The duality map is defined, according to Brezis (Functional Analysis, Sobolev Spaces and Partial Differential Equations) as follows, where $E$ is a Banach space and $x \in E$:

$$F(x) = \{f \in E^* \ : \ ||f|| = ||x|| \text{ and } \langle f, x \rangle = ||x||^2\}$$

It is one of the exercises to prove that

$$ F(x) = \{f \in E^* \ : \ ||f|| \leq ||x|| \text{ and } \langle f, x \rangle = ||x||^2\} = \{f \in E^* \ : \ \frac12 ||y||^2 - \frac12 ||x||^2 \geq \langle f, y - x \rangle \ \forall y \in E\}.$$

My question is:

Where does the duality map comes from? That, is, what motivates the definition, from a historical point of view? Why is it useful?

Thanks in advance.

  • 4
    $\begingroup$ I haven't seen this construction before, but for Hilbert spaces this "map" actually gives an anti-linear isomorphism with the dual. So depending on how big the set $F(x)$ is this thing could give you a way to see how far away from a Hilbert space a a given Banach space is. Particularly interesting could be cases where $F(x)$ is always a singleton set but the thus defined map $F: E\to E^*$ fails to be anti-linear or a bijective. $\endgroup$
    – s.harp
    Commented Apr 11, 2020 at 13:47

1 Answer 1


Suppose for a moment that $E$ is even a Hilbert-space and decompose a vector $y\in E$:

$$y=\alpha x+\beta z\text{ where }\langle z,x\rangle =0.$$ Then $y\in F(x)$ if and only if $$\langle y,x\rangle=\alpha\langle x,x\rangle=||x||^2\iff\alpha=1$$ for non-zero $x$, i.e $F(x)$ consists of all vectors that have as a "component in direction $x$" exactly $x$. For $\mathbb{R}^2$ it is easy to draw a picture and there are as well physical interpretations at hand. The definition in Brezi's book is a generalization to Banach spaces.


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