Norm of a function on a set of measure $0$ and convergence of a sequence on a set of finite measure
Let $Esubsetmathbb{R}^n$ be a measurable set with finite Lebesgue measure, ${f_n}_{n∈N}$ be a sequence of measurable functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$ for $p>1$, and pointwise converging to $f : E → mathbb{R}$ almost everywhere.
Since the sequence converge almost everywhere, consider the set $N_0$ of measure zero where the sequence does not converge, and $Esetminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $Esetminus N_0$.
Study the convergence in $L^q(E)$ for $q∈(1,p)$:
begin{align*}
||f_n-f||_{L^q(E)}&le||f_n-f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)} \ &le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)}
end{align*}
I'd say that the first two norms in the rhs are $0$ since they are integrals on a set of measure $0$, but wouldn't this mean that $f_n$ converge to $f$ on $N_0$? But it is a contraddiction because we defined it as the set where the sequence does not converge, isn'it?
Moreover how to compute or bound the last norm on the rhs?
functional-analysis convergence
asked Nov 22 '18 at 11:42
sound wave
1618
add a comment |
Let $Esubsetmathbb{R}^n$ be a measurable set with finite Lebesgue measure, ${f_n}_{n∈N}$ be a sequence of measurable functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$ for $p>1$, and pointwise converging to $f : E → mathbb{R}$ almost everywhere.
Since the sequence converge almost everywhere, consider the set $N_0$ of measure zero where the sequence does not converge, and $Esetminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $Esetminus N_0$.
Study the convergence in $L^q(E)$ for $q∈(1,p)$:
begin{align*}
||f_n-f||_{L^q(E)}&le||f_n-f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)} \ &le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)}
end{align*}
I'd say that the first two norms in the rhs are $0$ since they are integrals on a set of measure $0$, but wouldn't this mean that $f_n$ converge to $f$ on $N_0$? But it is a contraddiction because we defined it as the set where the sequence does not converge, isn'it?
Moreover how to compute or bound the last norm on the rhs?
functional-analysis convergence
asked Nov 22 '18 at 11:42
sound wave
1618
1
Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07
add a comment |
Let $Esubsetmathbb{R}^n$ be a measurable set with finite Lebesgue measure, ${f_n}_{n∈N}$ be a sequence of measurable functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$ for $p>1$, and pointwise converging to $f : E → mathbb{R}$ almost everywhere.
Since the sequence converge almost everywhere, consider the set $N_0$ of measure zero where the sequence does not converge, and $Esetminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $Esetminus N_0$.
Study the convergence in $L^q(E)$ for $q∈(1,p)$:
begin{align*}
||f_n-f||_{L^q(E)}&le||f_n-f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)} \ &le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)}
end{align*}
I'd say that the first two norms in the rhs are $0$ since they are integrals on a set of measure $0$, but wouldn't this mean that $f_n$ converge to $f$ on $N_0$? But it is a contraddiction because we defined it as the set where the sequence does not converge, isn'it?
Moreover how to compute or bound the last norm on the rhs?
functional-analysis convergence
asked Nov 22 '18 at 11:42
sound wave
1618
Let $Esubsetmathbb{R}^n$ be a measurable set with finite Lebesgue measure, ${f_n}_{n∈N}$ be a sequence of measurable functions $f_n : E → mathbb{R}$, bounded in $L^p(E)$ for $p>1$, and pointwise converging to $f : E → mathbb{R}$ almost everywhere.
Since the sequence converge almost everywhere, consider the set $N_0$ of measure zero where the sequence does not converge, and $Esetminus N_0$ where the sequence converge. By Egorov theorem the sequence converges to $f$ uniformly on $Esetminus N_0$.
Study the convergence in $L^q(E)$ for $q∈(1,p)$:
begin{align*}
||f_n-f||_{L^q(E)}&le||f_n-f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)} \ &le ||f_n||_{L^q(N_0)}+||f||_{L^q(N_0)}+||f_n-f||_{L^q(Esetminus N_0)}
end{align*}
I'd say that the first two norms in the rhs are $0$ since they are integrals on a set of measure $0$, but wouldn't this mean that $f_n$ converge to $f$ on $N_0$? But it is a contraddiction because we defined it as the set where the sequence does not converge, isn'it?
Moreover how to compute or bound the last norm on the rhs?
functional-analysis convergence
functional-analysis convergence
asked Nov 22 '18 at 11:42
sound wave
1618
asked Nov 22 '18 at 11:42
sound wave
1618
asked Nov 22 '18 at 11:42
sound wave
1618
asked Nov 22 '18 at 11:42
sound wave
1618
asked Nov 22 '18 at 11:42
sound wave
1618
1618
1
Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07
add a comment |
1
Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07
1
1
Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07
add a comment |
1 Answer
1
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oldest
votes
First, we note that Fatou's Lemma implies that $f in L^p(E)$. Now, we apply Egorov and obtain that for every $varepsilon > 0$, there exists a measurable set $M_varepsilon$ with $mu(M_varepsilon) le varepsilon$
such that $f_n$ converges uniformly to $f$ on $E setminus M_varepsilon$.
Now, we use the triangle inequality to get
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
In the first addend, we can use Hölders inequality
(with $1/q = 1/p + 1/r$ for some $r in (1,infty)$) to obtain
$$|f_n - f|_{L^q(M_varepsilon)} le |1|_{L^r(M_varepsilon)} , |f_n - f|_{L^p(M_varepsilon)} le varepsilon^{1/r} , (|f_n|_{L^p(M_varepsilon)}+|f|_{L^p(M_varepsilon)} le C , varepsilon^{1/r}.$$
Thus,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$
Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.
answered Nov 22 '18 at 12:23
gerw
19k11133
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
|
show 3 more comments
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1 Answer
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First, we note that Fatou's Lemma implies that $f in L^p(E)$. Now, we apply Egorov and obtain that for every $varepsilon > 0$, there exists a measurable set $M_varepsilon$ with $mu(M_varepsilon) le varepsilon$
such that $f_n$ converges uniformly to $f$ on $E setminus M_varepsilon$.
Now, we use the triangle inequality to get
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
In the first addend, we can use Hölders inequality
(with $1/q = 1/p + 1/r$ for some $r in (1,infty)$) to obtain
$$|f_n - f|_{L^q(M_varepsilon)} le |1|_{L^r(M_varepsilon)} , |f_n - f|_{L^p(M_varepsilon)} le varepsilon^{1/r} , (|f_n|_{L^p(M_varepsilon)}+|f|_{L^p(M_varepsilon)} le C , varepsilon^{1/r}.$$
Thus,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$
Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.
answered Nov 22 '18 at 12:23
gerw
19k11133
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
|
show 3 more comments
First, we note that Fatou's Lemma implies that $f in L^p(E)$. Now, we apply Egorov and obtain that for every $varepsilon > 0$, there exists a measurable set $M_varepsilon$ with $mu(M_varepsilon) le varepsilon$
such that $f_n$ converges uniformly to $f$ on $E setminus M_varepsilon$.
Now, we use the triangle inequality to get
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
In the first addend, we can use Hölders inequality
(with $1/q = 1/p + 1/r$ for some $r in (1,infty)$) to obtain
$$|f_n - f|_{L^q(M_varepsilon)} le |1|_{L^r(M_varepsilon)} , |f_n - f|_{L^p(M_varepsilon)} le varepsilon^{1/r} , (|f_n|_{L^p(M_varepsilon)}+|f|_{L^p(M_varepsilon)} le C , varepsilon^{1/r}.$$
Thus,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$
Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.
answered Nov 22 '18 at 12:23
gerw
19k11133
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
|
show 3 more comments
First, we note that Fatou's Lemma implies that $f in L^p(E)$. Now, we apply Egorov and obtain that for every $varepsilon > 0$, there exists a measurable set $M_varepsilon$ with $mu(M_varepsilon) le varepsilon$
such that $f_n$ converges uniformly to $f$ on $E setminus M_varepsilon$.
Now, we use the triangle inequality to get
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
In the first addend, we can use Hölders inequality
(with $1/q = 1/p + 1/r$ for some $r in (1,infty)$) to obtain
$$|f_n - f|_{L^q(M_varepsilon)} le |1|_{L^r(M_varepsilon)} , |f_n - f|_{L^p(M_varepsilon)} le varepsilon^{1/r} , (|f_n|_{L^p(M_varepsilon)}+|f|_{L^p(M_varepsilon)} le C , varepsilon^{1/r}.$$
Thus,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$
Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.
answered Nov 22 '18 at 12:23
gerw
19k11133
First, we note that Fatou's Lemma implies that $f in L^p(E)$. Now, we apply Egorov and obtain that for every $varepsilon > 0$, there exists a measurable set $M_varepsilon$ with $mu(M_varepsilon) le varepsilon$
such that $f_n$ converges uniformly to $f$ on $E setminus M_varepsilon$.
Now, we use the triangle inequality to get
$$|f_n - f|_{L^q(E)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^q(Esetminus M_varepsilon)}
le |f_n - f|_{L^q(M_varepsilon)} + |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
In the first addend, we can use Hölders inequality
(with $1/q = 1/p + 1/r$ for some $r in (1,infty)$) to obtain
$$|f_n - f|_{L^q(M_varepsilon)} le |1|_{L^r(M_varepsilon)} , |f_n - f|_{L^p(M_varepsilon)} le varepsilon^{1/r} , (|f_n|_{L^p(M_varepsilon)}+|f|_{L^p(M_varepsilon)} le C , varepsilon^{1/r}.$$
Thus,
$$|f_n - f|_{L^q(E)}
le
C , varepsilon^{1/r}
+ |f_n-f|_{L^infty(Esetminus M_varepsilon)} , mu(E)^{1/q}.$$
Hence, we can choose $N$ (depending on $varepsilon$) large enough such that
$$|f_n - f|_{L^q(E)}
le
2 , C , varepsilon^{1/r}
qquadforall n ge N.$$
Hence, $f_n to f$ in $L^q(E)$. Note that this argument also works for $q = 1$.
answered Nov 22 '18 at 12:23
gerw
19k11133
edited Nov 27 '18 at 14:56
answered Nov 22 '18 at 12:23
gerw
19k11133
answered Nov 22 '18 at 12:23
gerw
19k11133
answered Nov 22 '18 at 12:23
gerw
19k11133
19k11133
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
|
show 3 more comments
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
1
1
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
This proof is wrong. The hypothesis only gives pointwise convergence and boundedness of $L^{p}$ norms. How did you conclude that $f_n to f$ i n $L^{p}$
– Kavi Rama Murthy
Nov 22 '18 at 23:08
1
1
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@KaviRamaMurthy: Ups, I misread the assumptions. I will fix it.
– gerw
Nov 23 '18 at 7:43
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
@gerw In the first line why did you write that the $L^q(E)$ norm of $f_n-f$ is less or equal than the sum of $L^q$ and $L^p$ norms? Should not be both $L^q$ norms?
– sound wave
Nov 27 '18 at 12:28
1
1
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
@soundwave: Yes, both should be $L^q$ norms (changed). The second inequality using the simple bound $|g|_{L^q(C)}^q le int_C |g|^q , mathrm{d}x le |g|_{L^infty(C)}^q , mu(C)$ together with $mu(Esetminus M_varepsilon) le mu(E)$.
– gerw
Nov 27 '18 at 14:58
1
1
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
Because $|g(x)|^q le |g|_{L^infty(C)}^q$ for a.a. $x in C$. Now you integrate this inequality over $C$. The right-hand side is a constant, hence it is multiplied by the measure of $C$.
– gerw
Nov 28 '18 at 22:01
|
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Egorov's Theorem does not say that the sequence converges uniformly on $N$. Your attempt to prove $L^{q}$ convergence completely fails.
– Kavi Rama Murthy
Nov 22 '18 at 11:45
I didnt say that Egorov theorem implies convergence on $N_0$, I obtain that from my assumptions on the values of the norm. I know I'm wrong, that's why I'm here asking for help.
– sound wave
Nov 22 '18 at 11:57
The result you are trying ti prove is true but it has a lengthy argument. Are you familiar with uniform integrability? You have to show that ${|f_n|^{q}}$ (and hence ${|f_n-f|^{q}}$) is uniformly integrable from which $L^{q}$ convergence follows.
– Kavi Rama Murthy
Nov 22 '18 at 12:07