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In class we encountered the statement: $$H\geq C1\quad(C>0)\implies\|\mathrm{e}^{-\beta H}\|<1\quad(\beta>0)$$ How does one prove this?

Moreover, what about the weakened version: $$H\geq C1\quad(C\geq0)\implies\|\mathrm{e}^{-\beta H}\|<1\quad(\beta\geq0)$$

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  • $\begingroup$ Are you missing another condition on $C$? Because what if $T = 2I$? $\endgroup$ – Christopher A. Wong May 26 '14 at 10:36
  • $\begingroup$ Hmm, the situation is even worse since in class we're dealing with unbounded operators. The actual statement was: $$H\geq C\mathrm{Id},C>0\Rightarrow\|\mathrm{e}^{-\beta H}\|<1,\beta\in\mathbb{R}$$ $\endgroup$ – C-Star-W-Star May 26 '14 at 10:47
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    $\begingroup$ Must be $\beta > 0$. For $\beta = 0$, you have $e^{-0H} = I$ with norm $1$, and for $\beta < 0$, the norm is $> 1$. $\endgroup$ – Daniel Fischer May 26 '14 at 10:55
  • $\begingroup$ Yes right I corrected it $\endgroup$ – C-Star-W-Star May 26 '14 at 11:08
  • $\begingroup$ Since $H$ is unbounded, how do you define $e^{-\beta H}$? If it's using the spectral theorem, then the observation that $|e^{-\beta t}|\le 1$ on the spectrum of $H$ should be enough. $\endgroup$ – user147263 May 26 '14 at 19:33
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You can get the result you want using only $C_{0}$-semigroup theory. Because $H \ge C > 0$, then $K=-H+CI \le 0$ is the generator of a contractive $C_{0}$-semigroup $e^{tK}=e^{tC}e^{-tH}$. So $\|e^{tC}e^{-tH}\|\le 1$, which implies $\|e^{-tH}\| \le e^{-tC}$ for all $t \ge 0$.

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  • $\begingroup$ By the semigroup did you mean the holomorphic functional calculus for selfadjoint operators $\Phi_K(e^{t\,\cdot}),t\geq 0$ and what is meant by $\mathcal{C}_0$ and why is it contractive? $\endgroup$ – C-Star-W-Star May 26 '14 at 21:58
  • $\begingroup$ You don't need anything as strong as holomorphic. It is enough to show that $\|\lambda(\lambda I-K)^{-1}\| \le 1$ for all real $\lambda > 0$. Such an estimate is straightforward for $K^{\star}=K \le 0$. Then the $C_{0}$ semigroup $S(t)$ is guaranteed to satisfy $\|S(t)\|\le 1$ for all $t$. en.wikipedia.org/wiki/… $\endgroup$ – DisintegratingByParts May 26 '14 at 22:27
  • $\begingroup$ Now I understand your answer. ^^ $\endgroup$ – C-Star-W-Star Dec 29 '14 at 4:14

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