1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.