When the Multiplier algebra of a Banach algebra is exactly equal to the operator algebra?

Question

Let A be a Banach algebra. B(A) and M(A) be the operator algebra and the multiplier algebra of A, respectively. When we have M(A)=B(A)?

Answer 1

It's false unless $A$ is one-dimensional.

First, it's false if $A$ has more than one complex homomorphism. Say $\phi$ and $\psi$ are distinct complex homomorphisms. Show that $\phi$ and $\psi$ are linearly independent.

Now choose $x_0\in A$ with $\psi(x_0)\ne0$ and define $T:A\to A$ by $$Tx=\hat x(\phi)x_0.$$So $$\widehat{Tx}(\psi)=\hat x(\phi)\psi(x_0)$$ If $T$ is given by the multiplier $m$ then $$\widehat{Tx}(\psi)=\hat x(\psi)m(\psi),$$showing that $\phi$ and $\psi$ are linearly dependent.

So now suppose that $A$ has only one complex homomorphism and $M(A)=B(A)$. Since the maximal ideal space consists of just one point, $M(A)$ has dimension $1$. Hence $B(A)$ has dimension $1$. Hence $A$ is one-dimensional (for example, if $X$ is any Banach space then there exists an injective linear map from the dual $X^*$ into $B(x)$: Fix $x_0\ne0$ in $X$, and map $x^*\in X^*$ to $T$, where $Tx=(x^*(x))x_0$.)