I was given this exercise in class today:
Using Yosida's Lemma, proof that there exists a Riesz-homomorphism $\varphi: BC(\mathbb{R}) \to \mathbb{R}$ such that $\varphi(\textbf{1}) = 1$ and $\varphi(f) = 0$ where $f(x) = e^{-|x|}$.
Now, I looked up Yosida's Lemma:
Let $E$ be an Archimedian Riesz space with a unit $u$. Let $a \in E^+$. Then there exists a Riesz-homomorphism $\varphi: E \to \mathbb{R}$ such that $\varphi(u) = 1$ and $\varphi(a) = ||a||_u$.
Therefore, take $E = BC(\mathbb{R})$ then we get the unit $\textbf{1}$ and of course $f \in BC(\mathbb{R})^+$ because $f(x) \geq 0$ for all $x$. We find $\varphi$ with $\varphi(\textbf{1}) = 1$ and $\varphi(f) = ||f||_{\textbf{1}}$. Nothing weird up to this point, but:
$$ ||f||_{\textbf{1}} := \inf \{ \lambda \in [0, \infty) : |f| \leq \lambda \textbf{1} \}$$
This was the definition, and $|f| := f \vee (-f)$, the absolute value of $f$. But $f(0) = 1$, therefore I assume $||f||_{\textbf{1}} \geq 1$? But then $\varphi(f) \geq 1 \neq 0$?
I have found what I did wrong. Of course, we can also apply Yosida's Lemma to another function $\textbf{1} - f$, which is also in $BC(\mathbb{R})^+$. Then we get $||\textbf{1} - f||_{\textbf{1}} = 1$ and therefore $\varphi(f) = \varphi(\textbf{1}) - \varphi(\textbf{1} - f) = 1 - 1 = 0$