## § It suffices to check for weak convergence on a spanning set.

- Theorem: suppose $x[i]$ is a bounded sequence in $X$. Then, to check that $x[i] \to_w L$, it suffices to check on a spanning set $A \subseteq X$ such that $closure(span(A)) = X$.
- Proof: first, it easily suffices for linear combinations by triangle inequality.
- Next, to show it suffices for closures, we wish to show that $h(x[n]) \to h(L)$ given that $g(x[n]) \to g(x)$for all $g \in span(A)$.
- Let $h = \lim_j g[j]$ for some $g[j] \in X^\star$.
- Let us bound $|h(x[n]) - h(L)|$.
- This is equal to $|h(x[n]) - g[j](x[n]) + g[j](x[n]) + g[j](L) - g[j](L) - h(L)$
- Rearranging: $|(h(x[n]) - g[j](x[n])) + (g[j](x[n]) - g[j](L)) + (g[j](L) - h(L))|$.
- We bound each pair: $|h(x[n]) - g[j](x[n])|$ can be made arbitrary because $g[j] \to h$, and thus they are bounded pointwise since these are bounded linear functionals.
- $|g[j](x[n]) - g[j](L)$ can be made arbitrarily small because we know that $x[n] \to_w L$ on the set $A$.
- The third term $|g[j](L) - h(L))|$ can be made arbitrarily small because $g[j] \to h$ and these are bounded linear functionals.
- Thus we have shown that we can make stuff arbitrarily small, and we are done!