$u_n$ weakly converges to $u$ in $H_0^1(I)$, $u_n'$ converges pointwisely to $u'$












1












$begingroup$


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







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$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 16:49









      Nate Eldredge

      63.2k682171




      63.2k682171










      asked Dec 8 '18 at 16:08









      lxnllllxnlll

      464




      464






















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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.






          share|cite|improve this answer









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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.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 8 '18 at 17:10









                Nate EldredgeNate Eldredge

                63.2k682171




                63.2k682171






























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