Let $X$ be a locally compact space, $M := C_0(X)'$ the space of finite signed Radon measures on $X$, equipped with the weak$^*$ topology and $M_+ \subseteq M$ the set of finite positive measures. Assume that $X$ is infinite. Then $M$ is not locally compact (since it is an infinite-dimensional locally convex space). Is $M_+$ locally compact?
Case 1: If $X$ is compact then $M_+$ is locally compact.
Proof: $f = 1_X \in C_0(X) = C(X)$. For the zero measure $0 \in M_+$ consider the set $V := \{ \mu \in M \mid \mu(X) \leq 1 \}$. Then $V$ is weakly$^*$ compact (because it is closed in the unit ball, which is weakly$^*$ compact) and $V = \{ \mu \in M \mid |\mu 1_X| \leq 1 \} \cap M_+$ is a weak$^*$ neighborhood of $0$, so that $V$ is a compact neighborhood of $0$. Well, $0 \in M_+$ is a special point (e.g. it is an extreme point of $M_+$), but a similar argument seems to apply to any $\nu \in M_+$. Set $r := \nu(X) + 1$. Then $r V$ is compact. The set $U_\nu := \{ \mu \in M \mid |(\mu - \nu) 1_X| \leq 1 \} \cap M_+$ is a neighborhood of $\nu$ in $M_+$ and $U_\nu \subseteq rV$: if $|(\mu - \nu)1_X| = |\mu(X) - \nu(X)| \leq 1$ then $\mu(X) \leq \nu(X) + 1$, so that $\mu \in rV$. Therefore, $rV$ is a compact neighborhood of $\nu$.
Case 2: If $X$ is locally compact and not compact then $M_+$ is not locally compact. Here it is shown that the set of probability measures is not locally compact. Therefore, $M_+$ cannot be locally compact either. However, if we replace $C_0(X)$ by $C_b(X)$ then $1_X \in C_b(X)$ and by copying the argument in case 1 we get local compactness (for a different weak$^*$ topology).
Is there any flaw in the arguments, I did oversee? Does anyone know a reference to the literature, where these and related problems are discussed?
Your argument in Case 1 is correct.
In Case 2, your argument is wrong because the set of probability measures is not weak* closed in $M_+$ (if $x_i\to \infty$ then $\delta_{x_i}\to 0$ in the weak* topology!). Here is a correct argument. Let $U$ be any basic open neighborhood of $0$ in $M_+$, which we can take to have the form $\{\mu\in M_+:\mu f\leq\epsilon\}$ for some $\epsilon>0$ and some nonnegative $f\in C_0(X)$ (note that we can consider just a single $f$ since given finitely many we can replace them with the sum of their absolute values). If there exists $x\in X$ such that $f(x)=0$, then $U$ contains $c\delta_x$ for all $c\geq 0$ and letting $c\to\infty$ we see that $U$ is not precompact. If no such $x$ exists, then there exists a sequence $(x_n)$ in $X$ such that $f(x_n)\to 0$, since $X$ is not compact and $f$ vanishes at $\infty$. For each $n$, then, the measure $\mu_n=\frac{\epsilon}{2f(x_n)}\delta_{x_n}$ is in $U$. However, the sequence $(\mu_n)$ has no accumulation point in $M_+$, since $\mu_n\sqrt{f}\to\infty$ as $n\to\infty$ and $\sqrt{f}\in C_0(X)$.