Let $\mathcal A$ be a unital C* algebra and $A\in \mathcal A$ a positive element $A\geq \mathbb 0$, i.e. A is self-adjoint and the spectrum satisfies $\sigma(A)\subset [0,\infty)$.

Let $\alpha \in [0,1]$ Is it true that

$$\vert\vert\, \alpha A - \vert\vert A \vert\vert \,\vert\vert \leq \vert\vert A \vert\vert $$

I just know the triangle inequality which is not sharp enough. What other means are there to look for such an inequality?

  • 2
    $\begingroup$ the answer was given, but let me give another pointer. if you already know Gelfand duality or spectral theorem, just think of $A$ as a positive continuous function over some space (its spectrum) and the inequality comes out intuitively. $\endgroup$ – Yul Otani May 11 '13 at 14:29

Let $B:=\alpha A-\|A\| 1$. Then $A$ and $B$ are both self-adjoint.

The spectrum of $B$ is $$\sigma(B)=\alpha\sigma(A)-\|A\|=\{\alpha \lambda-\|A\|\;;\; \lambda\in \sigma(A)\}$$ as a special case of the spectral mapping theorem: $\sigma(f(A))=f(\sigma(A))$, which is trivial in the case of a degree $1$ polynomial $f$ like here.

Since $\sigma(A)\subseteq[0,\|A\|]$ we get $$ \sigma(B)\subseteq [-\|A\|,(\alpha-1)\|A\|]\subseteq [-\|A\|,0]. $$ Therefore the spectral radius is $\rho(B)\leq \|A\|$.

Now for a normal element, whence for a self-adjoint element, $\|B\|=\rho(B)$.

It follows that $\|B\|\leq \|A\|$.

Alternative: it is convenient to use the partial order on self-adjoint elements $C\leq D$ if $D-C$ is positive. Then $$ 0\leq \alpha A\leq A\leq \|A\|1\quad\Rightarrow \quad-\|A\|1\leq \alpha A-\|A\|1\leq 0. $$ Finally, for $B$ self-adjoint, we have $\|B\|\leq m$ if and only if $-m\cdot 1\leq B\leq m\cdot 1$.

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