# Evaluation Functional

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?

• Why do you introduce $W$? Linear functionals must take values in the underlying field, so if you have a function which takes you somewhere other than the underlying field, evaluation won't take you to the field. Of course, it is fine to look at linear maps from $V$ into some other vector space.
– lulu
Commented Dec 19, 2020 at 21:07
• fixed it; I at first wanted to refer to $W$ what I later called $B$. sorry for the confusion
– Syd
Commented Dec 19, 2020 at 21:10
• There is no evaluation functional on $L^p(X,\mathbb R)$ because elements of this space are equivalence classes of functions rather than actual functions.
– Ruy
Commented Dec 19, 2020 at 22:12

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 $$V^*$$.
• is there any chance to do the same for general function spaces $B$?
• Yes. Simply replace $\mathbb{R}^k$ with whatever the functions in $B$ map to and $\mathbb{R}^{k*}$ with the corresponding dual space. Commented Dec 19, 2020 at 21:13