If $v_n rightharpoonup v$ and $phi in C_c^infty(mathbbR^N)$, do we have $v_n phi rightharpoonup v phi$?

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I'm stuck trying to solve this question. If I have a sequence in the Sobolev space $D^1,vecp(mathbbR^N)$ (or $W_0^1,p(mathbbR^N)$ for simplicity) which converges weakly to $v$ and $phi in C_c^infty(mathbbR^N)$, do we necessarily have that $v_n phi$ converges weakly to $v phi$? It seems intuitive but I can't figure out a way to prove it.




sobolev-spaces weak-convergence




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    Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
    – user357151
    Aug 2 at 1:33










  • You're right! I didn't think of it that way, that works perfectly!
    – Sambo
    Aug 2 at 11:09














up vote
2
down vote

favorite
2












I'm stuck trying to solve this question. If I have a sequence in the Sobolev space $D^1,vecp(mathbbR^N)$ (or $W_0^1,p(mathbbR^N)$ for simplicity) which converges weakly to $v$ and $phi in C_c^infty(mathbbR^N)$, do we necessarily have that $v_n phi$ converges weakly to $v phi$? It seems intuitive but I can't figure out a way to prove it.




sobolev-spaces weak-convergence




share|cite|improve this question















  • 1




    Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
    – user357151
    Aug 2 at 1:33










  • You're right! I didn't think of it that way, that works perfectly!
    – Sambo
    Aug 2 at 11:09












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





I'm stuck trying to solve this question. If I have a sequence in the Sobolev space $D^1,vecp(mathbbR^N)$ (or $W_0^1,p(mathbbR^N)$ for simplicity) which converges weakly to $v$ and $phi in C_c^infty(mathbbR^N)$, do we necessarily have that $v_n phi$ converges weakly to $v phi$? It seems intuitive but I can't figure out a way to prove it.




sobolev-spaces weak-convergence




share|cite|improve this question











I'm stuck trying to solve this question. If I have a sequence in the Sobolev space $D^1,vecp(mathbbR^N)$ (or $W_0^1,p(mathbbR^N)$ for simplicity) which converges weakly to $v$ and $phi in C_c^infty(mathbbR^N)$, do we necessarily have that $v_n phi$ converges weakly to $v phi$? It seems intuitive but I can't figure out a way to prove it.





sobolev-spaces weak-convergence





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 22:38









Sambo

1,2771427




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  • 1




    Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
    – user357151
    Aug 2 at 1:33










  • You're right! I didn't think of it that way, that works perfectly!
    – Sambo
    Aug 2 at 11:09












  • 1




    Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
    – user357151
    Aug 2 at 1:33










  • You're right! I didn't think of it that way, that works perfectly!
    – Sambo
    Aug 2 at 11:09







1




1




Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
– user357151
Aug 2 at 1:33




Multiplication by a smooth function is a bounded linear operator on the Sobolev space, isn't it?
– user357151
Aug 2 at 1:33












You're right! I didn't think of it that way, that works perfectly!
– Sambo
Aug 2 at 11:09




You're right! I didn't think of it that way, that works perfectly!
– Sambo
Aug 2 at 11:09










1 Answer
1






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$newcommandDprD^1,vecp(mathbbR^N)$
An insightful comment by user357151 solves the question easily. Let $X$ be $Dpr$, $W^1,p(mathbbR^N)$, or even $L^p(mathbbR^N)$.



Remark that the $S: X rightarrow X : u mapsto phi u$ is a bounded linear operator, since $|phi u| leq C |u|$. Then for any $T in X^*$, $T circ S in X^*$, and so:
$$
lim limits_n rightarrow infty
T(S(v_n))
= T(S(v))
$$
And hence $S(v_n)=phi v_n$ converges weakly to $S(v) = phi v$.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    $newcommandDprD^1,vecp(mathbbR^N)$
    An insightful comment by user357151 solves the question easily. Let $X$ be $Dpr$, $W^1,p(mathbbR^N)$, or even $L^p(mathbbR^N)$.



    Remark that the $S: X rightarrow X : u mapsto phi u$ is a bounded linear operator, since $|phi u| leq C |u|$. Then for any $T in X^*$, $T circ S in X^*$, and so:
    $$
    lim limits_n rightarrow infty
    T(S(v_n))
    = T(S(v))
    $$
    And hence $S(v_n)=phi v_n$ converges weakly to $S(v) = phi v$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      $newcommandDprD^1,vecp(mathbbR^N)$
      An insightful comment by user357151 solves the question easily. Let $X$ be $Dpr$, $W^1,p(mathbbR^N)$, or even $L^p(mathbbR^N)$.



      Remark that the $S: X rightarrow X : u mapsto phi u$ is a bounded linear operator, since $|phi u| leq C |u|$. Then for any $T in X^*$, $T circ S in X^*$, and so:
      $$
      lim limits_n rightarrow infty
      T(S(v_n))
      = T(S(v))
      $$
      And hence $S(v_n)=phi v_n$ converges weakly to $S(v) = phi v$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        $newcommandDprD^1,vecp(mathbbR^N)$
        An insightful comment by user357151 solves the question easily. Let $X$ be $Dpr$, $W^1,p(mathbbR^N)$, or even $L^p(mathbbR^N)$.



        Remark that the $S: X rightarrow X : u mapsto phi u$ is a bounded linear operator, since $|phi u| leq C |u|$. Then for any $T in X^*$, $T circ S in X^*$, and so:
        $$
        lim limits_n rightarrow infty
        T(S(v_n))
        = T(S(v))
        $$
        And hence $S(v_n)=phi v_n$ converges weakly to $S(v) = phi v$.






        share|cite|improve this answer













        $newcommandDprD^1,vecp(mathbbR^N)$
        An insightful comment by user357151 solves the question easily. Let $X$ be $Dpr$, $W^1,p(mathbbR^N)$, or even $L^p(mathbbR^N)$.



        Remark that the $S: X rightarrow X : u mapsto phi u$ is a bounded linear operator, since $|phi u| leq C |u|$. Then for any $T in X^*$, $T circ S in X^*$, and so:
        $$
        lim limits_n rightarrow infty
        T(S(v_n))
        = T(S(v))
        $$
        And hence $S(v_n)=phi v_n$ converges weakly to $S(v) = phi v$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 11:52









        Sambo

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