Jordan Functionals on a Banach Algebra

# Jordan Functionals on a Banach Algebra

Definition: Let $\mathfrak{A}$ be an algebra. A linear functional $f$ on $\mathfrak{A}$ is said to be a Jordan Functional if $f$ is not identically the zero functional and $f(x^2) = f(x)^2$ for all $x \in \mathfrak{A}$. |

The following lemma and theorem shows us that every Jordan functional in a Banach algebra is multiplicative.

Lemma 1 (Zelasko): Let $\mathfrak{A}$ be a Banach algebra. If $f$ is a Jordan functional on $\mathfrak{A}$ then $f(xyx) = f(x)^2f(y)$ for all $x, y \in \mathfrak{A}$. |

**Proof:**Let $f$ be a Jordan functional on $\mathfrak{A}$. Then for all $x, y \in \mathfrak{A}$ we have that $f((x + y)^2) = [f(x+y)]^2$. That is:

\begin{align} \quad f(x^2 ) + f(xy) + f(yx) + f(y^2) = f(x^2 + xy + yx + y^2) = f((x + y)^2) = [f(x + y)]^2 = [f(x) + f(y)]^2 = f(x)^2 + f(x)f(y) + f(y)f(x) + f(y)^2 \end{align}

- Since $f$ is a Jordan functional on $\mathfrak{A}$ we have that $f(x^2) = [f(x)]^2$ and $f(y^2) = [f(y)]^2$, and so the above equality becomes:

\begin{align} \quad f(xy + yx) = 2f(x)f(y) \end{align}

- Define an operation $*$ on $\mathfrak{A}$ by $x * y = \frac{1}{2} (xy + yx)$. Observe from the above equation that then $f(x * y) = f(x)f(y)$. If $x, y, z \in \mathfrak{A}$ then we also have that:

\begin{align} \quad f((x * y) * z) = f(x * y)f(z) = f(x)f(y)f(z) \quad (\star) \end{align}

- Expanding the lefthand side of the equation above gives us:

\begin{align} \quad f((x*y)*z) &= f \left ( \frac{1}{2} (xy + yx) * z \right ) \\ &= \frac{1}{2} f \left ( \frac{1}{2} \left [ (xy + yx)z + z(xy + yx) \right ] \right ) \\ & \frac{1}{4} f (xyz + yxz + zxy + zyx) \\ &= f(x)f(y)f(z) \end{align}

- Therefore, for all $x, y, z \in \mathfrak{A}$ we have that:

\begin{align} \quad f(xyz + yxz + zxy + zyx) = 4f(x)f(y)f(z) \quad (\star \star) \end{align}

- Now also observe by $\star$ that for all $x, y, z \in \mathfrak{A}$ we have that:

\begin{align} \quad f(y * (z * x) - (y * z)*x) = f(y * (z * x)) - f((y * z) * x) = f(y)f(z * x) - f(y * z)f(x) = f(y)f(z)f(x) - f(y)f(z)f(x) = 0 \end{align}

- Expanding the lefthand side of this equation gives us that:

\begin{align} \quad f(y * (z * x) - (y * z)*x) &= \frac{1}{2} f(y * (zx + xz) - (yz + zy)* x) \\ &= \frac{1}{4}f(y(zx + xz) + (zx + xz)y - (yz + zy)x - x(yz + zy)) \\ &= \frac{1}{4}f(yzx + yxz + zxy + xzy - yzx - zyx - xyz -xzy) \\ &= \frac{1}{4} f(yxz + zxy - zyx -xyz) \\ &= 0 \end{align}

- Therefore:

\begin{align} \quad f(xyz + zyx) = f(yxz + zxy) \quad (\star \star \star) \end{align}

- Comparing $(\star \star)$ and $(\star \star \star)$ we see that:

\begin{align} \quad 2f(xyz + zyx) &= f(xyz + zyx) + f(yxz + zxy) \\ &= f(xyz + zyx + yxz + zxy) \\ & \overset{(\star \star)} = 4f(x)f(y)f(z) \end{align}

- Let $z = x$. Then for all $x, y \in \mathfrak{A}$ we have that:

\begin{align} \quad 2f(xyx + xyx) = 4f(x)f(y)f(x) \\ \quad f(xyx) = f(x)^2f(y) \quad \blacksquare \end{align}

Theorem 2 (Zelasko): Let $\mathfrak{A}$ be a Banach algebra with unit. If $f$ is a Jordan functional on $\mathfrak{A}$ then $f$ is multiplicative. |

*The above theorem also applies when $\mathfrak{A}$ is a Banach algebra without unit.*

**Proof:**From Proposition 1, since $f$ is a Jordan functional on $\mathfrak{A}$ we have that for all $x, y \in \mathfrak{A}$:

\begin{align} \quad f(xy + yx) = 2f(x)f(y) \quad (\star) \end{align}

- We claim that $f(xy) = f(yx)$. Suppose otherwise, i.e., suppose there exists $x_0, y_0 \in \mathfrak{A}$ such that

\begin{align} \quad f(x_0y_0 - y_0x_0) = f(x_0y_0) - f(y_0x_0) = C \neq 0 \end{align}

- Note that if $x_0, y_0 \in \mathfrak{A}$ are such that $f(x_0y_0) \neq f(y_0x_0)$ then for all $\alpha \neq 0$, $x_0, \alpha^{-1}y_0$ are such that:

\begin{align} \quad f(x_0 \alpha^{-1} y_0) = \alpha^{-1}f(x_0y_0) \neq \alpha^{-1}f(y_0x_0) = f(\alpha^{-1}y_0x_0) \end{align}

- So without loss of generality we may assume that $C = 1$.

- Furthermore, note that if $x_0, y_0 \in \mathfrak{A}$ are such that $f(x_0y_0) \neq f(y_0x_0)$ then for all $\alpha \in \mathbf{F}$ we have that $x_0 + \alpha 1_{\mathfrak{A}}, y_0$ is such that:

\begin{align} \quad f((x_0 + \alpha1_{\mathfrak{A}})y_0) = f(x_0y_0 + \alpha y_0) = f(x_0y_0) + \alpha f(y_0) \neq f(y_0x_0) + \alpha f(y_0) = f(y_0x_0 + \alpha y_0) = f(y_0(x_0 + \alpha 1_{\mathfrak{A}})) \end{align}

- But note that $f(x_0 + \alpha 1_{\mathfrak{A}}) = f(x_0) + \alpha$. So without loss of generality we may also assume that $f(x_0) = 0$. Thus suppose there exists $x_0, y_0 \in \mathfrak{A}$ with $f(x_0y_0 - y_0x_0) = 1$ and $f(x_0) = 0$. By the equality at $(\star)$ we also have that $f(x_0y_0 + y_0x_0) = 2f(x_0)f(y_0) = 0$. Thus;

\begin{align} \quad 2f(x_0y_0) = 1 \end{align}

- So $f(x_0y_0) = \frac{1}{2}$ which implies that $f(y_0x_0) = \frac{1}{2}$. Finally, since $f$ is a Jordan functional and by Lemma 1 we have that:

\begin{align} \quad 1 &= [f(x_0y_0 - y_0x_0)]^2 \\ &= f([x_0y_0 - y_0x_0]^2) \\ &= f([x_0y_0]^2 - [x_0y_0][y_0x_0] - [y_0x_0][x_0y_0] + [y_0x_0]^2) \\ &= f([x_0y_0]^2) - f(x_0y_0^2x_0) - f(y_0x_0^2y_0) + f([y_0x_0]^2) \\ &= [f(x_0y_0)]^2 - f(x_0)^2f(y_0^2) - f(y_0)^2f(x_0^2) + [f(y_0x_0)]^2 \\ &= \left [ \frac{1}{2} \right ]^2 - 0 - f(y_0)^2[f(x_0)]^2 + \left [ \frac{1}{2} \right ]^2 \\ &= \frac{1}{4} - 0 - 0 + \frac{1}{4} \\ &= \frac{1}{2} \end{align}

- So we have arrived at a contradiction. So the assumption that $f(xy) \neq f(yx)$ is false. Thus $f(xy) = f(yx)$ for all $x, y \in \mathfrak{A}$. So by $(\star)$ we have that for all $x, y \in \mathfrak{A}$:

\begin{align} \quad 2f(xy) = f(xy) + f(yx) = f(xy + yx) = 2f(x)f(y) \end{align}

- Thus for all $x, y \in \mathfrak{A}$, $f(xy) = f(x)f(y)$. Thus $f$ is multiplicative. $\blacksquare$

Observe the equality $(*)$ in the above proof. Note that if $\mathfrak{A}$ is a commutative Banach algebra then the Jordan functional $f$ being multiplicative comes immediately since then $xy = yx$, so $f(xy + yx) = f(2xy) = 2f(xy)$, so $2f(xy) = 2f(x)f(y)$, i.e., $f(xy) = f(x)f(y)$ for all $x, y \in \mathfrak{A}$.