Let $A$ be a unital $C^*$-algebra, $a\in A$. Why is $a^*a\le \|a\|^21_A$?
I think this is a functional calculus argument. The functional calculus to $a^*a$ is an isomorphism of $C^*$-algebras $$C(\sigma(a^*a))\to C^*(a^*a,1_A),\; f\mapsto f(a^*a).$$ We have that $\|a\|^2=\|a^*a\|$ and $\sigma(a^*a)\subseteq [0,\|a\|^2]$. My first guess was to consider the function $f(x)=|x|-x$, but $f$ is zero on $\sigma(a^*a)$, so it doesn't work. So, how to prove $a^*a\le \|a\|^21_A$? Thank you.
On
You can indeed prove this using the Gelfand transform, but at this point you will probably have proven the following much more general lemma (using the Gelfand transform) that gives you a very useful characterisation for positive elements.
Let $x$ be a self-adjoint element of $A$ and $r \in \mathbb{R}$. If $\lVert x-t\mathbb{1} \rVert \leq t$ then $x$ is positive.
This allows you to avoid repeating the above argument all the time while also simplifying it. Note that $$ \lVert (\lVert a \rVert^2 \mathbb{1}- a^* a) - \lVert a \rVert^2 \mathbb{1} \rVert = \lVert a^* a \rVert = \lVert a \rVert^2, $$ so $\lVert a \rVert^2 \mathbb{1}- a^* a \geq 0$ or $a^*a \leq \lVert a \rVert^2 \mathbb{1}$.
On
Demophilus has given a nice answer, but there is also a way to see it that follows from elementary considerations.
First note that $a^*a$ is positive and $\|a^*a\|=\|a\|^2$ by the C* property. Now if $p$ is a positive element in a $C^*$ algebra (that is self-adjoint and $\sigma(p)\subset[0,\infty)$ ), then you have that $\|p\|\Bbb1-p$ is self-adjoint and $$\sigma(\|p\|\Bbb1-p) = \|p\|-\sigma(p)$$ which follows immediately from the definition of the spectrum. One of the first results one sees in C* algebras is that for self-adjoint operators $a$ you have $\|a\|=\sup_{x\in\sigma(a)}|x|$ so $\sigma(\|p\|\Bbb1 -p)$ is a subset of $[0,\infty)$ and thus $\|p\|\Bbb1-p$ is positive. Since the partial ordering on the positives is defined via $a≥b$ iff $a-b$ is positive you see $\|p\|\Bbb1≥p$, thus $\|a\|^2\Bbb1≥a^*a$.
Consider $f(x)=\lVert a\rVert^2-x$.