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.



Weak continuity under Laplace transform
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 $\sup\_{s>0} [L(u)-L(u\_n)](s)=\int\_n^\infty u(x) \, dx=\infty$.


