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DualSpace

Stephen Crowley edited this page Jun 29, 2023 · 4 revisions

Dual Space

In linear algebra and functional analysis, given a vector space $V$ over a field $F$ (which is typically the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$), the dual space of $V$, denoted by $V^{\ast}$, is defined as the set of all linear functionals on $V$.

A linear functional is a linear map from $V$ to the field $F$. In other words, a linear functional is a function $f: V \rightarrow F$ that satisfies the following properties:

  • Additivity: For all $x, y$ in $V$, we have $f(x + y) = f(x) + f(y)$
  • Homogeneity: For all $x$ in $V$ and all $\alpha$ in $F$, we have $f(\alpha x) = \alpha f(x)$

If $V$ is a finite-dimensional vector space, then $V$ and $V^{\ast}$ are isomorphic (they have the same structure), although there is no natural isomorphism between them: an isomorphism depends on a choice of basis in $V$.

Now, when $V$ is an infinite-dimensional topological vector space (such as a normed space or a Hilbert space), the situation is a bit more complicated. In this case, the continuity of linear functionals becomes an important issue. The dual space $V^{\ast}$ usually refers to the set of continuous linear functionals on $V$.

In particular, if $V$ is a normed space, we can define the norm of a linear functional $f$ in $V^{\ast}$ as follows:

$$|f| = \sup { |f(x)| : x \in V, |x| \leq 1}$$

With this norm, $V^{\ast}$ becomes a normed space itself. Moreover, if $V$ is complete (i.e., if $V$ is a Banach space), then $V^{\ast}$ is also complete. This implies that $V^{\ast}$ is a Banach space too. This space $V^{\ast}$ is often called the continuous dual space or topological dual space of $V$.

The concept of a dual space is a fundamental one in functional analysis and it leads to many important results, such as the Hahn-Banach theorem, the Riesz representation theorem, and the Banach-Alaoglu theorem.

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