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?

  • 1
    $\begingroup$ No, but why would you want it? The numerical range condition is easier to check than the spectral condition. And it has a nice generalization to Banach spaces: en.wikipedia.org/wiki/… $\endgroup$ Commented Sep 11, 2017 at 7:24

1 Answer 1


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 $\|(A - \lambda) x\| = 1 < \sqrt{2} = \lambda \|x\|$.

  • $\begingroup$ I see, I need a self-adjoint operator in this case, but there is probably no replacement for that on Banach spaces, right? $\endgroup$
    – Zinkin
    Commented Sep 11, 2017 at 7:13

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