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Let $(H, (\cdot , \cdot), \Vert \cdot \Vert)$ be a Hilbert-space and $A \in \mathcal L(H)$ an accretive operator, that means $-A$ is dissipative. How can I show that the Operator $R_{\lambda}:=(I + \lambda A)^{-1}$ (where $I:H \to H$) is the identical mapping) exists in $H$ as a linear and bounded operator for all $\lambda \geq 0$ and is nonexpansive, that means that for all $u,v \in H$ the inequality $$\Vert R_\lambda u-R_\lambda v\Vert \leq \Vert u-v\Vert$$ holds?

For $\lambda \neq 0$ one should consider the recursion $$\frac{u^{n+1}-u^n}{\tau} +(I+\lambda A)u^n=f$$ to the solution of $(I+\lambda A)u=f$ for an arbitrary $u^0 \in H$ and a suitably chosen $\tau > 0$.

Some help to get things going is much appreciated.

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    Are you using the definition of dissipative operator as found on the Wikipedia page? If so, it can't be guaranteed that $R_\lambda$ is bounded unless you assume $-A$ is maximally dissipative – 2017-01-26

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