Let $E= C([0,1])$ and $E^* $ it's dual, the relative Borel measures on [0,1], and $E_+^* $ it's positive cone (the positive measures). The constraint is the w* closed convex subset C of $E^* $ obtained as intersection of $E^*_+ $ with the w* closed affine subspace of E

{ $m\in E^*: \langle m,f_k \rangle=1\; \forall k=1\dots n $ },$ \qquad $

It's not completely clear to me under what conditions on the $f_1,\dots ,f_n$ the convex C is not empty (e.g if $f_3=f_1+f_2$ the constraint is empty). I will assume therefore that (1) C is not empty.

Clearly, a necessary condition for the existence of the minimizer is also

$\mathrm{supp}(f_0) \subset \mathrm{supp}(f_1)\cup\dots \cup \mathrm{supp}(f_n).\qquad(2)$

Otherwise the functional to maximize is unbounded from above on the constraint C since e.g. C contains a whole half-line $t \delta_x +\mu$, with $t\geq0, $ $\ \mu\in C $, $f_0(x)>0$ and

$ x\notin \mathrm{supp}(f_1)\cup\dots \cup \mathrm{supp}(f_n). $

Assuming both necessity condition (1) and (2) (say wlog $f_0>0$ everywhere) C is non-empty, bounded, in fact w* compact by the Banach -Alaoglu theorem, and the functional to be maximized $m\mapsto \langle m,f_0 \rangle$ is linear and w* continuous (indeed it's the evaluation at $f_0$). So by compactness it attains a maximum. Moreover, *any* maximum point $\mu$ is attained at an extremal point of C.

The only non standard part is to recognize that in fact all extremal points of C are positive linear combinations of at most *n* measures $\delta_x.$ Indeed, if $\mu$ is an extremal point of C then for any $k=1,\dots ,n$ the restirction of $\mu$ to

$\mathrm{supp}(f_k) \setminus \bigcup_{j\neq k} \mathrm{supp}(f_j)$

is either zero or atomic, which implies $\mathrm{card}( \mathrm{supp}(\mu))\le n. $

**Rmk**: as a consequence the set C is not empty if and only if it containe a positive linear combination of n deltas. The case of probability measures, that apparently was not required by the initial question, is covered adding as n+1 th function the constant 1 (this authomatically satisfies (2)). So incidentally this is the proof of the above quoted theorem (with continuous $f_k$; the case of Borel measurable shouldn't be different).