$u_n$ weakly converges to $u$ in $H_0^1(I)$, $u_n'$ converges pointwisely to $u'$
$begingroup$
I is a bounded interval, like (0,1).
$u_n$ is a sequence weakly converges to $u$ in $H_0^1(I)$, do we have $u_n'$ converges pointwisely to $u'$? If it is not true, can we pass it to a subsequence and then have the pointwise convergence?
Moreover, if we have the pointwise convergence of $u_n'$, do we have even better convergence, like $L^2(I)$ convergence?
functional-analysis
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
$endgroup$
add a comment |
$begingroup$
I is a bounded interval, like (0,1).
$u_n$ is a sequence weakly converges to $u$ in $H_0^1(I)$, do we have $u_n'$ converges pointwisely to $u'$? If it is not true, can we pass it to a subsequence and then have the pointwise convergence?
Moreover, if we have the pointwise convergence of $u_n'$, do we have even better convergence, like $L^2(I)$ convergence?
functional-analysis
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
$endgroup$
add a comment |
$begingroup$
I is a bounded interval, like (0,1).
$u_n$ is a sequence weakly converges to $u$ in $H_0^1(I)$, do we have $u_n'$ converges pointwisely to $u'$? If it is not true, can we pass it to a subsequence and then have the pointwise convergence?
Moreover, if we have the pointwise convergence of $u_n'$, do we have even better convergence, like $L^2(I)$ convergence?
functional-analysis
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
$endgroup$
I is a bounded interval, like (0,1).
$u_n$ is a sequence weakly converges to $u$ in $H_0^1(I)$, do we have $u_n'$ converges pointwisely to $u'$? If it is not true, can we pass it to a subsequence and then have the pointwise convergence?
Moreover, if we have the pointwise convergence of $u_n'$, do we have even better convergence, like $L^2(I)$ convergence?
functional-analysis
functional-analysis
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
edited Dec 8 '18 at 16:49
Nate Eldredge
63.2k682171
63.2k682171
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
asked Dec 8 '18 at 16:08
lxnllllxnlll
464
464
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
No.
A complex example is a little easier to give, so let's take $I = (0, 2pi)$ and $f_n(x) = frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero. This follows from the fact that $f_n to 0$ strongly in $L^2$, and $f_n' to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.) So if you write down the inner product of $f_n$ with any $g in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x in (0,2pi)$. If you pass to a subsequence, you can get it to converge at a few points, but not a.e. For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$. Then by dominated convergence, it would also converge to $h$ strongly in $L^2$. Now since $f_{n_k}' to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' to 0$ strongly in $L^2$. But this is absurd because $|f_{n_k}'|_{L_2} = 2pi$ for every $k$.
If you want an example over $mathbb{R}$, take imaginary parts, so $f_n(x) = frac{1}{n} sin (nx)$ and $f_n'(x) = cos(nx)$. (Real parts would also work.)
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
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add a comment |
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$begingroup$
No.
A complex example is a little easier to give, so let's take $I = (0, 2pi)$ and $f_n(x) = frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero. This follows from the fact that $f_n to 0$ strongly in $L^2$, and $f_n' to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.) So if you write down the inner product of $f_n$ with any $g in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x in (0,2pi)$. If you pass to a subsequence, you can get it to converge at a few points, but not a.e. For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$. Then by dominated convergence, it would also converge to $h$ strongly in $L^2$. Now since $f_{n_k}' to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' to 0$ strongly in $L^2$. But this is absurd because $|f_{n_k}'|_{L_2} = 2pi$ for every $k$.
If you want an example over $mathbb{R}$, take imaginary parts, so $f_n(x) = frac{1}{n} sin (nx)$ and $f_n'(x) = cos(nx)$. (Real parts would also work.)
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
$endgroup$
add a comment |
$begingroup$
No.
A complex example is a little easier to give, so let's take $I = (0, 2pi)$ and $f_n(x) = frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero. This follows from the fact that $f_n to 0$ strongly in $L^2$, and $f_n' to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.) So if you write down the inner product of $f_n$ with any $g in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x in (0,2pi)$. If you pass to a subsequence, you can get it to converge at a few points, but not a.e. For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$. Then by dominated convergence, it would also converge to $h$ strongly in $L^2$. Now since $f_{n_k}' to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' to 0$ strongly in $L^2$. But this is absurd because $|f_{n_k}'|_{L_2} = 2pi$ for every $k$.
If you want an example over $mathbb{R}$, take imaginary parts, so $f_n(x) = frac{1}{n} sin (nx)$ and $f_n'(x) = cos(nx)$. (Real parts would also work.)
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
$endgroup$
add a comment |
$begingroup$
No.
A complex example is a little easier to give, so let's take $I = (0, 2pi)$ and $f_n(x) = frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero. This follows from the fact that $f_n to 0$ strongly in $L^2$, and $f_n' to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.) So if you write down the inner product of $f_n$ with any $g in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x in (0,2pi)$. If you pass to a subsequence, you can get it to converge at a few points, but not a.e. For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$. Then by dominated convergence, it would also converge to $h$ strongly in $L^2$. Now since $f_{n_k}' to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' to 0$ strongly in $L^2$. But this is absurd because $|f_{n_k}'|_{L_2} = 2pi$ for every $k$.
If you want an example over $mathbb{R}$, take imaginary parts, so $f_n(x) = frac{1}{n} sin (nx)$ and $f_n'(x) = cos(nx)$. (Real parts would also work.)
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
$endgroup$
No.
A complex example is a little easier to give, so let's take $I = (0, 2pi)$ and $f_n(x) = frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero. This follows from the fact that $f_n to 0$ strongly in $L^2$, and $f_n' to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.) So if you write down the inner product of $f_n$ with any $g in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x in (0,2pi)$. If you pass to a subsequence, you can get it to converge at a few points, but not a.e. For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$. Then by dominated convergence, it would also converge to $h$ strongly in $L^2$. Now since $f_{n_k}' to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' to 0$ strongly in $L^2$. But this is absurd because $|f_{n_k}'|_{L_2} = 2pi$ for every $k$.
If you want an example over $mathbb{R}$, take imaginary parts, so $f_n(x) = frac{1}{n} sin (nx)$ and $f_n'(x) = cos(nx)$. (Real parts would also work.)
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
answered Dec 8 '18 at 17:10
Nate EldredgeNate Eldredge
63.2k682171
63.2k682171
add a comment |
add a comment |
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