weak topology and dual pairs I'm reading a note on functional analysis and the following statement is given without a proof: 

Let $(X,Y,\langle,\rangle)$ be a dual pair and $\tau$ a topology on $X$. Let $\sigma(X,Y)$ be the weak topology on $X$, which is defined as the weakest topology on $X$ such that the linear maps $x\mapsto\langle x,y\rangle$ are continuous for all $y\in Y$. If the continuous dual of $(X,\tau)$ is $Y$ then trivially $\tau$ is stronger than $\sigma(X,Y)$. 

Here is my question: 
In this statement, in what sense can $Y$ be thought as the continuous dual of $(X,\tau)$ (and why does it trivially imply that $\tau$ is stronger than $\sigma(X,Y)$)? 
 A: Let us denote the algebraic dual of a vector space $V$ with $V^{\ast}$, and the topological dual (or continuous dual) of a topological vector space $W$ with $W'$. Clearly, we have $W' \subset W^{\ast}$, since a continuous linear functional is in particular a linear functional.
The dual pairing $\langle\,\cdot\,,\,\cdot\,\rangle \colon X\times Y \to \mathbb{C}$ (or $\mathbb{R}$) gives an injection $\iota \colon Y \hookrightarrow X^{\ast}$ by $\iota(y) \colon x \mapsto \langle x,y\rangle$.
If we have a vector space topology $\tau$ on $X$, then it may be the case that $\iota(Y) = X'$. That is what is meant by "the continuous dual of $(X,\tau)$ is $Y$", that the image of the injection $\iota$ given by the dual pairing $\langle\,\cdot\,,\,\cdot\,\rangle$ is just the space of continuous linear functionals on $X$ (with respect to the topology $V$).
If we have "$Y = X'$" in that sense, then the given topology $\tau$ is automatically stronger than $\sigma(X,Y)$ because the latter is by definition the weakest topology on $X$ which makes all the $\iota(y)$ for $y\in Y$ continuous. It is perhaps not obvious, but a general topological fact that such a weakest topology exists (it's the initial topology with respect to the maps $\iota(y)$).
Note that "stronger" and "weaker" do not exclude equality, so it is possible that $\tau = \sigma(X,Y)$ in that case.
