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Prove that for $E$ of finite measure, $T(f) = int_Ephicirc f$ is continuous on $L^p(E)$ if $phi(x)$ in continuous on $R$ and $phi(x)<a+b|x|^p$.
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In Chapter 8 of Real analysis, 4th edition by Royden, a functional is continuous if $f_n rightarrow f$ in $L^p$ implies $T(f_n)rightarrow T(f)$. Royden had given a proof of this proposition in Collary 18 in that Chapter but there seems to be a mistake in it.
Let $f_n$ be a sequence in $L^p$ that convergences strongly to $f$ in $L^p(E)$. By taking a subsequence if necessary and relabelling, we suppose $f_n$ is rapidly Cauchy. Therefore, according to Theorem 6 of Chapter 7, $f_n$ convergences pointwise a.e. on $E$ to $f$. Since $phi$ is continuous, $phicirc f_n$ convergences pointwise a.e. on E to $phicirc f$. Moreover, by the completeness of $L^p(E)$, since $f_n$ is rapid Cauchy in $L^p(E)$, the function
$$g = |f_1| + sum_n=1^infty |f_n+1-f_n| $$
belongs to $L^p(E)$. It is clear that
$$ |f_n| le g text a.e. on E for all n. $$
and hence, by the inequality (33),
$$ |phicirc f_n| le a + b cdot |f_n|^p le a + bcdot g^p text a.e. on E for all n.$$
We infer from the Dominated Convergence Theorem that
$$ lim_nrightarrow infty int_E phicirc f_n = int_Ephicirc f. $$
Therefore T is continuous on $L^p(E)$.
But, why could one take a subsequence if necessary as in the boldface text above? If this was done, it occur to me that it is only proven that for the subsequence $T(f_n_k)rightarrow T(f)$ rather than for the original sequence $T(f_n)rightarrow T(f)$.
real-analysis functional-analysis lp-spaces
asked Jul 23 at 14:46
Ze Chen
182
add a comment |Â
up vote
3
down vote
favorite
In Chapter 8 of Real analysis, 4th edition by Royden, a functional is continuous if $f_n rightarrow f$ in $L^p$ implies $T(f_n)rightarrow T(f)$. Royden had given a proof of this proposition in Collary 18 in that Chapter but there seems to be a mistake in it.
Let $f_n$ be a sequence in $L^p$ that convergences strongly to $f$ in $L^p(E)$. By taking a subsequence if necessary and relabelling, we suppose $f_n$ is rapidly Cauchy. Therefore, according to Theorem 6 of Chapter 7, $f_n$ convergences pointwise a.e. on $E$ to $f$. Since $phi$ is continuous, $phicirc f_n$ convergences pointwise a.e. on E to $phicirc f$. Moreover, by the completeness of $L^p(E)$, since $f_n$ is rapid Cauchy in $L^p(E)$, the function
$$g = |f_1| + sum_n=1^infty |f_n+1-f_n| $$
belongs to $L^p(E)$. It is clear that
$$ |f_n| le g text a.e. on E for all n. $$
and hence, by the inequality (33),
$$ |phicirc f_n| le a + b cdot |f_n|^p le a + bcdot g^p text a.e. on E for all n.$$
We infer from the Dominated Convergence Theorem that
$$ lim_nrightarrow infty int_E phicirc f_n = int_Ephicirc f. $$
Therefore T is continuous on $L^p(E)$.
But, why could one take a subsequence if necessary as in the boldface text above? If this was done, it occur to me that it is only proven that for the subsequence $T(f_n_k)rightarrow T(f)$ rather than for the original sequence $T(f_n)rightarrow T(f)$.
real-analysis functional-analysis lp-spaces
asked Jul 23 at 14:46
Ze Chen
182
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In Chapter 8 of Real analysis, 4th edition by Royden, a functional is continuous if $f_n rightarrow f$ in $L^p$ implies $T(f_n)rightarrow T(f)$. Royden had given a proof of this proposition in Collary 18 in that Chapter but there seems to be a mistake in it.
Let $f_n$ be a sequence in $L^p$ that convergences strongly to $f$ in $L^p(E)$. By taking a subsequence if necessary and relabelling, we suppose $f_n$ is rapidly Cauchy. Therefore, according to Theorem 6 of Chapter 7, $f_n$ convergences pointwise a.e. on $E$ to $f$. Since $phi$ is continuous, $phicirc f_n$ convergences pointwise a.e. on E to $phicirc f$. Moreover, by the completeness of $L^p(E)$, since $f_n$ is rapid Cauchy in $L^p(E)$, the function
$$g = |f_1| + sum_n=1^infty |f_n+1-f_n| $$
belongs to $L^p(E)$. It is clear that
$$ |f_n| le g text a.e. on E for all n. $$
and hence, by the inequality (33),
$$ |phicirc f_n| le a + b cdot |f_n|^p le a + bcdot g^p text a.e. on E for all n.$$
We infer from the Dominated Convergence Theorem that
$$ lim_nrightarrow infty int_E phicirc f_n = int_Ephicirc f. $$
Therefore T is continuous on $L^p(E)$.
But, why could one take a subsequence if necessary as in the boldface text above? If this was done, it occur to me that it is only proven that for the subsequence $T(f_n_k)rightarrow T(f)$ rather than for the original sequence $T(f_n)rightarrow T(f)$.
real-analysis functional-analysis lp-spaces
asked Jul 23 at 14:46
Ze Chen
182
In Chapter 8 of Real analysis, 4th edition by Royden, a functional is continuous if $f_n rightarrow f$ in $L^p$ implies $T(f_n)rightarrow T(f)$. Royden had given a proof of this proposition in Collary 18 in that Chapter but there seems to be a mistake in it.
Let $f_n$ be a sequence in $L^p$ that convergences strongly to $f$ in $L^p(E)$. By taking a subsequence if necessary and relabelling, we suppose $f_n$ is rapidly Cauchy. Therefore, according to Theorem 6 of Chapter 7, $f_n$ convergences pointwise a.e. on $E$ to $f$. Since $phi$ is continuous, $phicirc f_n$ convergences pointwise a.e. on E to $phicirc f$. Moreover, by the completeness of $L^p(E)$, since $f_n$ is rapid Cauchy in $L^p(E)$, the function
$$g = |f_1| + sum_n=1^infty |f_n+1-f_n| $$
belongs to $L^p(E)$. It is clear that
$$ |f_n| le g text a.e. on E for all n. $$
and hence, by the inequality (33),
$$ |phicirc f_n| le a + b cdot |f_n|^p le a + bcdot g^p text a.e. on E for all n.$$
We infer from the Dominated Convergence Theorem that
$$ lim_nrightarrow infty int_E phicirc f_n = int_Ephicirc f. $$
Therefore T is continuous on $L^p(E)$.
But, why could one take a subsequence if necessary as in the boldface text above? If this was done, it occur to me that it is only proven that for the subsequence $T(f_n_k)rightarrow T(f)$ rather than for the original sequence $T(f_n)rightarrow T(f)$.
real-analysis functional-analysis lp-spaces
asked Jul 23 at 14:46
Ze Chen
182
asked Jul 23 at 14:46
Ze Chen
182
asked Jul 23 at 14:46
Ze Chen
182
asked Jul 23 at 14:46
Ze Chen
182
182
add a comment |Â
add a comment |Â
1 Answer
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This is enough to prove the claim because we now know that any sequence $f_nto f$ has a subsequence such that $T(f_n_k)to T(f)$.
In particular, assume $T(f_n)notto T(f)$. Then there is a subsequence such that $T(f_n_k)-T(f)$ is uniformly large. However, $f_n_k$ has a subsequence $f_n_k_ell$ such that $T(f_n_k_ell)to T(f)$, which is a contradiction.
This is a standard argument (if every subsequence has a subsubsequence that converges, and they all converge to the same limit, then the entire sequence converges) that is rather useful quite often.
answered Jul 23 at 16:05
Kusma
1,132111
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is enough to prove the claim because we now know that any sequence $f_nto f$ has a subsequence such that $T(f_n_k)to T(f)$.
In particular, assume $T(f_n)notto T(f)$. Then there is a subsequence such that $T(f_n_k)-T(f)$ is uniformly large. However, $f_n_k$ has a subsequence $f_n_k_ell$ such that $T(f_n_k_ell)to T(f)$, which is a contradiction.
This is a standard argument (if every subsequence has a subsubsequence that converges, and they all converge to the same limit, then the entire sequence converges) that is rather useful quite often.
answered Jul 23 at 16:05
Kusma
1,132111
add a comment |Â
up vote
3
down vote
accepted
This is enough to prove the claim because we now know that any sequence $f_nto f$ has a subsequence such that $T(f_n_k)to T(f)$.
In particular, assume $T(f_n)notto T(f)$. Then there is a subsequence such that $T(f_n_k)-T(f)$ is uniformly large. However, $f_n_k$ has a subsequence $f_n_k_ell$ such that $T(f_n_k_ell)to T(f)$, which is a contradiction.
This is a standard argument (if every subsequence has a subsubsequence that converges, and they all converge to the same limit, then the entire sequence converges) that is rather useful quite often.
answered Jul 23 at 16:05
Kusma
1,132111
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is enough to prove the claim because we now know that any sequence $f_nto f$ has a subsequence such that $T(f_n_k)to T(f)$.
In particular, assume $T(f_n)notto T(f)$. Then there is a subsequence such that $T(f_n_k)-T(f)$ is uniformly large. However, $f_n_k$ has a subsequence $f_n_k_ell$ such that $T(f_n_k_ell)to T(f)$, which is a contradiction.
This is a standard argument (if every subsequence has a subsubsequence that converges, and they all converge to the same limit, then the entire sequence converges) that is rather useful quite often.
answered Jul 23 at 16:05
Kusma
1,132111
This is enough to prove the claim because we now know that any sequence $f_nto f$ has a subsequence such that $T(f_n_k)to T(f)$.
In particular, assume $T(f_n)notto T(f)$. Then there is a subsequence such that $T(f_n_k)-T(f)$ is uniformly large. However, $f_n_k$ has a subsequence $f_n_k_ell$ such that $T(f_n_k_ell)to T(f)$, which is a contradiction.
This is a standard argument (if every subsequence has a subsubsequence that converges, and they all converge to the same limit, then the entire sequence converges) that is rather useful quite often.
answered Jul 23 at 16:05
Kusma
1,132111
answered Jul 23 at 16:05
Kusma
1,132111
answered Jul 23 at 16:05
Kusma
1,132111
answered Jul 23 at 16:05
Kusma
1,132111
1,132111
add a comment |Â
add a comment |Â
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