An operator $A$ is called dissipative if for all $x \in D(A)$ and $\lambda >0$


$$ \left\lVert (A-\lambda)x \right\rVert \ge \lambda \left\lVert x \right\rVert.$$


On a Hilbert space this is equivalent to saying that $\Re\langle Ax,x\rangle \le 0.$


In particular, if the spectrum of $A$ fulfills $\sigma(A)\subset (-\infty,0]$, then $A$ is dissipative.


I ask: Is the same true on Banach spaces, i.e. is any operator $A$ on a Banach space with$\sigma(A)\subset (-\infty,0]$ dissipative?



Dissipative operator on Banach spaces
No, and it's not true on Hilbert space either.
For example, on $\mathbb C^2$ or $\mathbb R^2$ try
$$ A = \pmatrix{0 & 0\cr 1 & 0\cr},\ x = \pmatrix{1\cr -1\cr},\ \lambda = 1$$
The spectrum is $\{0\}$, but $\|(A - \lambda) x\| = 1 < \sqrt{2} = \lambda \|x\|$.


