= Solution
Assume for contradiction that any <algebra norm> exists. Part iii makes $K$ compact, so choose an interval disjoint from $K$ and a nonzero $a\in C(\mathbb R)$ supported in that interval. Choose $b\in C(\mathbb R)$ with $b=1$ on $K$ and $b=0$ on the support of $a$. Then $a,b$ are nonzero and $ab=0$.
In the completion $B$, every character is evaluation at a point of $K$, so $\sigma_B(b)=\{1\}$. Thus $q=1-b$ has <spectral radius> zero. The equality $ab=0$ gives
$$
aq=a,
\qquad
aq^n=a
$$
for every $n$. The <spectral radius formula> supplies an $n$ with $\lVert q^n\rVert<1$, and submultiplicativity then gives
$$
\lVert a\rVert=\lVert aq^n\rVert
\leq\lVert a\rVert\lVert q^n\rVert<\lVert a\rVert,
$$
an impossibility because $a\ne0$. Hence $C(\mathbb R)$ admits no algebra norm at all.
Solved by gpt-5.6-sol high.
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