Mixing
Discuss strong and weak convergence of a sequence in $W^{1,p}$ Sobolev space
$begingroup$
Discuss the strong and weak convergence of the sequence of functions
$$u_n(x)=frac{1}{n}sin nx+2sqrt{x}$$
in the $W^{1,p}(0,1)$ Sobolev space.
Pointwise limit is $u(x)=2sqrt{x}$ and can be easily shown that the sequence converge strongly in $L^p(0,1) forall 1le p le +infty$.
The derivative of the sequence is $u'_n(x)=cos nx +frac{1}{sqrt{x}}$ and the pointwise limit $u'(x)=frac{1}{sqrt{x}}$ (right?).
For $p=1$ I would say that
$$int_0^1 cos nx dx = frac{sin nx}{n}to 0text{ as }nto+infty$$
so the sequence would converge strongly in $W^{1,1}$, but actually
$$||u'_n(x)-u'(x)||_{L^1} = int_0^1 |cos nx| dx$$
and $cos nx$ does not have a fixed sign for $xin[0,1]$ since $n$ is going to infinity so how to solve it?
For $p=+infty$
$$||u'_n(x)-u'(x)||_{L^{infty}} = max|cos nx|=1nrightarrow 0$$
so the sequence doesn't converge strongly in $W^{1,+infty}$.
For $1<p<+infty$, since $max(cos nx)=1$
$$||u'_n(x)-u'(x)||_{L^p}^p le 1$$
so we don't have strong convergence, but since $L^p$ is reflexive for $1<p<+infty$, by Banach-Alaoglu we can extract a subsequence of $u'_n$ converging weakly (but where?). In particular in $L^2$ $cos nx$ is a subsequence of the trigonometric basis, which converges weakly to $0$ in $L^2$. But for others $p$ where does the subsequence extracted from $u'_n$ weakly converge?
functional-analysis convergence
asked Jan 18 at 10:08
sound wavesound wave
28619
$endgroup$
add a comment |
$begingroup$
Discuss the strong and weak convergence of the sequence of functions
$$u_n(x)=frac{1}{n}sin nx+2sqrt{x}$$
in the $W^{1,p}(0,1)$ Sobolev space.
Pointwise limit is $u(x)=2sqrt{x}$ and can be easily shown that the sequence converge strongly in $L^p(0,1) forall 1le p le +infty$.
The derivative of the sequence is $u'_n(x)=cos nx +frac{1}{sqrt{x}}$ and the pointwise limit $u'(x)=frac{1}{sqrt{x}}$ (right?).
For $p=1$ I would say that
$$int_0^1 cos nx dx = frac{sin nx}{n}to 0text{ as }nto+infty$$
so the sequence would converge strongly in $W^{1,1}$, but actually
$$||u'_n(x)-u'(x)||_{L^1} = int_0^1 |cos nx| dx$$
and $cos nx$ does not have a fixed sign for $xin[0,1]$ since $n$ is going to infinity so how to solve it?
For $p=+infty$
$$||u'_n(x)-u'(x)||_{L^{infty}} = max|cos nx|=1nrightarrow 0$$
so the sequence doesn't converge strongly in $W^{1,+infty}$.
For $1<p<+infty$, since $max(cos nx)=1$
$$||u'_n(x)-u'(x)||_{L^p}^p le 1$$
so we don't have strong convergence, but since $L^p$ is reflexive for $1<p<+infty$, by Banach-Alaoglu we can extract a subsequence of $u'_n$ converging weakly (but where?). In particular in $L^2$ $cos nx$ is a subsequence of the trigonometric basis, which converges weakly to $0$ in $L^2$. But for others $p$ where does the subsequence extracted from $u'_n$ weakly converge?
functional-analysis convergence
asked Jan 18 at 10:08
sound wavesound wave
28619
$endgroup$
add a comment |
$begingroup$
Discuss the strong and weak convergence of the sequence of functions
$$u_n(x)=frac{1}{n}sin nx+2sqrt{x}$$
in the $W^{1,p}(0,1)$ Sobolev space.
Pointwise limit is $u(x)=2sqrt{x}$ and can be easily shown that the sequence converge strongly in $L^p(0,1) forall 1le p le +infty$.
The derivative of the sequence is $u'_n(x)=cos nx +frac{1}{sqrt{x}}$ and the pointwise limit $u'(x)=frac{1}{sqrt{x}}$ (right?).
For $p=1$ I would say that
$$int_0^1 cos nx dx = frac{sin nx}{n}to 0text{ as }nto+infty$$
so the sequence would converge strongly in $W^{1,1}$, but actually
$$||u'_n(x)-u'(x)||_{L^1} = int_0^1 |cos nx| dx$$
and $cos nx$ does not have a fixed sign for $xin[0,1]$ since $n$ is going to infinity so how to solve it?
For $p=+infty$
$$||u'_n(x)-u'(x)||_{L^{infty}} = max|cos nx|=1nrightarrow 0$$
so the sequence doesn't converge strongly in $W^{1,+infty}$.
For $1<p<+infty$, since $max(cos nx)=1$
$$||u'_n(x)-u'(x)||_{L^p}^p le 1$$
so we don't have strong convergence, but since $L^p$ is reflexive for $1<p<+infty$, by Banach-Alaoglu we can extract a subsequence of $u'_n$ converging weakly (but where?). In particular in $L^2$ $cos nx$ is a subsequence of the trigonometric basis, which converges weakly to $0$ in $L^2$. But for others $p$ where does the subsequence extracted from $u'_n$ weakly converge?
functional-analysis convergence
asked Jan 18 at 10:08
sound wavesound wave
28619
$endgroup$
Discuss the strong and weak convergence of the sequence of functions
$$u_n(x)=frac{1}{n}sin nx+2sqrt{x}$$
in the $W^{1,p}(0,1)$ Sobolev space.
Pointwise limit is $u(x)=2sqrt{x}$ and can be easily shown that the sequence converge strongly in $L^p(0,1) forall 1le p le +infty$.
The derivative of the sequence is $u'_n(x)=cos nx +frac{1}{sqrt{x}}$ and the pointwise limit $u'(x)=frac{1}{sqrt{x}}$ (right?).
For $p=1$ I would say that
$$int_0^1 cos nx dx = frac{sin nx}{n}to 0text{ as }nto+infty$$
so the sequence would converge strongly in $W^{1,1}$, but actually
$$||u'_n(x)-u'(x)||_{L^1} = int_0^1 |cos nx| dx$$
and $cos nx$ does not have a fixed sign for $xin[0,1]$ since $n$ is going to infinity so how to solve it?
For $p=+infty$
$$||u'_n(x)-u'(x)||_{L^{infty}} = max|cos nx|=1nrightarrow 0$$
so the sequence doesn't converge strongly in $W^{1,+infty}$.
For $1<p<+infty$, since $max(cos nx)=1$
$$||u'_n(x)-u'(x)||_{L^p}^p le 1$$
so we don't have strong convergence, but since $L^p$ is reflexive for $1<p<+infty$, by Banach-Alaoglu we can extract a subsequence of $u'_n$ converging weakly (but where?). In particular in $L^2$ $cos nx$ is a subsequence of the trigonometric basis, which converges weakly to $0$ in $L^2$. But for others $p$ where does the subsequence extracted from $u'_n$ weakly converge?
functional-analysis convergence
functional-analysis convergence
asked Jan 18 at 10:08
sound wavesound wave
28619
asked Jan 18 at 10:08
sound wavesound wave
28619
asked Jan 18 at 10:08
sound wavesound wave
28619
asked Jan 18 at 10:08
sound wavesound wave
28619
asked Jan 18 at 10:08
sound wavesound wave
28619
28619
add a comment |
add a comment |
1 Answer
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$begingroup$
First notice that $u_n$ only belong to some of the $W^{1,p}$ spaces, as we have $u_n'=cos (nx)+frac{1}{sqrt{x}}$ and
$$cos(nx)+frac{1}{sqrt{x}}in L^p(0,1)iff frac{1}{sqrt{x}}in L^p(0,1)iff |x|^{-p/2}in L^1(0,1)iff p<2 $$
Therefore it is only meaningful to discuss convergence in $W^{1,p}(0,1)$ for $pin [1,2)$.
There is no pointwise limit for $u_n'(x)=cos (nx)+frac{1}{sqrt{x}}$ because there is no pointwise limit for $cos(nx)$ except for $x=0$. The sequence keeps on oscillating around $frac{1}{sqrt{x}}$.
Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $pin (1,infty)$. The issue is that it says nothing about the weak limit aside from existence.
To understand the weak limit, there is a general result (see my answer here) - if $g:mathbb{R}to mathbb{R}$ is a bounded periodic function such that its mean over a period is $alpha$, then if $v_n(x):=g(nx)$ we have $v_nrightharpoonup alpha$ weakly-star in $L^{infty}(mathbb{R})$ for $p<infty$. In this case, we have $g(x)=cos x$ whose mean over a period is $alpha=int_0^{2pi}cos x, dx = 0$. Therefore, $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(mathbb{R})$, hence $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(0,1)$ and thus (since $L^{infty}(0,1)hookrightarrow L^p(0,1)$ for $pin [1,infty]$) we also have $cos (nx)rightharpoonup 0$ weakly-star (and hence weakly, since they are reflexive) in $L^p(0,1)$ for all $pin (1,infty)$. Finally, we can include $p=1$ as weak convergence in $L^p(0,1)$ for $p>1$ implies weak convergence in $L^1(0,1)$.
In conclusion, the above argument shows that the sequence $u_n$ is weakly convergent to $2sqrt{x}$ in $W^{1,p}(0,1)$ for all $pin [1,2)$.
To study strong convergence, notice that as we have proved, the weak limit of $cos (nx)$ is $0$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $u_n'to u$ strongly in $L^p(0,1)$, then we would have $cos (nx)to 0$ strongly in $L^p(0,1)$ and in particular in $L^1(0,1)$, but if $2(k+1)pigeq ngeq 2kpi$, then
begin{align*}int_0^1|cos (n x)|,dx&= frac{1}{n}int_0^n|cos (y)|,dygeq frac{1}{2(k+1)pi}int_0^{2kpi}|cos y|,dygeq \
&geq frac{k}{2(k+1)pi}int_0^{2pi}|cos y|,dy=frac{2}{(1+1/k)pi}not to 0
end{align*}
a contradiction.
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
$endgroup$
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
|
show 7 more comments
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1 Answer
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1 Answer
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$begingroup$
First notice that $u_n$ only belong to some of the $W^{1,p}$ spaces, as we have $u_n'=cos (nx)+frac{1}{sqrt{x}}$ and
$$cos(nx)+frac{1}{sqrt{x}}in L^p(0,1)iff frac{1}{sqrt{x}}in L^p(0,1)iff |x|^{-p/2}in L^1(0,1)iff p<2 $$
Therefore it is only meaningful to discuss convergence in $W^{1,p}(0,1)$ for $pin [1,2)$.
There is no pointwise limit for $u_n'(x)=cos (nx)+frac{1}{sqrt{x}}$ because there is no pointwise limit for $cos(nx)$ except for $x=0$. The sequence keeps on oscillating around $frac{1}{sqrt{x}}$.
Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $pin (1,infty)$. The issue is that it says nothing about the weak limit aside from existence.
To understand the weak limit, there is a general result (see my answer here) - if $g:mathbb{R}to mathbb{R}$ is a bounded periodic function such that its mean over a period is $alpha$, then if $v_n(x):=g(nx)$ we have $v_nrightharpoonup alpha$ weakly-star in $L^{infty}(mathbb{R})$ for $p<infty$. In this case, we have $g(x)=cos x$ whose mean over a period is $alpha=int_0^{2pi}cos x, dx = 0$. Therefore, $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(mathbb{R})$, hence $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(0,1)$ and thus (since $L^{infty}(0,1)hookrightarrow L^p(0,1)$ for $pin [1,infty]$) we also have $cos (nx)rightharpoonup 0$ weakly-star (and hence weakly, since they are reflexive) in $L^p(0,1)$ for all $pin (1,infty)$. Finally, we can include $p=1$ as weak convergence in $L^p(0,1)$ for $p>1$ implies weak convergence in $L^1(0,1)$.
In conclusion, the above argument shows that the sequence $u_n$ is weakly convergent to $2sqrt{x}$ in $W^{1,p}(0,1)$ for all $pin [1,2)$.
To study strong convergence, notice that as we have proved, the weak limit of $cos (nx)$ is $0$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $u_n'to u$ strongly in $L^p(0,1)$, then we would have $cos (nx)to 0$ strongly in $L^p(0,1)$ and in particular in $L^1(0,1)$, but if $2(k+1)pigeq ngeq 2kpi$, then
begin{align*}int_0^1|cos (n x)|,dx&= frac{1}{n}int_0^n|cos (y)|,dygeq frac{1}{2(k+1)pi}int_0^{2kpi}|cos y|,dygeq \
&geq frac{k}{2(k+1)pi}int_0^{2pi}|cos y|,dy=frac{2}{(1+1/k)pi}not to 0
end{align*}
a contradiction.
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
$endgroup$
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
|
show 7 more comments
$begingroup$
First notice that $u_n$ only belong to some of the $W^{1,p}$ spaces, as we have $u_n'=cos (nx)+frac{1}{sqrt{x}}$ and
$$cos(nx)+frac{1}{sqrt{x}}in L^p(0,1)iff frac{1}{sqrt{x}}in L^p(0,1)iff |x|^{-p/2}in L^1(0,1)iff p<2 $$
Therefore it is only meaningful to discuss convergence in $W^{1,p}(0,1)$ for $pin [1,2)$.
There is no pointwise limit for $u_n'(x)=cos (nx)+frac{1}{sqrt{x}}$ because there is no pointwise limit for $cos(nx)$ except for $x=0$. The sequence keeps on oscillating around $frac{1}{sqrt{x}}$.
Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $pin (1,infty)$. The issue is that it says nothing about the weak limit aside from existence.
To understand the weak limit, there is a general result (see my answer here) - if $g:mathbb{R}to mathbb{R}$ is a bounded periodic function such that its mean over a period is $alpha$, then if $v_n(x):=g(nx)$ we have $v_nrightharpoonup alpha$ weakly-star in $L^{infty}(mathbb{R})$ for $p<infty$. In this case, we have $g(x)=cos x$ whose mean over a period is $alpha=int_0^{2pi}cos x, dx = 0$. Therefore, $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(mathbb{R})$, hence $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(0,1)$ and thus (since $L^{infty}(0,1)hookrightarrow L^p(0,1)$ for $pin [1,infty]$) we also have $cos (nx)rightharpoonup 0$ weakly-star (and hence weakly, since they are reflexive) in $L^p(0,1)$ for all $pin (1,infty)$. Finally, we can include $p=1$ as weak convergence in $L^p(0,1)$ for $p>1$ implies weak convergence in $L^1(0,1)$.
In conclusion, the above argument shows that the sequence $u_n$ is weakly convergent to $2sqrt{x}$ in $W^{1,p}(0,1)$ for all $pin [1,2)$.
To study strong convergence, notice that as we have proved, the weak limit of $cos (nx)$ is $0$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $u_n'to u$ strongly in $L^p(0,1)$, then we would have $cos (nx)to 0$ strongly in $L^p(0,1)$ and in particular in $L^1(0,1)$, but if $2(k+1)pigeq ngeq 2kpi$, then
begin{align*}int_0^1|cos (n x)|,dx&= frac{1}{n}int_0^n|cos (y)|,dygeq frac{1}{2(k+1)pi}int_0^{2kpi}|cos y|,dygeq \
&geq frac{k}{2(k+1)pi}int_0^{2pi}|cos y|,dy=frac{2}{(1+1/k)pi}not to 0
end{align*}
a contradiction.
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
$endgroup$
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
|
show 7 more comments
$begingroup$
First notice that $u_n$ only belong to some of the $W^{1,p}$ spaces, as we have $u_n'=cos (nx)+frac{1}{sqrt{x}}$ and
$$cos(nx)+frac{1}{sqrt{x}}in L^p(0,1)iff frac{1}{sqrt{x}}in L^p(0,1)iff |x|^{-p/2}in L^1(0,1)iff p<2 $$
Therefore it is only meaningful to discuss convergence in $W^{1,p}(0,1)$ for $pin [1,2)$.
There is no pointwise limit for $u_n'(x)=cos (nx)+frac{1}{sqrt{x}}$ because there is no pointwise limit for $cos(nx)$ except for $x=0$. The sequence keeps on oscillating around $frac{1}{sqrt{x}}$.
Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $pin (1,infty)$. The issue is that it says nothing about the weak limit aside from existence.
To understand the weak limit, there is a general result (see my answer here) - if $g:mathbb{R}to mathbb{R}$ is a bounded periodic function such that its mean over a period is $alpha$, then if $v_n(x):=g(nx)$ we have $v_nrightharpoonup alpha$ weakly-star in $L^{infty}(mathbb{R})$ for $p<infty$. In this case, we have $g(x)=cos x$ whose mean over a period is $alpha=int_0^{2pi}cos x, dx = 0$. Therefore, $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(mathbb{R})$, hence $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(0,1)$ and thus (since $L^{infty}(0,1)hookrightarrow L^p(0,1)$ for $pin [1,infty]$) we also have $cos (nx)rightharpoonup 0$ weakly-star (and hence weakly, since they are reflexive) in $L^p(0,1)$ for all $pin (1,infty)$. Finally, we can include $p=1$ as weak convergence in $L^p(0,1)$ for $p>1$ implies weak convergence in $L^1(0,1)$.
In conclusion, the above argument shows that the sequence $u_n$ is weakly convergent to $2sqrt{x}$ in $W^{1,p}(0,1)$ for all $pin [1,2)$.
To study strong convergence, notice that as we have proved, the weak limit of $cos (nx)$ is $0$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $u_n'to u$ strongly in $L^p(0,1)$, then we would have $cos (nx)to 0$ strongly in $L^p(0,1)$ and in particular in $L^1(0,1)$, but if $2(k+1)pigeq ngeq 2kpi$, then
begin{align*}int_0^1|cos (n x)|,dx&= frac{1}{n}int_0^n|cos (y)|,dygeq frac{1}{2(k+1)pi}int_0^{2kpi}|cos y|,dygeq \
&geq frac{k}{2(k+1)pi}int_0^{2pi}|cos y|,dy=frac{2}{(1+1/k)pi}not to 0
end{align*}
a contradiction.
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
$endgroup$
First notice that $u_n$ only belong to some of the $W^{1,p}$ spaces, as we have $u_n'=cos (nx)+frac{1}{sqrt{x}}$ and
$$cos(nx)+frac{1}{sqrt{x}}in L^p(0,1)iff frac{1}{sqrt{x}}in L^p(0,1)iff |x|^{-p/2}in L^1(0,1)iff p<2 $$
Therefore it is only meaningful to discuss convergence in $W^{1,p}(0,1)$ for $pin [1,2)$.
There is no pointwise limit for $u_n'(x)=cos (nx)+frac{1}{sqrt{x}}$ because there is no pointwise limit for $cos(nx)$ except for $x=0$. The sequence keeps on oscillating around $frac{1}{sqrt{x}}$.
Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $pin (1,infty)$. The issue is that it says nothing about the weak limit aside from existence.
To understand the weak limit, there is a general result (see my answer here) - if $g:mathbb{R}to mathbb{R}$ is a bounded periodic function such that its mean over a period is $alpha$, then if $v_n(x):=g(nx)$ we have $v_nrightharpoonup alpha$ weakly-star in $L^{infty}(mathbb{R})$ for $p<infty$. In this case, we have $g(x)=cos x$ whose mean over a period is $alpha=int_0^{2pi}cos x, dx = 0$. Therefore, $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(mathbb{R})$, hence $cos (nx)rightharpoonup 0$ weakly-star in $L^{infty}(0,1)$ and thus (since $L^{infty}(0,1)hookrightarrow L^p(0,1)$ for $pin [1,infty]$) we also have $cos (nx)rightharpoonup 0$ weakly-star (and hence weakly, since they are reflexive) in $L^p(0,1)$ for all $pin (1,infty)$. Finally, we can include $p=1$ as weak convergence in $L^p(0,1)$ for $p>1$ implies weak convergence in $L^1(0,1)$.
In conclusion, the above argument shows that the sequence $u_n$ is weakly convergent to $2sqrt{x}$ in $W^{1,p}(0,1)$ for all $pin [1,2)$.
To study strong convergence, notice that as we have proved, the weak limit of $cos (nx)$ is $0$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $u_n'to u$ strongly in $L^p(0,1)$, then we would have $cos (nx)to 0$ strongly in $L^p(0,1)$ and in particular in $L^1(0,1)$, but if $2(k+1)pigeq ngeq 2kpi$, then
begin{align*}int_0^1|cos (n x)|,dx&= frac{1}{n}int_0^n|cos (y)|,dygeq frac{1}{2(k+1)pi}int_0^{2kpi}|cos y|,dygeq \
&geq frac{k}{2(k+1)pi}int_0^{2pi}|cos y|,dy=frac{2}{(1+1/k)pi}not to 0
end{align*}
a contradiction.
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
edited Jan 20 at 20:02
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
answered Jan 18 at 11:57
Lorenzo QuarisaLorenzo Quarisa
3,690623
3,690623
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
|
show 7 more comments
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
How can we say that $cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $cos nx$ does not converge a.e. ...
$endgroup$
– Song
Jan 18 at 12:00
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $left{cos (nx)right}$ is bounded in $mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $xin [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 12:11
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
$begingroup$
It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer?
$endgroup$
– Song
Jan 18 at 12:14
1
1
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
$begingroup$
1. The issue is that there is no pointwise limit for $cos (nx)$, the $frac{1}{sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $lim_{xto 0^+}x^{-p/2+1}=+infty$ which shows that the singularity is non-integrable.
$endgroup$
– Lorenzo Quarisa
Jan 18 at 14:06
1
1
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
$begingroup$
3. Yes that's a mistake, what is correct is that there is no limit for $cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $left{u_nright}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is.
$endgroup$
– Lorenzo Quarisa
Jan 20 at 20:10
|
show 7 more comments
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