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What is the large class of operators for which one can define fractional powers? For example, we can consider an operator $A: D(A) \subset X \rightarrow X$, generator of an analytic semigroup on a Banach space $X$. Can we define the powers $(-A)^\alpha$ for $\alpha>0$ without additional assumptions? I found some references with some restrictions on the spectrum of $A$ or on the associated semigroup. I'm wondering if there is a general definition without further assumption. Any reference would be helpful.

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    $\begingroup$ Bochner's subordination works for general $C_0$ semigroups, and Balakrishnan's formula is, I think, even more general. [C. Martínez, M. Sanz, The Theory of Fractional Powers of Operators. North-Holland Math. Studies 187, Amsterdam, 2001] is an outstanding reference. $\endgroup$ Jan 2 '20 at 22:20
  • $\begingroup$ Thank you! For the restrictions I meant, the analytic semigroup needs to be exponentially stable. I'm wondering if I can omit this assumption. An other question: are all these definitions equivalent? $\endgroup$
    – Migalobe
    Jan 3 '20 at 14:41
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    $\begingroup$ For Bochner's subordination, it is sufficient to assume that $A$ generates a $C_0$ semigroup of contractions (no need for exponential stability). I think "$\exp(tA)$ uniformly bounded" is fine, too. Bochner's subordination and Balakrishnan's formula indeed coincide in this setting. I do not think there is a general approach under significantly weaker assumptions; if $A = \varepsilon \operatorname{Id}$ for any $\varepsilon > 0$, there is no meaningful way to define $(-A)^\alpha$. $\endgroup$ Jan 3 '20 at 15:07
  • $\begingroup$ Thank you so much! $\endgroup$
    – Migalobe
    Jan 3 '20 at 21:21
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This question was considered by Functional Analysis specialists around 1960 or earlier. In the case of Banach Algebras $C(X)$ (for Hausdorff compact $X$) this gets reduced often to studying the auto-homeomorphisms of $X$, and it is extra interesting when the compact space is nice. In this context, in 1961, I have rediscovered that orientation preserving homeomorphisms of $\,X:=[0;1]\,$ admit square root (and quite a bit more but I didn't know at the time that the group of homeomorphisms was described already in full by someone else, sometimes earlier). This means that there are root squares (and similar) of the respective operators. I also constructed an orientation preserving homeomorphism of $\,\Bbb S^1\,$ which does not admit a square root hence the respective operator didn't either. The later result was presented in a monograph by Marek Kuczma, on functional equations, (which made me feel good, especially that I was a student at the time).

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    $\begingroup$ Thank you for your answer. I'm interested especially to operators in semigroup framework. $\endgroup$
    – Migalobe
    Jan 3 '20 at 14:38

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