Let $X$ be a compact space. Let $\mathcal{M}(X)$ be the space of all probability measures on $X$. Denote by $C(X)$ and $C(\mathcal{M}(X))$ the real continuous function on $X$ and $\mathcal{M}(X)$ respectively. Naturally, $C(X)$ is a subspace of $C(\mathcal{M}(X))$ since for $f\in C(X)$, the function $\mu\mapsto \int fd\mu$ is continuous.
Question: is $C(X)$ is dense in $C(\mathcal{M}(X))$?
Thanks.
As suggested by Jochen Glueck, let us consider a two point space, $X = \{0,1\}$. Then $\mathcal{M}(X) = \{(1-\lambda)\delta_0 + \lambda \delta_1\mid \lambda \in [0,1]\}$ where $\delta_x$ denotes the Dirac mass concentrated at $x$. Now, consider the functional $$ F: \mathcal{M}(X) \longrightarrow \mathbb{R},\quad F((1-\lambda)\delta_0 + \lambda \delta_1) = \lambda^2. $$ Assume that $(f_n)\subseteq \mathcal{C}(X)$ is a sequence which converges to $F$, i.e. $$ \sup_{\mu\in \mathcal{M}(X)}\left|\int_X f_nd\mu - F(\mu)\right| = \sup_{\lambda \in [0,1]} \left|(1-\lambda)f_n(0) + \lambda f_n(1) - \lambda^2\right| \rightarrow 0. $$ Then, $\left|(1-\lambda)f_n(0) + \lambda f_n(1) - \lambda^2\right| \rightarrow 0$ for each fixed $\lambda \in [0,1]$. By taking $\lambda = 0$ we see that $f_n(0) \rightarrow 0$ and by taking $\lambda = 1$ we see that $f_n(1)\rightarrow 1$. But then we get that $\left|(1-\lambda)f_n(0) + \lambda f_n(1) - \lambda^2\right| \rightarrow 1/4$ for $\lambda = 1/2$ which is a contradiction. Hence the answer is no!
