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 $\sigma(X,Y)$)?


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.

  • $\begingroup$ Yes, I did. Thanks. $\endgroup$ Oct 21 '15 at 20:59
  • $\begingroup$ So in general, there can be any relationships between $\iota(Y)$and $X'$? $\endgroup$
    – user9464
    Oct 21 '15 at 21:08
  • $\begingroup$ Pretty much. Given $(X,\tau)$, we can take any subspace of $X^{\ast}$ that separates points on $X$ as $Y$, with the natural pairing $\langle x,\lambda\rangle = \lambda(x)$. If $X$ is finite-dimensional, then the only possibility is $Y = X^{\ast} = X'$, and if $\tau$ is fine enough we have $X' = X^{\ast}$, so then we always have $\iota(Y)\subset X'$. But for an infinite-dimensional $X$ with $X' \neq X^{\ast}$, we can typically have both, $\iota(Y) \subsetneq X'$ and $\iota(Y) \not\subset X'$. Often (if not always, I don't know off hand), we can then even have $\iota(Y) \cap X' = \{0\}$. $\endgroup$ Oct 21 '15 at 21:16

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