Convergence of functionals and weak convergence I consider a Banach space $V$ with its dual $V'$.
I had a sequence of functionals $\{f_k\}_{k\in \mathbb N} \subset V'$, and I wanted to show (strong or norm) convergence of $f_k \to f \in V'$.
I thought that $f=g \in V'$, iff $f(v)= g(v)$ for all $v\in V$.
So I showed that for any $v \in V$, I have $\lim_{k\to\infty}f_k(v)=f(v)$, to conclude that $f_k \to f$ in $V'$.
However, since this is the definition of weak-$*$ convergence of $f_k$, I was wondering: 
For sequences in a dual space, do I have to distinguish between weak and norm convergence?
If yes, what is wrong in my thoughts?
 A: Strong convergence in the dual $V'$ of a topological vector space $V$ means uniform convergence on all bounded subsets of $V$. For normed spaces $V$, the set $\{B\}$, where $B$ is the unit ball (open or closed, doesn't matter) in $V$, is a basis of the bounded sets, i.e. every bounded set in $V$ is contained in a multiple of $B$, and $B$ is bounded, so it is sufficient to check uniform convergence on $B$. Thus the strong topology on $V'$ is normable,
$$\lVert \lambda\rVert = \sup_{v\in B} \lvert \lambda(v)\rvert,$$
and that norm makes $V'$ a Banach space. So for strong - or norm - convergence of $f_k$ to $f$ in $V'$, we must check that for every $\varepsilon > 0$ there exists an index $k_\varepsilon \in \mathbb{N}$ such that
$$\bigl(\forall k \geqslant k_\varepsilon\bigr)\bigl(\forall v\in B\bigr)(\lvert f_k(v) - f(v)\rvert \leqslant \varepsilon),$$
or, in terms of the norm, $\bigl(\forall k \geqslant k_\varepsilon\bigr)(\lVert f_k -f \rVert \leqslant \varepsilon)$.
Weak$^\ast$ convergence is pointwise convergence, or uniform convergence on all finite subsets of $V$. When $\dim V = \infty$, that is strictly weaker than strong convergence.
