Counter example for Riesz representation theorem when $p=\infty$.

Question

The Riesz representation theorem holds for $1\leq p\leq\infty$. I wonder whether there is a counterexample.

Here is my attempt:

Consider $C_{\mathbb{R}}([-1,1])\subset L^{\infty}([-1,1])$. Define a bounded linear functional $\phi:C_{\mathbb{R}}([-1,1])\rightarrow \mathbb{R}$. By Hahn-Banach theorem, it can be extended to $L^{\infty}([-1,1])$. Define $\phi(f)=f(0)$.

Since $\{f:f(0)=0\}$ is dense in $C_{\mathbb{R}}$, $\phi=0$. However, $f(0)\neq 0$ in general.

Please comment on my attempt and, if possible, provide me with a better counterexample. Thanks!

Answer 1

$\{f: f(0) = 0\}$ is not dense in $C_{\mathbb R}([-1,1])$, if you're talking about the norm topology. In fact, any $f$ with $f(0) \ne 0$ has distance $|f(0)|$ from that set.

However, it is true that this set is dense in the topology generated by the linear functionals corresponding to $L^1$. That is, for any $\phi_1, \ldots, \phi_n \in L^1$ and $\epsilon > 0$, and $f \in C_{\mathbb R}$, there is $g \in C_{\mathbb R}$ with $g(0) = 0$ and $\left|\int \phi_i(x) (f(x) - g(x))\; dx\right| < \epsilon$ for $i=1 \ldots n$.