${u_m}$ is bounded in $W^{1, p}(U)$, but does not possess a (norm-)convergent sequence in $L^{p^∗}(U)$
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Let $U=B_1(0)$ be the unit ball in $Bbb{R}^n$, $1≤p<n$, $p^∗=frac{np}{n−p}$. Consider the sequence:
begin{equation}
label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande}
u_m = left{
begin{array}{ll}
m^{frac{n}{p}-1}(1-m|x|) & mathrm{if } x <1/m \
0 &mathrm{if } xgeq 1/m\
end{array}
right.
end{equation}
Prove that ${u_m}$ is bounded in $W^{1, p}(U)$, but does not possess a (norm-)convergent sequence in $L^{p^∗}(U)$.
I've managed to prove that ${u_m}$ is bounded in $W^{1, p}(U)$, and I've tried to prove the non-existence of a convegent subsequence via completeness of $L^{p^∗}(U)$ and taking a Cauchy series. So far I have gotten nothing. Any hints?
functional-analysis pde sobolev-spaces complete-spaces
asked yesterday
Ajafca
62
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up vote
0
down vote
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Let $U=B_1(0)$ be the unit ball in $Bbb{R}^n$, $1≤p<n$, $p^∗=frac{np}{n−p}$. Consider the sequence:
begin{equation}
label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande}
u_m = left{
begin{array}{ll}
m^{frac{n}{p}-1}(1-m|x|) & mathrm{if } x <1/m \
0 &mathrm{if } xgeq 1/m\
end{array}
right.
end{equation}
Prove that ${u_m}$ is bounded in $W^{1, p}(U)$, but does not possess a (norm-)convergent sequence in $L^{p^∗}(U)$.
I've managed to prove that ${u_m}$ is bounded in $W^{1, p}(U)$, and I've tried to prove the non-existence of a convegent subsequence via completeness of $L^{p^∗}(U)$ and taking a Cauchy series. So far I have gotten nothing. Any hints?
functional-analysis pde sobolev-spaces complete-spaces
asked yesterday
Ajafca
62
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $U=B_1(0)$ be the unit ball in $Bbb{R}^n$, $1≤p<n$, $p^∗=frac{np}{n−p}$. Consider the sequence:
begin{equation}
label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande}
u_m = left{
begin{array}{ll}
m^{frac{n}{p}-1}(1-m|x|) & mathrm{if } x <1/m \
0 &mathrm{if } xgeq 1/m\
end{array}
right.
end{equation}
Prove that ${u_m}$ is bounded in $W^{1, p}(U)$, but does not possess a (norm-)convergent sequence in $L^{p^∗}(U)$.
I've managed to prove that ${u_m}$ is bounded in $W^{1, p}(U)$, and I've tried to prove the non-existence of a convegent subsequence via completeness of $L^{p^∗}(U)$ and taking a Cauchy series. So far I have gotten nothing. Any hints?
functional-analysis pde sobolev-spaces complete-spaces
asked yesterday
Ajafca
62
Let $U=B_1(0)$ be the unit ball in $Bbb{R}^n$, $1≤p<n$, $p^∗=frac{np}{n−p}$. Consider the sequence:
begin{equation}
label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande}
u_m = left{
begin{array}{ll}
m^{frac{n}{p}-1}(1-m|x|) & mathrm{if } x <1/m \
0 &mathrm{if } xgeq 1/m\
end{array}
right.
end{equation}
Prove that ${u_m}$ is bounded in $W^{1, p}(U)$, but does not possess a (norm-)convergent sequence in $L^{p^∗}(U)$.
I've managed to prove that ${u_m}$ is bounded in $W^{1, p}(U)$, and I've tried to prove the non-existence of a convegent subsequence via completeness of $L^{p^∗}(U)$ and taking a Cauchy series. So far I have gotten nothing. Any hints?
functional-analysis pde sobolev-spaces complete-spaces
functional-analysis pde sobolev-spaces complete-spaces
asked yesterday
Ajafca
62
asked yesterday
Ajafca
62
edited yesterday
asked yesterday
Ajafca
62
asked yesterday
Ajafca
62
asked yesterday
Ajafca
62
62
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1 Answer
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The sequence converges pointwise a.e. to zero. But $|u_m|_{L^{p^*}} notto0$. Hence $(u_m)$ cannot contain a subsequence that strongly converges in $L^{p^*}$, since pointwise a.e. limits and strong limits coincide (if they exist).
answered yesterday
daw
23.6k1544
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The sequence converges pointwise a.e. to zero. But $|u_m|_{L^{p^*}} notto0$. Hence $(u_m)$ cannot contain a subsequence that strongly converges in $L^{p^*}$, since pointwise a.e. limits and strong limits coincide (if they exist).
answered yesterday
daw
23.6k1544
add a comment |
up vote
1
down vote
The sequence converges pointwise a.e. to zero. But $|u_m|_{L^{p^*}} notto0$. Hence $(u_m)$ cannot contain a subsequence that strongly converges in $L^{p^*}$, since pointwise a.e. limits and strong limits coincide (if they exist).
answered yesterday
daw
23.6k1544
add a comment |
up vote
1
down vote
up vote
1
down vote
The sequence converges pointwise a.e. to zero. But $|u_m|_{L^{p^*}} notto0$. Hence $(u_m)$ cannot contain a subsequence that strongly converges in $L^{p^*}$, since pointwise a.e. limits and strong limits coincide (if they exist).
answered yesterday
daw
23.6k1544
The sequence converges pointwise a.e. to zero. But $|u_m|_{L^{p^*}} notto0$. Hence $(u_m)$ cannot contain a subsequence that strongly converges in $L^{p^*}$, since pointwise a.e. limits and strong limits coincide (if they exist).
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
23.6k1544
add a comment |
add a comment |
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