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?

  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Dec 19, 2020 at 21:07
  • $\begingroup$ fixed it; I at first wanted to refer to $W$ what I later called $B$. sorry for the confusion $\endgroup$
    – Syd
    Commented Dec 19, 2020 at 21:10
  • $\begingroup$ 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. $\endgroup$
    – Ruy
    Commented Dec 19, 2020 at 22:12

2 Answers 2


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 $V^*$.

  • $\begingroup$ is there any chance to do the same for general function spaces $B$? $\endgroup$
    – Syd
    Commented Dec 19, 2020 at 21:11
  • $\begingroup$ Yes. Simply replace $\mathbb{R}^k$ with whatever the functions in $B$ map to and $\mathbb{R}^{k*}$ with the corresponding dual space. $\endgroup$ Commented Dec 19, 2020 at 21:13

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".


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