# Proof involving strongly continuous semigroups.

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?

• 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 $(I + T_{\lambda})^{-1}$. – Berrick Caleb Fillmore May 4 '15 at 18:16

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 $A = 0_{B(\mathcal{H})}$.