An estimate involving exponentials of a self-adjoint operator

Question

Let $H$ be a complex Hilbert space, and let $T$ be a bounded self-adjoint operator on $H$.

I've seen it written in a book that for $x,y\in[0,1]$, we have $$\|e^{ixT}-e^{iyT}\|\leq\|T\|\|x-y\|.$$

Question: How would one prove this, and where are the hypotheses (self-adjointness of $T$ and $x,y\in[0,1]$) needed?

Answer 1

Note that

\begin{align} e^{ixT} - e^{iyT} &= \int_0^1 (e^{i(y+ t(x-y)) T})' dt \\ &= \int_0^1 e^{i(y+t(x-y)T} (i(x-y)T)dt\end{align}

So $$\|e^{ixT} - e^{iyT}\| \le \|T\| |x-y| \int_0^1 \| e^{i(y+t(x-y)T}\|dt$$

Since $T$ is self adjoint, $e^{irt}$ is unitrary:

$$ (e^{irT})^* = e^{-ir T^*} = e^{-irT} = (e^{irT})^{-1}, $$ we have $$ \| e^{irT}\|=1$$ and thus

$$\|e^{ixT} - e^{iyT}\| \le \|T\| |x-y|. $$

The fact that $x, y\in [0,1]$ is not used, AFAICT.

Answer 2

Assuming WLOG that $x≤y$, one has by the mean value Theorem that $$ \|e^{ixT}-e^{iyT}\| ≤ \int_x^y\left\|{d\over dt} e^{itT}\right\| dt = \int_x^y\|iTe^{itT}\| dt = \int_x^y\|T\| dt = \|T\| |x-y| $$