# Sum of closed operator and linear operator is closed under additional constraint

Let $X$ be a Banach space, $A:\operatorname{Dom}\subset X\to X$ a closed operator and $B:\operatorname{Dom}\to X$ a linear operator. If there exist constants $a\in[0,1)$ and $b\geqslant 0$ such that $$\|Bx\|\leq a\|Ax\|+b\|x\| \quad \forall x\in \operatorname{Dom}$$ then the operator $C=A+B:\operatorname{Dom}\to X$ is closed.

Closedness of $C$ means that if $(x_n)_n\subset \operatorname{Dom}$ with $\lim_n x_n=x$ and $\lim_n Cx_n=y$, then $x\in \operatorname{Dom}$ and $Cx=y$.

By closedness of $A$, we have $x\in \operatorname{Dom}$. So we still need to show that $Cx=y$. I tried doing this as follows:

$$\begin{eqnarray} \|Cx-y\|&\leq& \|Cx-Cx_n\|+\|Cx_n-y\| \\&\leq&\|Ax-Bx-Ax_n+Bx_n\| + \|Cx_n-y\| \\&\leq& \|A(x-x_n)\| + \|B(x-x_n)\| + \|Cx_n-y\| \\&\leq& \|A(x-x_n)\| + a \|A(x-x_n)\| + b \|x-x_n\| + \|Cx_n-y\| \\&=& (1+a)\|A(x-x_n)\|+b \|x-x_n\| + \|Cx_n-y\| \end{eqnarray}$$

The last two terms are fine, they tend to $0$. Can we say anything about the first term?

In particular, we haven't used the full force of closedness of $A$ yet. Nor am I sure how the fact that $a<1$ comes into play.

I really appreciate help.

## 1 Answer

Closedness

Characterization: $$A=\overline{A}\iff\mathcal{D}_A=\hat{\mathcal{D}_A}$$ Operatornorm: $$\mathcal{D}_A:=\mathcal{D}(A):\quad\|\varphi\|_A:=\|A\varphi\|+\|\varphi\|$$

Equivalence

Equivalent Norms: $$\alpha_x\vee\alpha_y\left(\|\varphi\|_x+\|\varphi\|_y\right)\leq\alpha_x\|\varphi\|_x+\alpha_y\|\varphi\|_y\leq\alpha_x\wedge\alpha_y\left(\|\varphi\|_x+\|\varphi\|_y\right)$$

Suppose one has: $$\Delta A:=A-A_0:\quad\|\Delta A\varphi\|\leq\alpha\|A_0\varphi\|+\beta\|\varphi\|$$

Then one has estimates: $$\|A\varphi\|+\|\varphi\|\leq\|\Delta A\varphi\|+\|A_0\varphi\|+\|\varphi\|\leq(1+\alpha)\|A_0\varphi\|+(1+\beta)\|\varphi\|$$ $$\|A\varphi\|+(1+\beta)\|\varphi\|\geq-\|\Delta A\varphi\|+\|A_0\varphi\|+(1+\beta)\|\varphi\|\leq(1-\alpha)\|A_0\varphi\|+\|\varphi\|$$

Concluding equivalence.