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.
 A: 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.
A: I was assigned this problem on a homework, and came across this question. I had much the same approach as you, and ran into the same stumbling blocks. I'll leave my solution, which can then be completed with your work.
Suppose $\{f_n\}\subseteq \operatorname{Dom}$ so that $$f_n \to f \text{  and  } (A+B)f_n \to g$$
To conclude that $(A+B)$ is closed, we need to show that $f \in \operatorname{Dom}$ and that $(A+B)f=g$. We use both the closedness of $A$ and $a \in [0,1)$ to show this. Define $g_{n,m}:=f_n-f_m$ Then
$$\|Ag_{n,m}\|\leq \|(A+B)g_{n,m} \|+ \|Bg_{n,m}\|\leq \|(A+B)g_{n,m} \|+ a\|Ag_{n,m}\|+b\|g_{n,m}\|$$
Subtracting $a\|Ag_{n,m} \|$ from both sides, we have
$$(1-a)\|Ag_{n,m}\|\leq\|(A+B)g_{n,m} \|+ b\|g_{n,m}\|$$
Note that $(1-a)>0$. The right hand side approaches $0$ as $n,m \to \infty$, so we also have that the left hand side also approaches $0$. Therefore $\{Af_n\}$ is Cauchy, and converges by completeness of $X$. By applying closedness of $A$, we have that $f \in \operatorname{Dom}$ and $Af_n \to Af$, which is the last bit needed for your proof.
