# Convex analysis: possible misunderstanding about dual space

From Wikipedia:https://en.wikipedia.org/wiki/Convex_conjugate

Let $X$ be a real topological vector space, and let $X^{*}$ be the dual space to $X$. Denote the dual pairing by $\langle \cdot ,\cdot \rangle :X^{*}\times X\to \mathbb {R} .$ For a functional $f:X\to \mathbb {R} \cup \{+\infty \}$ taking values on the extended real number line, the convex conjugate $f^{\star }:X^{*}\to \mathbb {R} \cup \{+\infty \}$ is defined in terms of the supremum by $f^{\star }\left(x^{*}\right):=\sup \left\{\left.\left\langle x^{*},x\right\rangle -f\left(x\right)\right|x\in X\right\}}$

I am still bothered by the idea of a "dual space", because from the point of view of convex analysis, it seems that the idea of a dual space is tied in with the idea of a convex conjugate, whereas the dual space seems to be a much more general concept (i.e. the space of linear functional). For instance, we know that $\mathbb{R}^n$ is a Hilbert space, hence $(\mathbb{R}^n, \|\cdot\|_2)$ is the dual of itself. However, convex analysis is giving me some un-intuitive results.

For instance, in example 5.1 of http://www.doc.ic.ac.uk/~ahanda/lfreport.pdf

We are given $f(y) = \|y\|$, presumably the $2$-norm and the function is defined over all of $\mathbb{R}^n$, we find the conjugate to be $f^*(z) = 0, \|z\|\leq 1$.

Now if you had told me that the dual space to $X = (\mathbb{R}^n, \|\cdot\|_2)$ is the set $X^* = \{z\in \mathbb{R}^n | \|z\|_2\leq 1\}$, I think I will have a difficulty understanding you, but isn't this what the example above is showing?

• Are you sure that you are not confusing the terms "dual space" and "convex conjugate"? They are completely different. The superscript star on a convex function $f$ does NOT mean that $f^\star$ is a dual vector. Instead, it is a function defined on the dual space (or on a subset of it). – Giuseppe Negro Mar 22 '17 at 17:38
• @GiuseppeNegro In wikipedia's definition, it says $X^*$ is the dual space of $X$. I am confused as to why those $X^*$ in the examples are the dual spaces of $X$. – Carlos - the Mongoose - Danger Mar 22 '17 at 17:40
• $X^*$ is the dual space of $X$. In your examples, $f^\star$ is given by a certain formula on a certain subset of $X^*$, and is $+\infty$ on the complement of that subset. – Robert Israel Mar 22 '17 at 17:43
$$f^*(z) = \cases{0 & if \|z\| \le 1\cr +\infty & otherwise}$$
That doesn't mean $\{z: \|z\| \le 1\} = X^*$. It is just the subset of $X^*$ on which this particular conjugate functional is finite.