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Let the sequence $u_n\in L^2(0,\infty)$ weakly converges to $u\in L^2(0,\infty)$. What can we say about the corresponding Laplace transforms $U_n(s)$ and $U(s)$?

  1. $U_n(s)$ converges point-wise to $U(s)$ for almost all $s>0$.

  2. The convergence in (1) but also uniform.

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The notation in the question for Laplace transform should be improved for clarity.

  1. The Laplace transforms $L(u_n)(s)=\int_0^\infty e^{-sx}u_n(x)\, dx$ converge to $L(u)(s)$ pointwise for each $s>0$ by the definition of weak convergence.

  2. The convergence need not be uniform. E.g. take $u(x)=x/(1+x^2) \in L^2(0,\infty)$ and $u_n(x)=u(x)$ for $x \in (0,n)$, with $u_n(x)=0$ for $x \ge n$. Then $u_n \to u $ in $L^2(0,\infty)$, yet $\sup_{s>0} [L(u)-L(u_n)](s)=\int_n^\infty u(x) \, dx=\infty$.

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  • $\begingroup$ Is the converse also true? That is, if $L(u_n)(s)$ converges pointwise to $L(u)(s)$, then $u_n(t)$ converges weakly to $u(t)$? $\endgroup$
    – Saj_Eda
    Commented Nov 14, 2019 at 16:24
  • $\begingroup$ If the above answered your question, you can indicate that by accepting the answer. meta.mathoverflow.net/questions/3724/how-to-accept-answer $\endgroup$
    – Yuval Peres
    Commented Nov 14, 2019 at 21:38
  • $\begingroup$ The comments are not really designed for asking additional questions. Without further assumptions the converse does not hold, consider $u_n(x)=n/x$ for $x>n$ and $u_n(x)=0$ elsewhere. Then $L(u_n)$ tends to $L(0)=0$ pointwise for $s>0$, but $\int_0^\infty u_n u_1\, dx=1$ for all $n$. $\endgroup$
    – Yuval Peres
    Commented Nov 14, 2019 at 21:45

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