# spectrum of Banach algebras

Let $G$ is a locally compact group (non-Abelian)

Why $sp(L^1(G))$ , i.e. the set of all nonzero bounded multiplicative functionals on $L^1(G)$ is a locally compact group.

Even for any noncommutative Banach algebra A, why $sp(A)$ is a locally compact space?

Can you give me a reference?

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If $A$ is a (noncommutative) Banach algebra, a character (i.e. a multiplicative linear functional) is automatically bounded. Then consider the set $\Sigma(A) \subseteq A^*$ of nonzero characters with the weak-$*$ topology inherited from $A^*$. Then $\Sigma(A) \cup \{ 0 \}$ is a weak-$*$ closed subset of the unit ball in $A^*$, which itself is weak-* compact. Thus $\Sigma(A) \cup \{ 0 \}$ is weak-$*$ compact, so $\Sigma(A)$ is locally compact.
Just to add: the fact that $0$ compactifies this space is the Riemann Lebesgue lemma. –  Phil Isett Mar 8 '12 at 4:45