= Solution
Let $\varphi$ be a character of $C(\mathbb R)$. Put $q(t)=1/(t^2+1)$. Since
$$
(t^2+1)q=1,
$$
we have $(\varphi(t)^2+1)\varphi(q)=1$, so the restriction of $\varphi$ to $C_0(\mathbb R)$ is nonzero. Part e gives a point $x$ such that this restriction is $\delta_x$. For $g(t)=t/(t^2+1)\in C_0(\mathbb R)$, multiplicativity gives
$$
\varphi(t)\frac1{x^2+1}
=\varphi(t)\varphi(q)
=\varphi(g)
=\frac{x}{x^2+1},
$$
and hence $\varphi(t)=x$.
If $h\geq0$ is continuous, then
$$
k(t)=\frac1{h(t)+t^2+1}
$$
belongs to $C_0(\mathbb R)$. Applying $\varphi$ to $k(h+t^2+1)=1$ yields
$$
\frac{\varphi(h)+x^2+1}{h(x)+x^2+1}=1,
$$
so $\varphi(h)=h(x)$. Every complex-valued continuous function is a complex linear combination of nonnegative continuous functions, obtained from the positive and negative parts of its real and imaginary components. Therefore $\varphi=\delta_x$, and
$$
\Phi_{C(\mathbb R)}=\{\delta_x:x\in\mathbb R\}.
$$
Solved by gpt-5.6-sol high.
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