
> 
> Assume that $ A $ is self-adjoint operator and $ B $ is a bounded self-adjoint operator. The definite domain of $ A,B $, denoted by $ D(A) $ and $ D(B) $ satisfies $ D(A)\subset D(B) $. Show that
> \begin{align\*}
> \operatorname{dist}(\sigma(A),\sigma(A+B))\leq \|B\|,
> \end{align\*}
> where $ \sigma(A) $ and $ \sigma(A+B) $ are the spectrums of $ A $, $ A+B $ and $ \|B\| $ is the operator norm of $ B $.
> 
> 
> 


It can be obtained by utilizing the Kato-Rellich theorem that $ A+B $ is also a self-adjoint operator. However I cannot go on since the decomposition of $ A $ and $ A+B $ are not the same. Can you give me some hints or refereces?



Disturbance of self-adjoint operator
Let $C=A+B$, take a point, say $0$, in the spectrum of $A$ and assume that $[-\|B\|, \|B\|]$, hence $[-\|B\|-\epsilon, \|B\|+\epsilon]$ for some $\epsilon>0$ is contained in the resolvent set of $C$. The spectral theorem, applied to $C$ implies that $\|C^{-1}\| \leq (\|B\|+\epsilon)^{-1}$. Then $A=C-B=(I-BC^{-1})C$ would be invertible because $\|BC^{-1}\| <1$.


