Evaluation Functional

Question

Let $V$ be a vector space over $\mathbb R$. The algebraic dual of $V$, denoted $V^*$, is the space of all linear functions from $V$ to $\mathbb R$. These functions are called functionals. Most textbooks list the following as examples for functionals:

• The integration operator $f\mapsto \int f$ is a functional.
• The evaluation functional (at the query point $x$) $f\mapsto f(x)$

Obviously, both examples are only meaningful in the case $V$ is a function space of real-valued functions, e.g. $L^p(X,\mathbb R)$, or $C^p(X,\mathbb R)$.

Now, suppose we are looking at spaces of functions which map the domain $X$ to, say, $\mathbb R^k$, or even more general to an arbitrary (Banach) spaces $B$. The integral operator remains a functional as integration, by definition, yields eventually a real number. However, the evaluation functional is no longer a functional as the function in place no longer takes values in $\mathbb R$, right? So I was wondering whether an evaluation "functional" is also defined for such cases? Or do I get something fundamentally wrong?

Answer 1

Evaluation operator $\Phi_x: f \mapsto f(x)$ defined on the space $V$ of functions from $X$ to $\mathbb{R}^k$ for $k > 1$ is not a functional for the reason you stated: its codomain is not the scalar field.

However, you can compose the evaluation operator with a linear functional on $\mathbb{R}^k$ to obtain a linear functional on $V$. For an example, consider a basis $e_i$ of $\mathbb{R}^k$ and the corresponding dual basis $\xi_i$ of $\mathbb{R}^{k*}$. The composition $\xi_i \circ \Phi_x$ is a linear functional on $V$ and hence a member of the dual space $V^*$.

Answer 2

The word "functional" is just short for "linear function which codomain is the scalar field". So no, the evaluation is not a functional if the codomain is a vector space different from the field. In that case is just called "linear function".