Is it possible to extend this functional to a linear functional?

Question

Let $X$ be an arbitrary set, and let $\mathcal{F} = [0,1]^X$. Let $\phi:\mathcal{F} \to [0,1]$ have the property that $$\phi(q_1f_1 +...+q_nf_n) = q_1\phi(f_1)+...+q_n\phi(f_n)$$ whenever $n \in \mathbb{N}$, $q_1,...,q_n \in \mathbb{Q} \cap [0,1]$ and $f_1,...,f_n,q_1f_1 +...+ q_nf_n \in \mathcal{F}$.

Does there exist a linear functional $\phi'$ on the space $B(X)$ of bounded functions on $X$ such that $\phi'(f) = \phi(f)$ for all $f \in \mathcal{F}$?

I haven't made much progress with this because (a) $\mathcal{F}$ is not a linear subspace of $B(X)$ to begin with, so it doesn't seem like standard extension results (Hahn-Banach) or minor variations of them will apply, and (b) $\phi$ is only "linear" for rational scalars, so, without a continuity assumption, again, it seems like standard results will not apply. It seems likely to me that there are off-the-shelf results that apply to this problem, but I haven't found anything in the books that I own, so any references would be appreciated.

Answer 1

Lemma. (1) Let $f \in [0,1]^X$. Then as $r\to 0^+$, we have $\phi(rf)\to 0$.

(2) $\phi(rf) = r\phi(f)$ for all $r\in[0,1]$ and $f\in[0,1]^X$.

(3) The condition is true if we replace rational $q_1,\cdots,q_n$ by reals.

Proof. (1) Write $n=\lfloor 1/r\rfloor$. Then $nrf \in [0,1]^X$ and hence

$$|\phi(rf)| = \left| \frac{1}{n} \phi(nrf)\right| \leq \frac{1}{n} \xrightarrow[r\to0^+]{} 0.$$

(2) Pick $q_n\in[0,r]\cap\mathbb{Q}$ so that $q_n \uparrow r$. Then $\phi(rf)-r\phi(f) = \phi((r-q_n)f) + (q_n-r)\phi(f) \to 0$ as $n\to\infty$.

(3) Assume that $r_1,\cdots,r_n \in [0,1]$ and $f_1,\cdots,f_n\in[0,1]^X$ satisfy $r_1f_1+\cdots+r_nf_n\in [0,1]^X$. Then

$$ \phi(r_1f_1+\cdots+r_nf_n) = \phi(r_1f_1)+\cdots+\phi(r_nf_n) = r_1\phi(f_1)+\cdots+r_n\phi(f_n).$$

The first equality follows by applying the condition for $q_1=\cdots=q_n=1$ and $r_1f_1,\cdots,r_nf_n\in[0,1]^X$, and the second equality follows from (2). ////

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For each $f \in B(X)$, introduce $f_+$ and $f_-$ in $[0,\infty)^X$ which are defined by

$$ f_+(x) = \max\{f(x), 0\}, \qquad f_-(x) = \max\{-f(x), 0\}. $$

Since $f$ is bounded, we may pick a bound $M > 0$ of $f$. Then $f_{\pm}/M \in [0,1]^X$ and we may define $\phi'(f)$ by

$$ \phi'(f) = M \phi\left(\frac{f_+}{M}\right) - M \phi\left(\frac{f_-}{M}\right). $$

We need several things to check:

1. Well-definedness. The definition invokes an arbitrary bound $M$ of $f$. So we have to check that this does not depend on the choice of $M$. Indeed, let $0 < M_1 \leq M_2$ be bounds of $f$. Then $r = M_1/M_2 \in [0, 1]$ and thus

   \begin{align*} M_2 \phi\left(\frac{f_+}{M_2}\right) - M_2 \phi\left(\frac{f_-}{M_2}\right) &= M_2 \phi\left(r\frac{f_+}{M_1}\right) - M_2 \phi\left(r\frac{f_-}{M_1}\right) \\ &= M_1 \phi\left(\frac{f_+}{M_1}\right) - M_1 \phi\left(\frac{f_-}{M_1}\right) \end{align*}

   proving that the $\phi'(f)$ does not depend on the choice of $M$.

2. Linearity. We first check that if $f = f_1 - f_2$ for some $f_1, f_2 \in B(X) \cap [0,\infty)^X$, then we have $\phi'(f) = \phi'(f_1) - \phi'(f_2)$. Indeed, let $M > 0$ be sufficiently large. Then

   \begin{align*} \phi'(f) + \phi'(f_2) &= M \phi(f_+/M) - M \phi(f_-/M) + M \phi(f_2/M) \\ &= M \phi((f_+ + f_2)/M) - M\phi(f_-/M) \\ &= M \phi((f_- + f_1)/M) - M\phi(f_-/M) \\ &= M \phi(f_1/M) \\ &= \phi'(f_1) \end{align*}

   and rearranging the equality proves the claim. Now for any $f, g \in B(X)$ and for a sufficiently large $M > 0$,

   \begin{align*} \phi'(f+g) &= \phi'((f_+ + g_+) - (f_- + g_-)) \\ &= \phi'(f_+ + g_+) - \phi'(f_- + g_-) \\ &= M \phi((f_+ + g_+)/M) - M \phi((f_- + g_-)/M) \\ &= M \phi(f_+/M) + M \phi(g_+/M) - M \phi(f_-/M) - M \phi(g_-/M) \\ &= \phi'(f) + \phi'(g). \end{align*}

   Moreover, for any $f \in B(X)$ and $\alpha > 0$, choose $M > 0$ sufficiently large and write $M' = M/\alpha$. Then using the fact that $(\alpha f)_+ = \alpha (f_+)$ and $(\alpha f)_- = \alpha (f_-)$, we have

   \begin{align*} \phi'(\alpha f) &= M \phi( \alpha f_+ / M) - M \phi( \alpha f_- / M) \\ &= \alpha ( M' \phi(f_+/M') - M'\phi(f_-/M') ) \\ &= \alpha \phi'(f) \end{align*}

   Finally, $\phi'(0) = 0$ is trivial and

   \begin{align*} \phi'(-f) = M \phi( f_- / M) - M \phi( f_+ / M) = -\phi'(f) \end{align*}

   Therefore $\phi'$ is linear.

3. Boundedness. For each $f\in B(X)$, let us write the supremum norm of $f$ by

   $$\|f\|=\sup\{|f(x)|:x\in X\}.$$

   Then $\|f\|$ is a bound of $f$. So if $f \neq 0$,

   $$|\phi'(f)| = \|f\| \cdot |\phi(f_+/\|f\|)-\phi(f_-/\|f\|)| \leq \|f\|. $$

   Therefore $\phi'$ is bounded.