showing $f=g$ a.e. given $f_nto f$ uniformly and $f_nto g$ in $L^p(Bbb R)$
$begingroup$
Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.
Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.
I need to prove that $f=g$ a.e.
So it's enough to show that $||f-g||_p =0$.
By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.
so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.
So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?
I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.
Thanks for helping!
real-analysis functional-analysis measure-theory lp-spaces
$endgroup$
add a comment |
$begingroup$
Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.
Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.
I need to prove that $f=g$ a.e.
So it's enough to show that $||f-g||_p =0$.
By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.
so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.
So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?
I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.
Thanks for helping!
real-analysis functional-analysis measure-theory lp-spaces
$endgroup$
$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20
$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21
$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43
add a comment |
$begingroup$
Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.
Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.
I need to prove that $f=g$ a.e.
So it's enough to show that $||f-g||_p =0$.
By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.
so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.
So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?
I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.
Thanks for helping!
real-analysis functional-analysis measure-theory lp-spaces
$endgroup$
Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.
Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.
I need to prove that $f=g$ a.e.
So it's enough to show that $||f-g||_p =0$.
By triangle inequality we have $||f-g||_p le ||f-f_n||_p +||f_n-g||$.
so it's enough to show that $||f-f_n||_p to 0$ (as ,by assumption $||f_n-g||to 0)$.
So I reduced the problem to the problem of showing $||f_n-f||_p to 0$ , any ideas how to do that?
I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $Bbb R$ which is $infty$.
Thanks for helping!
real-analysis functional-analysis measure-theory lp-spaces
real-analysis functional-analysis measure-theory lp-spaces
edited Dec 2 '18 at 15:54
qbert
22.1k32561
22.1k32561
asked Dec 2 '18 at 15:05
user123user123
1,362316
1,362316
$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20
$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21
$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43
add a comment |
$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20
$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21
$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43
$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20
$begingroup$
Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
$endgroup$
– Rhys Steele
Dec 2 '18 at 15:20
$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21
$begingroup$
I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
$endgroup$
– gangrene
Dec 2 '18 at 15:21
$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43
$begingroup$
@RhysSteele now i know :)
$endgroup$
– user123
Dec 2 '18 at 19:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$.
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set
$$
E_varepsilon = {|f-g| > varepsilon}
$$
is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives
$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$
and taking measures,
$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$
Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).
$endgroup$
add a comment |
$begingroup$
Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
$endgroup$
– user123
Dec 2 '18 at 19:40
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
$begingroup$
Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$.
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$.
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$.
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.
$endgroup$
Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$.
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.
edited Dec 2 '18 at 15:41
answered Dec 2 '18 at 15:31
qbertqbert
22.1k32561
22.1k32561
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @qbert
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set
$$
E_varepsilon = {|f-g| > varepsilon}
$$
is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives
$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$
and taking measures,
$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$
Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).
$endgroup$
add a comment |
$begingroup$
I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set
$$
E_varepsilon = {|f-g| > varepsilon}
$$
is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives
$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$
and taking measures,
$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$
Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).
$endgroup$
add a comment |
$begingroup$
I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set
$$
E_varepsilon = {|f-g| > varepsilon}
$$
is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives
$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$
and taking measures,
$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$
Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).
$endgroup$
I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $varepsilon > 0$, the set
$$
E_varepsilon = {|f-g| > varepsilon}
$$
is of measure zero. Since $|f-g| leq |f-f_n| + |f_n-g|$, then $|f-g| > varepsilon$ implies either $|f-f_n| > varepsilon/2$ or $|f_n-g| > varepsilon/2$. This gives
$$
E_varepsilon subset {|f-f_n| > varepsilon/2} cup {|f_n-g| > varepsilon/2}
$$
and taking measures,
$$
|E_varepsilon| leq |{|f-f_n| > varepsilon/2}| + |{|f_n-g| > varepsilon/2}|.
$$
Now you can use that $f_n to f$ uniformly and $f_n to g$ in $L^p$ (think of Tchevyscheb's inequality).
answered Dec 2 '18 at 15:20
Guido A.Guido A.
7,4271730
7,4271730
add a comment |
add a comment |
$begingroup$
Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.
$endgroup$
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
$begingroup$
Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.
$endgroup$
Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $mathbb R$ that converges pointwise to $f$ a.e. If also $f_nto g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.
answered Dec 2 '18 at 18:32
zhw.zhw.
73.3k43175
73.3k43175
$begingroup$
Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
$endgroup$
– user123
Dec 2 '18 at 19:40
add a comment |
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Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
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– user123
Dec 2 '18 at 19:40
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Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
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– user123
Dec 2 '18 at 19:40
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Thank you very much! if you can take a look at my other question that would be great - math.stackexchange.com/questions/3022692/… @zhw.
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– user123
Dec 2 '18 at 19:40
add a comment |
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Do you know that $f_n to g$ in $L^p$ implies that there is a subsequence $f_{n_k}$ such that $f_{n_k} to g$ a.e.?
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– Rhys Steele
Dec 2 '18 at 15:20
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I guess you have to assume also $f_n in L^p (mathbb{R})$ for all but finite $n$, otherwise $f_n to g$ in $L^p$ does not quite make sense - unsing the reverse triangle inequality you would get $ |f_n -g|_{L^p} ge | |f_n|_{L^p} - |g|_{L^p}| = infty$ (possibly)...
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– gangrene
Dec 2 '18 at 15:21
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@RhysSteele now i know :)
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– user123
Dec 2 '18 at 19:43