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Let $\mathcal{A}$ be a $C^*$-algebra, then for $a\in \mathcal{A}$, there are $b,c \in \mathcal{A}_{SA}$ such that $a=b+ic$, where $\mathcal{A}_{SA}$ is the self adjoint part of $C^*$-algebra.

Question: Whether there exist any theory so that norm of $a$ can be recover from its norm of self adjoint parts.

I appreciate, if some one provide reference or any suggestion.

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  • $\begingroup$ I assume that $a$ being self-adjoint is a typo, otherwise $c=0$, and $\Vert a \Vert = \Vert b \Vert$. $\endgroup$ Commented Jan 8, 2017 at 13:28
  • $\begingroup$ You are right, It was typo. $\endgroup$
    – Andy
    Commented Jan 8, 2017 at 14:12

2 Answers 2

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The norm of the self adjoint part is not sufficient: Even if $b$ and $c$ commutes (which should be the easiest case) then it corresponds (by looking at the commutative algebra generated by $b$ and $c$ and using gelfand duality) to the case where $a$ is a complexe valued functions from some compact space $X$ to $\mathbb{C}$ and $b$ and $c$ are its real and imaginary part. Then the norm of $a$ (i.e. the supremum of $|a(x)|$) cannot be determined only knowing the supremum of $b$ and $c$, it also depends for example on wheter $b$ and $c$ reach their supremum at the same point and it can be anything from $Max(\Vert b \Vert , \Vert c \Vert )$ if $bc=0$ to $(\Vert b \Vert^{2} + \Vert c \Vert^{2})^{1/2}$ if $b$ and $c$ reach their supremum at the same point.

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Your question has been settled in the negative already: $\|a\|$ cannot be determined if alone $\|b\|$ and $\|c\|$ are known.

Let me suggest a 'weaker' description to yield $\|a\|$, involving $\mathbb{R}$-linear combinations of $b$ and $c$ and valid for every normal $\,a\in\mathcal{A}\,$ (This is equivalent to $[b,c]=0$.) namely $$\|a\|\:=\:\sup_{t\in [0,2\pi]}\|b\cdot\cos(t)-c\cdot\sin(t)\|$$ Details incl. proof to be found in the post Is it the C*-norm in disguise? [Yes, within the realm of commutativity.] on MathSE.

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