In a C*-algebra $A$, $x$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$.

Question

The question I am having trouble with is the following:

Let $A$ be a C$^*$-algebra. Show that an element $x$ of $A$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$. (Hint: If $h\in A$ is self-adjoint, then $\exp(ith)=1+ith+o(t)$ is unitary for every $t\in\mathbb{R}$. If $k$ is another nonzero self-adjoint element of $A$, then $\Vert 1+ith-tk\Vert\geq\Vert 1-tk\Vert\neq 1+o(t)$.)

(In the book it is written $O(t)$ instead of $o(t)$, but this is the correct statement - see comment below)

This is Exercise 4.3 from Takesaki's "Theory of Operator Algebras I". I managed to solve the first implication ($x$ self-adjoint implying the limit being zero) using the hint, but I'm having problems with the converse.

In the hint, I belie he means we should decompose $x$ in the real and imaginary parts, $x=h+ik$, so the real part of $1+itx$ is $1-tk$, and then we work with it to show that $k\neq 0$ gives a contradiction.

Answer 1

So suppose that $\inf \text{Sp}(k) = \alpha$ and $\sup \text{Sp}(k) = \beta$. If $\beta \ge 0$, then $\|1+tk\| = 1+t\beta$ for sufficiently small $t > 0$. If $\alpha \le 0$ then $\|1+tk\| = 1+t\alpha$ for sufficiently small $t < 0$. If $\alpha =\beta = 0$ then $k=0$. So enumerating all the cases, we see that $\|1+tk\| = 1 + o(t)$ fails as $t \to 0$.

Answer 2

Given an element $a$ of $A$, we denote its real and imaginary parts $a_r$ and $a_i$, respectively. Given a self-adjoint $a$, we denote it's positive and negative parts by $a_+$ and $a_-$, respectively.

Suppose $\lim_{t\to 0}\frac{\Vert 1+itx\Vert-1}{t}=0$. We will first show that $(x_i)_-=0$. As you noted, the real part of $1+itx$ is $1-tx_i$. Given $\epsilon>0$, let $t>0$ such that $$\epsilon>\frac{\Vert 1+itx\Vert-1}{t}\geq\frac{\Vert1-tx_i\Vert-1}{t}.$$ For a general C*-algebra, if $a$ is self-adjoint and $\Vert a\Vert<1$, then $\Vert1-a\Vert=1+\Vert a_-\Vert$. This can be easily seen by considering the Gelfand transform. Applying this to the inequality above, where we choose $t$ positive and small, we obtain $$t\epsilon\geq 1+t\Vert(x_i)_-\Vert-1=t\Vert(x_i)_-\Vert$$ i.e., $\Vert (x_i)_-\Vert\leq\epsilon$ for all $\epsilon$ and therefore $(x_i)_-=0$.

Now consider the limit $\lim_{t\to 0}\frac{\Vert 1+itx\Vert-1}{t}=0$ once more. Substituting $t$ by $-t$ and taking an adjoint inside the norm, we obtain $\lim_{t\to 0}\frac{\Vert 1+itx^*\Vert-1}{t}=0$. The argument above implies $0=((x^*)_i)_-=(-x_i)_-=(x_i)_+$.

We conclude that $x_i=(x_i)_+-(x_i)_-=0$ and therefore $x=x_r$ is self-adjoint.

Answer 3

Assume first that $x$ is selfadjoint. Then $e^{-itx}$ is a unitary for all real $t$. So, using the Taylor expansion $e^{-itx}=1-itx+o(t^2)$, for $t>0$ we have $$ \frac{\|1-itx\|-1}t=\frac{\|e^{-itx}+o(t^2)\|-1}t=\frac{\|1+o(t^2)e^{itx}\|-1}t \leq\frac{1+o(t^2)-1}t=o(t)\to0. $$ Similarly, $$ \frac{\|1-itx\|-1}t=\frac{\|e^{-itx}+o(t^2)\|-1}t=\frac{\|1+o(t^2)e^{itx}\|-1}t \geq\frac{1-o(t^2)-1}t=-o(t)\to0. $$ Analog computations deal with $t<0$, and thus $$\tag{1}\lim_{t\to0}(1/t)(\|1-itx\|-1)=0.$$

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Conversely, assume that the limit in $(1)$ is zero. Write $x=a+ib$, in terms of its real and imaginary parts. So, by hypothesis, $$ \lim_{t\to0}\frac{\|1-ita+tb\|-1}t=0, $$ while by the first part of the proof $$ \lim_{t\to0}\frac{\|1-ita\|-1}t=0, $$ So \begin{align} 0&=\lim_{t\to0}\frac{\|1-ita+tb\|-\|1-ita\|}t =\lim_{t\to0}\frac{\|e^{-ita}+o(t^2)+tb\|-\|e^{-ita}+o(t^2)\|}t\\ &=\lim_{t\to0}\frac{\|e^{-ita}+tb\|-\|e^{-ita}\|}t =\lim_{t\to0}\frac{\|1+tbe^{ita}\|-1}t\\ \end{align} In particular, for $t<0$, $$ \frac{\|1+tbe^{ita}\|-1}t\geq\frac{1+\|tbe^{-ita}\|-1}t=\|be^{-ita}\|=\|b\|; $$ so making $t\to0$ we get $0\geq\|b\|$, and $b=0$. So $x$ is selfadjoint.