How to show that $(I+T/n)^{-1}$ converges strongly to the identity operator?

Question

Let $T$ be an m-accretive operator on a Hilbert space. Then how can I show that $(I+T/n)^{-1}$ converges in the strong operator topology to the identity operator $I$?

It seems trickier than I expected...Could anyone help me?

Answer 1

Hint: Use $\operatorname{Real}\langle Tx,x\rangle \geq 0$ and $T+\lambda I$ is surjective for every $\lambda >0$ to show that $$\|(T+\lambda I)^{-1}\|\leq \frac{1}{\lambda}\quad\forall\; \lambda >0.$$

Answer 2

The main point of the argument has already been pointed out in the answer by Sahiba Arora. Since there seems to be a bit of confusion left, I'll add some of the details here:

Let $X$ be a complex Hilbert space, or more generally, a complex Banach space. Let $T: X \supseteq D(T) \to X$ be an $m$-accreative operator.

1. Then $\lambda \|(\lambda I + T)^{-1}\| \le 1$ for all scalars $\lambda \in (0,\infty)$. This follows, for instance, from [1, Lemma 3.4.2].

2. Now we use that, for each $x \in D(T)$, $$ \Big(\lambda (\lambda I + T)^{-1} - I\Big)x = \Big(\lambda (\lambda I + T)^{-1} - (\lambda I + T)^{-1}(\lambda I + T)\Big)x = -(\lambda I + T)^{-1} T x, $$ and the norm of the later vector is bounded by $\frac{\|Tx\|}{\lambda}$ for each $\lambda \in (0,\infty)$ according to Step 1. This proves that $\lambda(\lambda I + T)^{-1}$ converges to the identity operator strongly on $D(T)$ as $\lambda \to \infty$.

3. Since $\lambda(\lambda +T)^{-1}$ is uniformly norm bounded for $\lambda \in (0,\infty)$ according to Step 1, it follows from a simple approximation argument that $\lambda (\lambda + T)^{-1}$ converges strongly to $I$ even on the closure of $D(T)$ as $\lambda \to \infty$.

4. Now assume in addition that $X$ is reflexive (for instance, a Hilbert space). Then the estimate from Step 1 implies that $T$ is densely defined, see for instance [1, Proposition 3.3.8]. So Step 3) shows that $\lambda (\lambda + T)^{-1}$ converges strongly to $I$ on the whole space $X$ as $\lambda \to \infty$.

5. Finally, note that $(I + T/n)^{-1} = n(n+T)^{-1}$ for each integer $n$, so the assertion you are looking for follows from Step 4 if use set $\lambda = n$ (as indicated by Sahiba Arora).

Reference: [1] Arendt, Batty, Hieber, Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (second edition, 2011).

Remark. The above argument works for single-valued operators. The assertion is false for multi-valued operators.