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(I had never thought about this question before, and I want to record the basic answer here for future reference)

Let $A$ be a C$^*$-algebra, and $x,y\in A$. When do we have equality in the triangle inequality? $$\|x+y\|=\|x\|+\|y\|$$

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Given $x,y\in A$, it turns out that the following statements are equivalent:

  • (i) $\|x+y\|=\|x\|+\|y\|$

  • (ii) there exists a norm-one linear functional $f$ such that $f(x)=\|x\|$, $f(y)=\|y\|$

  • (iii) there exists a state $g$ such that $g(x^*y)=\|x\|\,\|y\|$

Proof.

(i)$\implies$(ii) Let $f$ be a norm-one linear functional such that $f(x+y)=\|x+y\|$. Then $$ f(x)+f(y)=f(x+y)=\|x+y\|=\|x\|+\|y\|.$$As we also have $|f(x)|\leq\|x\|$ and $|f(y)|\leq\|y\|$, it follows that $|f(x)|=\|x\|$, $|f(y)|=\|y\|$. As $$f(x)+f(y)=f(x+y)=\|x+y\|=\|x\|+\|y\|=|f(x)|+|f(y)|,$$ we deduce that $f(x)=|f(x)|=\|x\|$, $f(y)=|f(y)|=\|y\|$.

(ii)$\implies$(iii) See below for the existence of the absolute value $g$ of the functional $f$. We have \begin{align} \|x\|^2+\|y\|^2+2\|x\|\,\|y\|&=(\|x\|+\|y\|)^2=f(x+y)^2=|f(x+y)|^2\\ \ \\ &=|g(v(x+y))|^2\\ \ \\ &\leq g((x+y)^*v^*v(x+y))\leq g((x+y)^*(x+y))\\ \ \\ &=g(x^*x)+g(y^*y)+2\operatorname{Re}g(x^*y)\\ \ \\ &\leq \|x\|^2+\|y\|^2+2|g(x^*y)|. \end{align} We then have, using the Schwarz inequality, $$ \|x\|\,\|y\|\leq|g(x^*y)|\leq \|x\|\,\|y\|, $$ impliying equality. From the inequalities above we also get $|g(x^*y)|=\operatorname{Re}g(x^*y)$, so $|g(x^*y)|=g(x^*y)$.

(iii)$\implies$(i) We have, using Cauchy-Schwarz, \begin{align} \|x\|^2\|y\|^2=|g(x^*y)|^2\leq g(x^*x)g(y^*y)\leq\|x\|^2\,\|y\|^2. \end{align} Thus $g(x^*x)=\|x\|^2$, $g(y^*y)=\|y\|^2$. Then \begin{align} (\|x\|+\|y\|)^2 &=\|x\|^2+\|y\|^2+2\|x\|\,\|y\| =g(x^*x)+g(y^*y)+2g(y^*x) =g((x+y)^*(x+y))\\ \ \\ &\leq \|(x+y)^*(x+y)\| =\|x+y\|^2. \end{align} So $\|x\|+\|y\|\leq\|x+y\|$, and the reverse inequality is the triangle inequality.


Absolute value of a functional

Given a bounded linear functional $f$ on $A$, we may see $f$ as a normal functional in the double dual $A''$ and perform the polar decomposition of $f$. In other words, there exists a positive functional $g$ on $A$ (a state, since $\|f\|=1$) such that $f=g(v\cdot)$ for a partial isometry $v\in A''$.

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  • $\begingroup$ @Pro Argerami,I was quite confused about what my teacher said.If $\phi$ is a state on $C^*$ algebra $A$,I assume that $\phi(a^*a) \leq 1$.He told me it is not correct.Is this definition of of norm of positive linear functional:$\|\phi\|=sup_{a\in A}|(\phi(a))|$? $\endgroup$ – math112358 Oct 15 '18 at 3:35
  • $\begingroup$ That sup you wrote is infinite for any nonzero linear functional. What does hold is that $$\|\phi\|=\sup\{|\phi(a)|/\|a\|:\ a\ne0\}=\sup\{|\phi(a)|:\ \|a\|=1\}.$$ $\endgroup$ – Martin Argerami Oct 15 '18 at 3:44
  • $\begingroup$ It is true if $\|a^*a\|=1$,thanks! $\endgroup$ – math112358 Oct 15 '18 at 3:52
  • $\begingroup$ I wonder whether the sup can be attained(there exists a positive element $a$ in $A$ with norm 1 such that $\phi(a)=1$)? $\endgroup$ – math112358 Oct 15 '18 at 5:58
  • $\begingroup$ If $\phi $ is a state and $A $ is unital, then $\phi (1)=1$. When $A $ is not unital, the answer in general is no. $\endgroup$ – Martin Argerami Oct 15 '18 at 13:43

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