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Please I need help with an example I cant figure out and which will hopefully help me to get the theory:

Let $X$ be Banach space and $A, B$ general operators. Furthermore $A$ is closed, $\operatorname{dom}(A) \subset \operatorname{dom}(B)$, there are $a,b; \quad a\ge 0, \; 0 \lt b \lt 1$
$\forall x \in \operatorname{dom} (A); \quad \lVert Bx \rVert \le a\lVert x\rVert +b\lVert Ax\rVert $

I need to:

  • find $\operatorname{dom}(A+B)$
  • show that $\forall x \in \operatorname{dom}(A), \quad \lVert Ax\rVert \le \frac{1}{1-b}\lVert (A+B)x\rVert +\frac{a}{1-b}\lVert x\rVert$
  • show that $A+B$ is closed

Thanks a lot, i would like to finally get it, but i probably cant do it alone.

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You can find a discussion of this and many examples in Chapter IV, §1, sections 1 and 2 in Kato, Perturbation theory for linear operators. (I think you want to assume $0 \lt b \lt 1$). –  Martin Jun 23 '13 at 21:57
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up vote 3 down vote accepted

$\DeclareMathOperator{dom}{dom}$The domain of a sum $A + B$ of unbounded operators is by definition the intersection of the domains of $A$ and $B$: $\dom(A+B) = \dom A \cap \dom B$ and by hypothesis $\dom A \subseteq \dom B$, so $\dom(A + B) = \dom A$.


For all $x \in \dom{A}$ we have $$\tag{$\ast$} \lVert (A+B)x \rVert \leq a \lVert x \rVert + (1+b)\lVert Ax\rVert $$ by the triangle inequality and the hypothesis that $\lVert Bx\rVert \leq a\lVert x\rVert + b\lVert Ax\rVert$. On the other hand, $$ -a \lVert x \rVert + (1-b)\lVert Ax\rVert = \lVert Ax\rVert - (a\lVert x\rVert + b\lVert Ax\rVert) \leq \lVert Ax\rVert - \lVert -Bx\rVert \leq \lVert (A + B)x\rVert $$ which immediately yields for all $x \in \dom A$ the inequality $$\tag{${\ast}{\ast}$} \lVert Ax \rVert \leq \frac{a}{1-b}\lVert x\rVert + \frac{1}{1-b}\lVert (A+B)x\rVert $$ of your second bullet since $1 -b \gt 0$.


Let $x_n \in \dom A = \dom(A+B)$ be a Cauchy sequence. Then $Ax_n$ is a Cauchy sequence if and only if $(A+B)x_n$ is a Cauchy sequence by $(\ast)$ and by $({\ast}{\ast})$.

The same estimates show that for a sequence $x_n \to 0$ with $x_n \in \dom A$ we have $Ax_n \to 0$ if and only if $(A+B)x_n \to 0$. In particular, $A$ is closable if and only if $A+B$ is closable.

If we now assume that $A$ is closed, then we know that $A+B$ is closable. Let us show that $A+B$ is closed.

Suppose $x_n \in \dom A$ is such that $x_n \to x$ and $(A+B)x_n \to y$. Then $(A+B)x_n$ is Cauchy, so $Ax_n$ is Cauchy, too. Therefore $Ax_n \to z$. It follows that $x \in \dom A$ and $Ax = z$ because $A$ is closed. But as $\dom A = \dom(A+B)$, we know that $x \in \dom(A+B)$ and since we already know that $(A+B)$ is closable we must have $(A+B)x = y$ (otherwise $(x_n -x)$ would be a null sequence in $\dom(A+B)$ with $(A+B)(x_n - x) \nrightarrow 0$ contradicting closability of $A+B$).


Further results in this direction and many examples can be found in Kato's Perturbation theory for linear operators, Chapter IV, §1, sections 1 and 2.

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