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I am new to functional analysis and therefore will appreciate help on the below, perhaps very simple question.

Let $A$ be a bounded linear self-adjoint operator on a Hilbert space $X$. Suppose that $0 \le A \le I$ where $T\le W$ if $W-T$ is non-negative, i.e., $\langle(T-W)x, x\rangle \ge 0$. I am told that in this case $\| I - A\| \le 1$ is obvious. I am not able to show this using the definition of norm. Is there some fact from which this follows.

Martin Argerami
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  • It is not that obvious. See the inequality in the first answer here the one after "In particular. This gives $$|Px|^2\le |P|, \langle Px,x\rangle $$ – Ryszard Szwarc Dec 02 '23 at 07:26

1 Answers1

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First you notice that you have $$\tag10\leq I -A\leq I$$(the first inequality is $A\leq I$ and the second one is $A\geq0$). Next one shows that $$\tag2 0\leq T\leq S\implies \|T\|\leq\|S\|. $$ There are several ways to show this, none entirely straightforward. It's a bit easier when $S=I$.

When $S=I$ one can do, with $\|x\|=1$, $$ \|Tx \|^2=\langle T^2x,x\rangle=\langle TT^{1/2}x,T^{1/2}x \rangle \leq\langle T^{1/2}x,T^{1/2}x \rangle=\langle Tx,x\rangle\leq 1. $$

One path to show $(2)$ in general is to use the fact that for a selfadjoint operator $T$, $$ \|T\|=\sup\{|\langle Tx,x\rangle|: \|x \|=1\}. $$ Then $\|T\|\leq\|S\|$ follows from $$ \langle Tx,x\rangle\leq\langle Sx,x\rangle. $$

Another path to show $(2)$ is to use the fact that if $0\leq T\leq S$ then there exists a contraction $C$ such that $T^{1/2}=CS^{1/2}$. For then you have, via the C$^*$-identity $\|X\|^2=\|X^*X\|$, $$ \|T\|=\|T^{1/2}\|^2=\|CS^{1/2}\|^2\leq\|S^{1/2}\|^2=\|S\|. $$

In the case $S=I$ another approach is to use functional calculus. The C$^*$-algebra generated by $T$ is isomorphic to $C(\sigma(T))$ in a way that $T$ gets mapped to the identity function. Then $(1)$ becomes the number inequality $0\leq 1-t\leq 1$ which easily implies $|1-t|\leq1$.

Martin Argerami
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