Let $A$ and $B$ be Banach algebras. A linear map $\phi: A\longrightarrow B$ is called a Jordan Homomorphism if $\phi(a^2)=\phi(a)^2$ for every $a\in A$. And $\phi$ is a ring Homomorphism if $\phi(ab)=\phi(a)\phi(b)$ for every $a,b\in A$.
Is every Jordan Homomorphism a ring Homomorphism? I know that every Jordan Homomorphism is a ring Homomorphism if $B$ is a commutative semisimple Banach algebra.
It appear that commutativity of both $A$ and $B$ is sufficient, as can be seen from the following computation. Let $a,b\in A$ be arbitrary, then one sees \begin{align} \phi(a)^{2} + \phi(b)^{2} + \phi(a)\phi(b) + \phi(b) \phi(a) &= (\phi(a) + \phi(b))^{2} \\ &= \phi((a+b)^{2}) \\ &= \phi(a)^{2} + \phi(b)^{2} + \phi(ab) + \phi(ba). \end{align} In general, we see that the equation \begin{equation} \phi(a)\phi(b) + \phi(b) \phi(a) = \phi(ab) + \phi(ba), \end{equation} holds.
Let us thus try to find a counter-example where both $A$ and $B$ are non-commutative. A simple choice would be $A = B = \text{Mat}_{2}(\mathbb{R})$. Now define the map $\phi$ as follows: \begin{align} \phi:\text{Mat}_{2}(\mathbb{R}) &\rightarrow \text{Mat}_{2}(\mathbb{R}),\\ a & \mapsto a^{T}, \end{align} where $a^{T}$ is the transpose of the matrix $a$. A simple verification shows that this map is a Jordan homomorphism, but not a ring homomorphism.