# Showing that $\lambda - (A + B)$ has dense range

Let $$A$$ be the generator of a $$C_0$$-semigroup $$(T(t))_{t \geq 0}$$ of contractions on a Banach space $$X$$ and $$B \in \mathcal L(X)$$ a bounded operator. To apply some approximation formula I want to show that there exist $$\lambda > 0$$ such that $$\lambda - (A + B)$$ has dense range, i.e., $$X= \overline{\operatorname{ran}}(\lambda - (A + B))$$ and I am not quite sure how to do that.

To begin with, I know that since $$(T(t))_{t \geq 0}$$ is a contraction semigroup, $$\lambda - A$$ is invertible for each $$\lambda > 0$$. Moreover, it holds $$R(\lambda, A + B) = R(\lambda, A) (I - BR(\lambda, A + B)) \tag{1}$$ for each $$\lambda \in \rho(A + B)$$. Since $$\Vert R(\lambda, A + B)) \Vert \to 0$$ as $$\vert \lambda \vert \to \infty$$, one should be able to find $$\lambda > 0$$ such that $$\Vert BR(\lambda, A + B) \Vert < 1$$. Hence, both operators $$\lambda - A$$ and $$BR(\lambda, A + B)$$ should be invertible for some $$\lambda > 0$$ that is large enough. Hence, due to $$(1)$$ $$\lambda - (A + B)$$ should be invertible as long as I can find a sequence $$(\lambda_n)_{n \in \mathbb N}$$ in $$\rho(A + B)$$ with $$\lambda_n \to \infty$$ as $$n \to \infty$$. In this case, the range of $$\lambda - (A + B)$$ would be $$X$$ for some $$\lambda > 0$$ big enough. But I am not sure how I can deduce the existence of such a sequence from scratch.

On the other hand, I think my argument is to difficult. Because it is a consequence of the well-known Hille-Yosida theorem that $$A+ B$$ generates a semigroup of type $$(1, \Vert B \Vert)$$. In this case, one would know that $$\lambda - (A + B)$$ is invertible for each $$\lambda > \Vert B \Vert$$, and therefore has dense range, by standard semigroup theory. But I hoped to avoid this consequence of Hille-Yoshida in my proof for sake of simplicity.

So my question is, if there is any chance to show that $$\lambda - (A + B)$$ has dense range in this setting without Hille-Yosida, maybe by results from spectral theory, for example?

• Your attempted argument is a bit strange. Note that once you've found any $\lambda_n \in \rho(A+B)$, $\lambda_n - (A+B)$ is invertible and hence is surjective so this assumption would already imply your desired conclusion. – Rhys Steele Feb 16 at 23:43

Lemma: If $$\lambda \in \rho(A)$$ and $$|c| < \|BR(\lambda,A)\|^{-1}$$ then $$\lambda \in \rho(A+cB)$$.
Proof of Lemma: Let $$R = R(\lambda, A)(\operatorname{Id} - cBR(\lambda,A))^{-1}$$ (which exists by a von Neumann series argument). Then \begin{align*} (\lambda - (A+cB))R &= ((\lambda - A) - cB) R(\lambda, A)(\operatorname{Id} - cBR(\lambda,A))^{-1} \\ &= (\operatorname{Id} - cBR(\lambda,A))((\operatorname{Id} - cBR(\lambda,A))^{-1} \\&= \operatorname{Id} \end{align*} $$\square$$
To conclude, we'd like to be able to show that we can find $$\lambda \in \rho(A)$$ such that we can plug $$c = 1$$ into the above. Since $$B$$ is bounded, this is straightforward. Indeed, $$\|BR(\lambda,A)\| \leq \|B\| \|R(\lambda,A)\| \to 0$$ as $$\lambda \to \infty$$ with $$\lambda \in \mathbb{R}$$ (where I use that $$(0, \infty) \subseteq \rho(A)$$). So for $$\lambda \in \mathbb{R}$$ large enough, $$\|BR(\lambda,A)\|^{-1}>1$$ and so we can take $$c = 1$$ in the lemma to see that $$\lambda \in \rho(A+B)$$ and hence $$\lambda - (A+B)$$ is surjective.