Regarding self adjoint elements in a C*- algebra

Question

Let $A$ be a complex unital C*- algebra. Let $a\in A$, then $a$ can be decomposed into its real and imaginary parts say $a_1$ and $a_2$ such that $a=a_1+ia_2$. Now $a_1$ and $a_2$ are self adjoint elements. Let $|a|$ denote the positive square root $(a^*a)^\frac{1}{2}$.

I have a few questions:

1. Am I right when I say that since $a_1$ and $a_2$ are self adjoint, hence $|a_1|=a_1$ and $|a_2|=a_2$? Because $(a_1^*a_1)^\frac{1}{2}=(a_1a_1)^\frac{1}{2}=({a_1}^2)^\frac{1}{2}=a_1$.

2. Is $\left|{{\sqrt{|a_1|^2+|a_2|^2}}}\right|={\sqrt{|a_1|^2+|a_2|^2}}?$

3. If I consider the element $x=e^{\sqrt{|a_1|^2+|a_2|^2}}$, then what is the correct reason for $x$ to be self adjoint?

4. If $\phi:A\longrightarrow \mathbb{C}$ is a multiplicative linear functional, then is $\phi(e^{\sqrt{|a_1|^2+|a_2|^2}})=e^{\sqrt{|\phi(a_1)|^2+|\phi(a_2)|^2}}$?

Answer 1

1. No, this already fails at the level of $\mathbb C$. If you have $a=-1+2i$, then $a_1=-1$ while $|a_1|=1$. Positive elements are selfadjoint, but selfadjoint does not imply positive. I think you are making the typical mistake of thinking that $\sqrt{x^2}=x$; this is false even over the real numbers, the correct equality is $\sqrt{x^2}=|x|$. Try it with a few numbers.

2. Yes. The square root is positive by definition, and there is a unique positive square root of a positive element. So if $a$ is positive, then $|a|=(a^*a)^{1/2}=(a^2)^{1/2}=a$, the last equality justified by the uniqueness of the positive square root.

3. When you do functional calculus of a selfadjoint element $a$, you are (via the Gelfand transform) working on $C(\sigma(a))$. So when you apply a real-valued function to $a$, $f(a)$ corresponds to the real-valued function $f$ is $C(\sigma(a))$, which is selfadjoint. Alternatively, $f(a)$ is a norm limit of real-valued polynomials on $a$, and each of these is selfadjoint; and a norm limit of selfadjoint is selfadjoint.

4. Yes. First consider a positive element $a$. Because $\phi$ is also $*$-preserving, it is positive. Then $\phi(a^{1/2})=\phi(a)^{1/2}$ by the positivity, multiplicativity, and the uniqueness of the square root. Hence $$ \phi(e^{a^{1/2}})=e^{\phi(a^{1/2})}=e^{\phi(a)^{1/2}}, $$ the first inequality due to the continuity of $\phi$. Similarly, since $a_j$ is selfadjoint, $$ \phi(|a_j|)=\phi((a_j^2)^{1/2})=\phi(a_j^2)^{1/2}=(\phi(a_j)^2)^{1/2}=|\phi(a_j)|. $$

Now, with $a=|a_1|^2+|a_2|^2$, \begin{align} \phi\big(e^{(|a_1|^2+|a_2|^2)^{1/2}}\big) =e^{\phi((|a_1|^2+|a_2|^2)^{1/2})} =e^{(\varphi(|a_1|^2)+\varphi(|a_2|)^2)^{1/2}} =e^{(|\varphi(a_1)|^2+|\varphi(a_2)|^2)^{1/2}} \end{align}