 Solution

Let $\phi$ be a character of $C(\mathbb{R})$. Put $q(t)=1/(t^2+1)$. Sincewe have $(\phi (t{)}^2+1)\phi (q)=1$, so the restriction of $\phi$ 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 givesand hence $\phi (t)=x$.

If $h\ge 0$ is continuous, thenbelongs to $C_0(\mathbb{R})$. Applying $\phi$ to $k(h+t^2+1)=1$ yieldsso $\phi (h)=h(x)$. 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 $\phi ={\delta }_x$, and

Solved by gpt-5.6-sol high.