monotone relations.

\[prop:pert\]Let $A\subseteq H\times H$ be maximal monotone and $B:H\to H$ Lipschitz continuous. Moreover, assume that $A+B\coloneqq\left\{ (x,y+Bx)\,;\,(x,y)\in A\right\} $ is monotone. Then $A+B$ is maximal monotone.

For $B=0$ there is nothing to show. So assume that $B\ne0.$ By it suffices to prove that for $0<\lambda<\frac{1}{|B|_{\mathrm{Lip}}}$ and each $z\in H$ there exists $u\in H$ with $(u,z)\in1+\lambda(A+B).$ The latter is equivalent to $$u=(1+\lambda A)^{-1}(z-\lambda Bu).$$ Since $(1+\lambda A)^{-1}$ is Lipschitz-continuous with $|(1+\lambda A)^{-1}|_{\mathrm{Lip}}\leq1$ and $|\lambda B|_{\mathrm{Lip}}<1$ by the choice of $\lambda,$ the assertion follows from the contraction mapping theorem.

Let $B:H\to H$ be monotone and Lipschitz-continuous. Then $B$ is maximal monotone.

In particular, if $B\subseteq H\times H$ is maximal monotone, then its Yosida-approximation $B_{\lambda}$ is Lipschitz-continuous and monotone and hence, maximal monotone. Moreover, if $A\subseteq H\times H$ is maximal monotone, then so is $A+B_{\lambda}$ for each $\lambda>0$ and hence, $(1+A+B_{\lambda})^{-1}$ is a Lipschitz-continuous mapping. This observation can be used to prove the following perturbation result.

\[prop:pert-yosida\] Let $A,B\subseteq H\times H$ be two maximal monotone relations. Moreover, let $z\in H.$ Then there exists $u\in H$ such that $(u,z)\in1+A+B$ (i.e., there are $v,w\in H$ with $(u,v)\in A,(u,w)\in B$ and $u+v+w=z$) if and only if $$\sup_{\lambda>0}\left|B_{\lambda}(u_{\lambda})\right|<\infty$$ where $u_{\lambda}\coloneqq(1+A+B_{\lambda})^{-1}(z).$ In the latter case, $u_{\lambda}\to 