Extension of character in Banach algebras

Solution 1:

Since $\phi$ is non-zero you can choose $a\in A$ with $\phi(a)=1$. Then define $\psi(x)=\phi(ax)$. This is well-defined because $A$ is an ideal. It is clearly linear, and it is multiplicative because $\psi(xy)=\phi(axy)=\phi(axy)\phi(a) = \phi(axya)=\phi(ax)\phi(ya)=\psi(x)\phi(a)\phi(ya)=\psi(x)\phi(aya)=\psi(x)\phi(ay)\phi(a)=\psi(x)\psi(y).$

For $x\in A$ you have $\psi(x)=\phi(ax)=\phi(a)\phi(x)=\phi(x)$. Finally, $\psi$ is continuous because characters on Banach algebras are always continuous (or because of the continuity of $B\to A$, $x\mapsto ax$ which follows from the closed graph theorem if the inclusion $A\hookrightarrow B$ is continuous).

The extension is unique: If $\tilde \psi$ is any extension then $\tilde \psi(x)=\phi(a) \tilde \psi(x)=\tilde \psi(a)\tilde \psi(x)=\tilde \psi(ax)=\phi(ax)=\psi(x)$.