# Some contraction property for bounded self-adjoint operator

Problem: For bounded self-adjoint operator $$A,B$$ on a same Hilbert space $$\mathcal{H}$$, prove $$\left\|e^{iA} - e^{iB}\right\| \leq \|A-B\|$$.

This problem has a hint: use mean value theorem to the function $$t \to \left\langle e^{-itB}e^{itA}x,y\right\rangle$$ on $$[0,1]$$ for $$x,y$$ in $$\mathcal{H}$$. When I use mean value theorem, I got $$\left\langle e^{-iB}e^{iA}x,y\right\rangle - \langle x,y\rangle = \left\langle i(A-B)e^{-it_0B}e^{it_0A}x,y\right\rangle.$$ Then I have no idea to go further, thanks for any help.

That's not right, but you can write $$\dfrac{d}{dt} \langle e^{-itB} e^{itA} x, y \rangle = \langle (-i e^{-itB} B e^{itA} + i e^{-itB} A e^{itA}) x, y\rangle = \langle i e^{-itB} (A-B) e^{itA} x, y \rangle$$ which is different from what you wrote if $$A$$ and $$B$$ don't commute. Next, estimate the absolute value.