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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?

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  • $\begingroup$ 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. $\endgroup$ – Rhys Steele Feb 16 at 23:43
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We have the following small perturbations lemma (which I'll prove here in your special case, see e.g. Theorem 11.1.3 in "Linear Operators and their Spectra" by Davis for a more general result).

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.

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