Some questions about an exercise about $C^\infty \subset L^\infty$ Let 
$$ L^\infty (\mathbb R) = \{f : \mathbb R \to \mathbb C\mid \text{essential sup of } f < \infty \text{ and } f \text{ Borel measurable} \}$$
and
$$ C^\infty (\mathbb R ) = \{ f: \mathbb R \to \mathbb C \mid f \text{ continuous }, \lim_{|x|\to \infty}f(x) = 0 \}$$
I was going to solve the following exercise but then realised there were a few things that were not clear to me. Here is the exercise:
Prove that there exists an element $\lambda \in (L^\infty (\mathbb R))^\ast$ such that $\lambda (f) = f(0) $ for all $f \in C^\infty(\mathbb R)$.
Hint: Use consequences of the Hahn-Banach theorem.
Here are my thoughts and questions:
(1) It's clear to me that $C^\infty$ is a subspace of the space of bounded functions $B(\mathbb R)$ but here in this exercise $L^\infty$ is a space consisting of elements that are equivalence classes. 

Is this not a problem? Does $C^\infty$ embed into $L^\infty$? (I'm not
  sure embed is the correct word...)

(2) Say we use representatives of $L^\infty$ so that (1) is not a problem. Then to me it seems that the evaluation map $f \mapsto f(0)$ is an element of $(L^\infty)^\ast$. And this would answer the question but the hint suggests I am missing something. 

What am I missing?

 A: Let $\mathscr C^{\infty}$ be the vector space of almost-everywhere equivalence classes of elements of $C^{\infty}$. That is, $$\mathscr C^{\infty}\equiv\{[f]\,|\,f\in C^{\infty}\},$$ where, for each $f\in C^{\infty}$, $$[f]\equiv\left\{g:\mathbb R\to\mathbb C\,\big|\,g\text{ is Borel measurable and }g\overset{\text{a.e.}}{=}f\right\}.$$ Since almost-everywhere equivalence does not change the essential supremum norm, one has, for any $f\in C^{\infty}$, that $\|[f]\|_{\infty}=\|f\|_{\mathsf u}$, where $\|f\|_{\mathsf u}$ is the usual uniform norm (as you already observed, every $f\in C^{\infty}$ is bounded, so this makes sense). This is because the essential supremum norm of a bounded continuous function is actually equal to the uniform norm. In short, we just proved that $\mathscr C^{\infty}\subseteq L^{\infty}$, where $L^{\infty}$ is the space of equivalence classes of Borel-measurable functions with finite essential suprema.
Define the map $\mu: {\mathscr C}^{\infty}\to\mathbb R$ as $\mu([f])\equiv f(0)$ for each $f\in C^{\infty}$. This is well-defined, since any equivalence class in ${\mathscr C}^{\infty}$ contains exactly one continuous function (it contains at least one by definition, and it cannot contain two, because two continuous functions that are equal almost everywhere are, in fact, equal everywhere). Also, one has that $$|\mu([f])|=|f(0)|\leq\|f\|_{\mathsf u}=\|[f]\|_{\infty},$$ so that $\mu$ is a bounded linear functional on the subspace $\mathscr C^{\infty}$ of $L^{\infty}$. Now you can use the Hahn–Banach theorem to obtain a bounded linear functional $\lambda:L^{\infty}\to\mathbb R$ such that $\lambda([f])=f(0)$ for any $f\in C^{\infty}$; or, less rigorously speaking, “$\lambda(f)=f(0)$ for any $f\in C^{\infty}$.”
This less rigorous statement is formally incorrect (which is probably what caused your confusion), but, at the end of the day, it makes intuitive sense: After all, $\lambda$ is such a functional that picks the (unique) element that is in $ C^{\infty}$ from every equivalence class that contains such an element, and assigns the value of that function at $0$ to the equivalence class.
