yinmathbb R$ such that $lvert x-yrvert <delta$ then $lvert f(x)-f(y)rvert<frac{epsilon}{2}$. If $xin ]x_n-delta, x_n+delta[$ then
$$
f(x)geq f(x_n)-lvert f(x)-f(x_n)rvert>epsilon-frac{epsilon}{2}=frac{epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
int^{+infty}_{-infty}f(x)dxgeqsum^{+infty}_{k=0}frac{epsilon}{2}left(x_{n_k}+delta-x_{n_k}+deltaright)=sum^{+infty}_{k=0}epsilondelta=+infty
$$
absurd.
Your last assertion is true: let $f:mathbb Rrightarrow [0, +infty]$ such that integral is finite let $K={xinmathbb R : f(x)=+infty}$ then
$$
int^{+infty}_{-infty}f(x)dxgeqint_Kf(x)dx=begin{cases}
+infty & text{ if }lvert Krvert >0\
0 & text{ if }lvert K rvert =0
end{cases}
$$
so $f(x)<+infty$ for almost every $xinmathbb R$.
answered Nov 21 '18 at 0:41
P De Donato
4147
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
add a comment |
Neither if $f$ is continue, let $rho:mathbb Rrightarrow [0, +infty[$ a continue non identically zero function such that $rho(x)neq 0$ only inside $[-1, 1]$ and
$$
int^{+infty}_{-infty}rho(x)dx = 1
$$
Then let
$$
f(x)=sum^{+infty}_{i=1}2^irholeft(i+4^ixright)
$$
$f$ is continuous and
$$
int^{+infty}_{-infty}2^irholeft(i+4^ixright)dx=int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^irholeft(i+4^ixright)dx=2^i4^{-i}int^{1}_{-1}rho(y)dyleq 2^{-i}
$$
then $int fdx<+infty$ but $f$ isn't bounded almost everywhere: let $k=max_{xinmathbb R}rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $rho$ is continue exists an open set $Usubseteqmathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $xin U$
$$
M < 2^irho(i+4^ix)Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $mathbb R$ because statement
$$
lim_{xrightarrow infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $mathbb R$ and suppose exists $epsilon>0$ and increasing sequence $x_n$ such that $x_nrightarrow +infty$ and $f(x_n)>epsilon$.
From uniform continuity exists $delta>0$ such that for every $x, yinmathbb R$ such that $lvert x-yrvert <delta$ then $lvert f(x)-f(y)rvert<frac{epsilon}{2}$. If $xin ]x_n-delta, x_n+delta[$ then
$$
f(x)geq f(x_n)-lvert f(x)-f(x_n)rvert>epsilon-frac{epsilon}{2}=frac{epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
int^{+infty}_{-infty}f(x)dxgeqsum^{+infty}_{k=0}frac{epsilon}{2}left(x_{n_k}+delta-x_{n_k}+deltaright)=sum^{+infty}_{k=0}epsilondelta=+infty
$$
absurd.
Your last assertion is true: let $f:mathbb Rrightarrow [0, +infty]$ such that integral is finite let $K={xinmathbb R : f(x)=+infty}$ then
$$
int^{+infty}_{-infty}f(x)dxgeqint_Kf(x)dx=begin{cases}
+infty & text{ if }lvert Krvert >0\
0 & text{ if }lvert K rvert =0
end{cases}
$$
so $f(x)<+infty$ for almost every $xinmathbb R$.
answered Nov 21 '18 at 0:41
P De Donato
4147
Neither if $f$ is continue, let $rho:mathbb Rrightarrow [0, +infty[$ a continue non identically zero function such that $rho(x)neq 0$ only inside $[-1, 1]$ and
$$
int^{+infty}_{-infty}rho(x)dx = 1
$$
Then let
$$
f(x)=sum^{+infty}_{i=1}2^irholeft(i+4^ixright)
$$
$f$ is continuous and
$$
int^{+infty}_{-infty}2^irholeft(i+4^ixright)dx=int^{4^{-i}(1-i)}_{4^{-i}(-1-i)}2^irholeft(i+4^ixright)dx=2^i4^{-i}int^{1}_{-1}rho(y)dyleq 2^{-i}
$$
then $int fdx<+infty$ but $f$ isn't bounded almost everywhere: let $k=max_{xinmathbb R}rho(x)$ and $M>0$ generic then exists $i$ such that
$$
M<k2^i
$$
Because $rho$ is continue exists an open set $Usubseteqmathbb R$ (its Lebesgue measure is strictly greater than $0$) such that for every $xin U$
$$
M < 2^irho(i+4^ix)Rightarrow M<f(x)
$$
and $f$ isn't almost everywhere bounded.
You need uniform continuity on $mathbb R$ because statement
$$
lim_{xrightarrow infty}f(x)=0
$$
is true if $f$ is uniformly continue.
Let $f$ uniformly continue on $mathbb R$ and suppose exists $epsilon>0$ and increasing sequence $x_n$ such that $x_nrightarrow +infty$ and $f(x_n)>epsilon$.
From uniform continuity exists $delta>0$ such that for every $x, yinmathbb R$ such that $lvert x-yrvert <delta$ then $lvert f(x)-f(y)rvert<frac{epsilon}{2}$. If $xin ]x_n-delta, x_n+delta[$ then
$$
f(x)geq f(x_n)-lvert f(x)-f(x_n)rvert>epsilon-frac{epsilon}{2}=frac{epsilon}{2}
$$
and exists an extract $n_k$ such that
$$
int^{+infty}_{-infty}f(x)dxgeqsum^{+infty}_{k=0}frac{epsilon}{2}left(x_{n_k}+delta-x_{n_k}+deltaright)=sum^{+infty}_{k=0}epsilondelta=+infty
$$
absurd.
Your last assertion is true: let $f:mathbb Rrightarrow [0, +infty]$ such that integral is finite let $K={xinmathbb R : f(x)=+infty}$ then
$$
int^{+infty}_{-infty}f(x)dxgeqint_Kf(x)dx=begin{cases}
+infty & text{ if }lvert Krvert >0\
0 & text{ if }lvert K rvert =0
end{cases}
$$
so $f(x)<+infty$ for almost every $xinmathbb R$.
answered Nov 21 '18 at 0:41
P De Donato
4147
edited Nov 21 '18 at 11:08
answered Nov 21 '18 at 0:41
P De Donato
4147
answered Nov 21 '18 at 0:41
P De Donato
4147
answered Nov 21 '18 at 0:41
P De Donato
4147
4147
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
add a comment |
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
When you show that for continuous $f$ it’s integral is finite , you skip the term $i2^i$ , why can we do that ?
– user3503589
Nov 21 '18 at 1:42
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
I don't understand, note that $rho$ is non zero only inside the interval $[-1, 1]$ so we can reduce integration bounds to $4^{-i}(1-i)$ and $4^{-i}(-1-i)$ then we use substitution $i+4^ix=y$. Now I've rectified
– P De Donato
Nov 21 '18 at 11:06
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
Why I've downvoted?
– P De Donato
Nov 21 '18 at 23:35
add a comment |
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3
$f(x)=chi_{[-1,1]}(x),/,|x|^{1/2}$ seems like a counterexample, where $chi$ is an indicator function.
– user254433
Nov 21 '18 at 0:25
How is this a counter example when the function doesnt even exist at $x=0$? But perhaps we can exclude $0$ from your function
– user3503589
Nov 21 '18 at 0:40
1
$lim_{xto infty} f(x) = 0$ isn't necessarily true: you could make the graph of $f$ have triangles with base $[n, n + 4^{-n}]$ and height $2^n$ near each $n$. (And by replacing the triangles with $C^infty$ bump functions you can even have $f$ be $C^infty$.)
– Daniel Schepler
Nov 21 '18 at 0:47
@user3503589 Define $f(0)=0$ if you like. I don't see a requirement that $f$ be continuous.
– user254433
Nov 21 '18 at 1:09
$f$ should be uniformly continue, not only continue
– P De Donato
Nov 21 '18 at 1:11