Weak convergence in $X=L^p(0,1)$ for $1<p<\infty$ can be characterized as following: $f_n\rightharpoonup f$ if and only if $f_n$ is bounded in $X$ and $\int_{(0,t)}\;f_n\;\rightarrow \int_{(0,t)}\;f$ pointwise for every $0\leq t\leq 1$.
Is this true also for $X=L^p(0,\infty)$ for $1<p<\infty$ (if we replace the condition $0\leq t\leq 1$ by $0\leq t$) ? Thank you in advance for answer.
We have to prove that linear combinations of functions of the form $\mathbf 1_{(0,t)}$ ($t\gt 0$) form a dense subset in $\mathbb L^q(0,+\infty)$ for any $1\lt q\lt +\infty$. We then apply this for $q$ such that $1/p+1/q=1$ and use boundedness and the fact that $f$ belong to $\mathbb L^p$ (there is probably a more elementary argument, but to see this, we can use reflexivity of $\mathbb L^p$ for $1\lt p\lt +\infty$) to get the weak convergence.
Proving that linear combinations of functions of the form $\mathbf 1_{(0,t)}$ ($t\gt 0$) form a dense subset in $\mathbb L^q(0,+\infty)$ for any $1\lt q\lt +\infty$ is equivalent to prove that the collection linear combinations of indicator functions of bounded sets in dense in $\mathbb L^q$. This is due to the fact that sets of finite Lebesgue measure can be approximated with respect to Lebesgue measure by open sets which can be written themselves as a countable disjoint union of open intervals.
The fact that the collection linear combinations of indicator functions of bounded sets in dense in $\mathbb L^q$ follows from the definition of Lebesgue integral and the fact that if $A\subset (0,+\infty)$ has a finite measure and $\varepsilon \gt 0$, we can find a bounded set $A_\varepsilon $ such that $\left|\lambda(A)-\lambda\left(A_\varepsilon\right)\right|\lt \varepsilon$.