Weak convergence implies weak* convergence? In lecture my professor said that "weak convergence implies weak* convergence" but gave no explanation or proof, and ended class there.  I'm trying to make sense of this statement but can't even see what is really being claimed here.
So first of all, what is the assumption of the claim?  Clearly we have to be talking about a normed linear space $(X,\|\cdot\|)$.  Are we assuming that there exists a weakly convergent sequence $\{f_n\}\rightharpoonup f$ in $X$?  I feel like this must be the assumption, since any other quantifier seems silly.
Now what is the conclusion of the claim?  That there exists a weak* convergent sequence?  Or that $f_n$ determines a particular sequence $\varphi_n\in X^*$, and $f$ determines $\varphi$, such that $\varphi_n\overset w \to \varphi$ in $X^*$?  I know that we can define a natural element in $X^{**}$ corresponding to any $g\in X$ by the definition $\psi_{g}(\phi') = \phi'(g)$, and this has the property that $\|\psi_{g}\|_{**}=\|g\|$ but it's not clear to me how this would determine some particular sequence of $\{\varphi_n\}\subseteq X^*$.
 A: Your professor was just saying that if you have a sequence $(\varphi_n)$ in $X^*$ which converges weakly, then $(\varphi_n)$ also converges weak* (and both to the same limit). That is, weak convergence is stronger than weak* convergence.
Suppose that $\varphi_n\rightarrow\varphi$ weakly. Let $\Phi:X\rightarrow X^{**}$ be the identification map of $X$ and its double dual $X^{**},$ i.e. $\Phi(x)(\psi)=\psi(x).$ For any $x\in X,$
$$\varphi_n(x)=\Phi(x)(\varphi_n)\rightarrow \Phi(x)(\varphi)=\varphi(x),$$ using that $\varphi_n\rightarrow\varphi$ weakly. This shows that $\varphi_n\rightarrow\varphi$ weak*, as well.
A: *

*Coarser topologies have "more" convergent sequences.
A net (or a sequence) $(x_i)_{i \in I}$ is said to converge to a point $x$ if for every open set $U$ containing $x$ the net $(x_i)_{i \in I}$ is eventually in $U$, i.e. there exist an index $i_0$ such that $x_j \in U$ for all $j \geq i_0$. In other words, the "more" open sets you have in your topology, the "more difficult" you make it for a net to converge. In particular we have:

If we have an inclusion of topologies, $\tau' \subseteq \tau$, then for any net $(x_i)_{i \in I}$ we have
$$(x_i)_{i \in I} \text{ converges in } \tau \Rightarrow (x_i)_{i \in I} \text{ converges in } \tau' $$

As extreme examples you see that everything converges in the indiscrete topology, while only eventually constant nets converge in the discrete topology.


*Initial topologies based on "less" functions generate coarser topologies
The open sets of an initial topology generated by the functions $f_i: X \rightarrow Y_i$ for $i \in I$ are precisely the sets generated by the sets of the form $f_i^{-1}(U)$ for $U$ open in $Y_i$, i.e. you can express it as an arbitrary union of finite intersections of "cylinders" $f_i^{-1}(U)$. In other words, the "more" functions you want them to be continuous, the "more open sets" you can generate in your initial topology. In particular we have:

The initial topology of a subfamily of functions generates a coarser topology
$$I' \subseteq I \Rightarrow \big(X, (f_i)_{i \in I'}\big) \subseteq \big(X, (f_i)_{i \in I}\big)$$



*Weak convergence implies weak$*$ convergence
The weak topology of a topological vector space is defined to be the initial topology generated by the (continuous) functionals $X^*$. In particular, the weak topology on $X^*$ is the one generated by $X^{**}$.
The weak$*$ topology is defined to be the initial topology (on $X^*$) generated by the image of the canonical evaluation map $\Phi: X \rightarrow X^{**}$. The claim follows from $\Phi(X) \subseteq X^{**}$ and from point 1. and 2.

Weak convergence implies weak$*$ convergence

