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Let $ (T(t))_{t \geq 0} $ be a $ C_{0} $-semigroup on a Hilbert space $ X $ with an infinitesimal generator $ A $, and let $ \rho \in (0,1) $.

I want to prove that $ \displaystyle \sup_{t \geq 0} \| T(t) - I \| \leq \rho $ is equivalent to $ A = 0 $.

I have already proven that when the inequality holds, then the infinitesimal generator $ A $ satisfies $$ \forall \lambda > 0: \quad \left\| \lambda (\lambda I - A)^{-1} - I \right\| \leq \rho, $$ and I want to use this new inequality to show that $ A = 0 $ and thus $ T(t) = I $.

Define $ T_{\lambda} \stackrel{\text{df}}{=} \lambda (\lambda I - A)^{-1} - I $. Then it is clear that $ T \in \mathcal{L}(X) $ and $ \| T_{\lambda} \| \leq \rho $. Hence, $$ \| T_{\lambda} \| \leq \rho \iff 1 - \| T_{\lambda} \| \leq 1 - \rho \iff (1 - \| T_{\lambda} \|)^{-1} \leq (1 - \rho)^{-1}. $$ We also have that $ (1 - \| T_{\lambda} \|)^{-1} = \dfrac{1}{1 - \| \lambda (\lambda I - A)^{-1} - I \|} $.

How do I proceed from here?

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  • $\begingroup$ Hi user3482534. It turns out that you were on the right track. However, instead of considering the number $ (1 - \| T_{\lambda} \|)^{-1} $, you need to consider the operator $ (I + T_{\lambda})^{-1} $. $\endgroup$ – Berrick Caleb Fillmore May 4 '15 at 18:16
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As the OP has mentioned, we have $$ \forall \lambda \in \mathbb{R}_{> 0}: \quad T_{\lambda} \stackrel{\text{df}}{=} \lambda (\lambda I - A)^{-1} - I \in B(\mathcal{H}) \quad \text{and} \quad \| T_{\lambda} \|_{B(\mathcal{H})} \leq \rho. $$ This is a consequence of the following theorem, whose statement is found in Theorem $ 13.35 $ of Walter Rudin’s Functional Analysis:

Theorem. Let $ (T(t))_{t \geq 0} $ be a $ C_{0} $-semigroup on a Hilbert space $ \mathcal{H} $. Denote its infinitesimal generator by $ A $. If $ C \in \Bbb{R}_{> 0} $ and $ \gamma \in \Bbb{R} $ are constants such that $ \| T(t) \|_{B(\mathcal{H})} \leq C e^{\gamma t} $ for each $ t \in \Bbb{R}_{\geq 0} $, then for any $ \lambda \in \Bbb{C} $ satisfying $ \Re(\lambda) > \gamma $, the map $ R_{\lambda}: \mathcal{H} \to \mathcal{H} $ defined by $$ \forall x \in \mathcal{H}: \quad {R_{\lambda}}(x) \stackrel{\text{df}}{=} \int_{0}^{\infty} e^{- \lambda t} \cdot [T(t)](x) ~ \mathrm{d}{t} $$ is a well-defined bounded linear operator (called the resolvent of $ (T(t))_{t \geq 0} $ at $ \lambda $) that has the following two properties:

  • Its range is $ \text{Dom}(A) $.
  • It inverts $ \lambda I - A $.

The given assumption on our $ C_{0} $-semigroup $ (T(t))_{t \geq 0} $ is that $$ \forall t \in \Bbb{R}_{\geq 0}: \quad \| T(t) - I \|_{B(\mathcal{H})} \leq \rho < 1. $$ Hence, if we choose $ C = 1 + \rho $ and $ \gamma = 0 $, then $$ \forall t \in \Bbb{R}_{\geq 0}: \quad \| T(t) \|_{B(\mathcal{H})} \leq C e^{\gamma t}. $$ Fix $ \lambda \in \Bbb{R}_{> 0} $ for the present moment. By the theorem, we have $$ (\heartsuit) \qquad \forall x \in \mathcal{H}: \quad {(\lambda I - A)^{-1}}(x) = \int_{0}^{\infty} e^{- \lambda t} \cdot [T(t)](x) ~ \mathrm{d}{t}. $$ Using the fact that $ \displaystyle \int_{0}^{\infty} \lambda e^{- \lambda t} ~ \mathrm{d}{t} = 1 $, we also have $$ (\diamondsuit) \qquad \forall x \in \mathcal{H}: \quad x = \int_{0}^{\infty} \lambda e^{- \lambda t} \cdot x ~ \mathrm{d}{t}. $$ Multiplying $ (\heartsuit) $ by $ \lambda $ and then subtracting $ (\diamondsuit) $ from that yields \begin{align} \forall x \in \mathcal{H}: \quad {T_{\lambda}}(x) & = \left( \lambda (\lambda I - A)^{-1} - I \right) \! (x) \\ & = \int_{0}^{\infty} \lambda e^{- \lambda t} \cdot [T(t) - I](x) ~ \mathrm{d}{t}. \end{align} We thus get \begin{align} \forall x \in \mathcal{H}: \quad \| {T_{\lambda}}(x) \|_{\mathcal{H}} & \leq \int_{0}^{\infty} \left\| \lambda e^{- \lambda t} \cdot [T(t) - I](x) \right\|_{B(\mathcal{H})} ~ \mathrm{d}{t} \\ & \leq \int_{0}^{\infty} \lambda e^{- \lambda t} \cdot \| T(t) - I \|_{B(\mathcal{H})} \cdot \| x \|_{\mathcal{H}} ~ \mathrm{d}{t} \\ & \leq \left( \int_{0}^{\infty} \lambda e^{- \lambda t} ~ \mathrm{d}{t} \right) \rho \| x \|_{\mathcal{H}} \\ & = \rho \| x \|_{\mathcal{H}}. \end{align} Therefore, $ \| T_{\lambda} \|_{B(\mathcal{H})} \leq \rho $, as desired.

Claim: $ \text{Dom}(A) = \mathcal{H} $ and $ A \in B(\mathcal{H}) $.

Proof of Claim

As $ \| T_{\lambda} \| \leq \rho < 1 $, a standard result about Banach algebras tells us that $ I + T_{\lambda} $ is an invertible element of $ B(\mathcal{H}) $ whose inverse is the ‘geometric series’ $$ \sum_{k = 0}^{\infty} (-1)^{k} \cdot (T_{\lambda})^{k}. $$ As $ I + T_{\lambda} = \lambda (\lambda I - A)^{-1} $, we obtain \begin{align} \text{Dom}(A) & = \text{Range} \! \left( (\lambda I - A)^{-1} \right) \\ & = \text{Range} \! \left( \lambda (\lambda I - A)^{-1} \right) \\ & = \text{Range}(I + T_{\lambda}) \\ & = \mathcal{H}. \qquad (\text{As $ I + T_{\lambda} $ is invertible, hence surjective.}) \end{align} As $ A $ is a closed operator (this is a standard fact in the theory of $ C_{0} $-semigroups), it follows that $ A \in B(\mathcal{H}) $. (Note: By the Closed Graph Theorem, a closed operator that is defined everywhere is bounded.) $ \quad \blacksquare $

Observe now that $$ \frac{1}{\lambda} (\lambda I - A) = (I + T_{\lambda})^{-1} = \sum_{k = 0}^{\infty} (-1)^{k} \cdot (T_{\lambda})^{k}, $$ so $$ \left\| \frac{1}{\lambda} (\lambda I - A) \right\|_{B(\mathcal{H})} \leq \sum_{k = 0}^{\infty} \| T_{\lambda} \|_{B(\mathcal{H})}^{k} \leq \sum_{k = 0}^{\infty} \rho^{k} = \frac{1}{1 - \rho}. $$ It follows readily that $$ (\spadesuit) \qquad \| \lambda I - A \|_{B(\mathcal{H})} \leq \frac{\lambda}{1 - \rho}. $$ Let $ \lambda \to 0^{+} $. The left-hand side of $ (\spadesuit) $ goes to $ \| A \|_{B(\mathcal{H})} $, while the right-hand side goes to $ 0 $. By the Squeeze Theorem, $ \| A \|_{B(\mathcal{H})} = 0 $, so we arrive at the desired conclusion that $ A = 0_{B(\mathcal{H})} $.

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