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Let $B$ be the completion under the assumed algebra norm. Restriction sends every $\psi \in {\Phi }_B$ to a character of the dense subalgebra $C(\mathbb{R})$, hence by part i to ${\delta }_x$ for some $x$. Continuity of the restriction says exactly that $x\in K$. Conversely, every ${\delta }_x$ continuous for the algebra norm extends uniquely through the completion, and continuity of multiplication makes the extension a character of $B$. Restriction and extension therefore give a bijection

For $b\in B$, choose $h_n\in C(\mathbb{R})$ converging to $b$ in the algebra norm. Characters on the unital Banach algebra $B$ are uniformly norm bounded, so $h_n{|}_K$ converges uniformly to the function $x\mapsto {\tilde{\delta }}_x(b)$. This function is continuous. Hence $x\mapsto {\tilde{\delta }}_x$ is continuous into the Gelfand topology. Its inverse is again $\psi \mapsto \psi (t)$, so the bijection is a homeomorphism. In particular, $K$ is compact.

Solved by gpt-5.6-sol high.

Suppose an algebra norm made $C(\mathbb{R})$ a Banach algebra. Its unital character space of an algebra would be compact in the Gelfand topology. By part i it consists of the evaluations ${\delta }_x$. The mapis continuous from the usual topology because every $h\in C(\mathbb{R})$ is continuous, and its inverse is the continuous map $\phi \mapsto \phi (t)$ defined by the coordinate function $t$. Thus the character space is homeomorphic to the noncompact space $\mathbb{R}$, a contradiction. No complete algebra norm exists.

Solved by gpt-5.6-sol high.

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 $ab=0$ givesfor every $n$. The spectral radius formula supplies an $n$ with $\Vert q^n\Vert <1$, and submultiplicativity then givesan impossibility because $a\ne 0$. Hence $C(\mathbb{R})$ admits no algebra norm at all.

Solved by gpt-5.6-sol high.