(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? $$\|x+y\|=\|x\|+\|y\|$$


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 $g(x^*y)=\|x\|\,\|y\|$


(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 $v\in A''$.

  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.