Let $X$ be a vector space and $Y$ a space of linear functionals on $X$. Reed and Simon give the following definition:
The Mackey topology on $X$ is the topology of uniform convergence on weak* compact convex sets of $Y$; that is, $x_\alpha \rightarrow x$ if and only if $y(x_\alpha) \rightarrow y(x)$ uniformly as $y$ runs through any fixed weak* compact subsets of $Y$.
I am puzzled by their use of the word uniformly and what is meant by it.
The topology of pointwise convergence is defined as $x_n \rightarrow x$ if and only if $f(x_n) \rightarrow f(x)$ for all $f \in X^*$, so I assume the "uniform" part of their convergence definition means more than just $f(x_\alpha) \rightarrow f(x)$ for all $f \in Y$.
The usual definition of uniform convergence on metric spaces is $f_n \rightarrow f$ on a set $E$ if and only if $$\sup_{x \in E} |f_n(x) - f(x)| \rightarrow 0$$ as $n \rightarrow \infty$. But I do not see how this applies to our definition as it does not preserve any kind of order that we have in $\{x_\alpha\}$.
What is meant by uniform convergence here?
[Uniform convergence makes sense for a "uniform space" of course. This includes even topological groups. But in this case, the uniform space is $\mathbb R$.]
Let $E$ be a subset of $Y$. Let $(f_\alpha)_{\alpha \in D}$ be a net of functions, $f_\alpha : E \to \mathbb R$. Then we say $(f_\alpha)$, converges uniformly on $E$ to $f : E \to \mathbb R$ iff: for every $\varepsilon > 0$ there exists $\alpha_0$ so that for all $\alpha > \alpha_0$ and all $y \in E$ we have $|f_\alpha(y)-f(y)| \le \varepsilon$.
In the quote, elements of $X$ are considered to be linear functionals on $Y$. That is, maps from $Y$ to $\mathbb R$. We cannot see from your quote: maybe scalars $\mathbb C$ are used? Same definition.
NOTE: if our spaces are not metrizable, then in general it is not enough to use only sequences, but we must use more general nets to describe the topology of uniform convergence on given family of sets.