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I have been recently wondering what is a (or even the) "correct" generalization of the notion of an operator norm to nonlinear operators?

Please excuse the naivete of my question; if you think that question will benefit from being made more precise, then I will appreciate help towards making it so.

Because I lack formal education in mathematics, I might be missing something obvious or well-known here. Could somebody point me in the right direction, and let me know what are the key concepts to think about when defining operator norms for nonlinear operators?

Some vague ideas that occurred to me:

  1. Linearizing the operator (locally), so the essentially traditional operator norms of the linearized operator could be considered? This sounds very unsatisfactory though.

  2. If $A$ is a nonlinear operator for which we can sensibly define $\log A$, maybe that helps in tackling the nonlinearity.

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    $\begingroup$ Maybe the Lispschitz constant? $\endgroup$
    – Mikael de la Salle
    Commented Apr 6, 2011 at 14:41
  • $\begingroup$ Thanks Mikael; I guess linearizing takes us to Lipschitz? But for functions that are only locally Lipschitz? or which flavors of Lipschitz? Perhaps you'd care to expand a bit on your comment? $\endgroup$
    – Suvrit
    Commented Apr 6, 2011 at 15:36
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    $\begingroup$ A useful bit of perspective on this question is provided by recalling that for a linear operator $A$ between normed spaces, the following are all equivalent: $A$ is continuous, $A$ is uniformly continuous, $A$ is Lipschitz, $\| A \|$ is finite. Without linearity the equivalences all fail. As abatkai points out, how to generalize $\| A \|$ appropriately depends on the context, in particular which properties of $A$ are actually important for you. $\endgroup$
    – Mark Meckes
    Commented Apr 6, 2011 at 15:38
  • $\begingroup$ Usually a norm is used to turn a vector space into a normed space. Do you have a special vector space of nonlinear operators in mind? Probably the space of all nonlinear operators between two (normed?) spaces is too large... $\endgroup$
    – Dirk
    Commented Apr 6, 2011 at 19:39
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    $\begingroup$ Look at the book "Geometric nonlinear functional analysis" by Benyamini and Lindenstrauss. $\endgroup$
    – Bill Johnson
    Commented Apr 7, 2011 at 13:33

2 Answers 2

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Probably the answer depends on the context, where you need the generalized concept.

Unfortunately, usually the linearized operator is not bounded anymore, because it contains differential operators.

As pointed out by Mikael, the Lipschitz constant may be one possibility. For example, in the Crandall-Liggett theory of nonlinear semigroups the Hille-Yosida generation theorem on linear contraction semigroups is generalized to nonlinear contractions. Here, definitely the Lipschitz constant replaces the role of the operator norm. Or in the Banach fixed point theorem the convergence of the geometric series is generalized to iterations of nonlinear maps.

But there might be other answers, I am really curious.

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If $A:X\to Y$ is an operator that maps between the normed vector spaces $X$ and $Y$, with norms $\|\cdot\|_X$ and $\|\cdot\|_Y$, respectively, one can define $\|A\| \equiv \sup_{x\neq 0}\frac{\|A(x)\|_Y}{\|x\|_X}$. If the supremum is infinite, then the operator is unbounded, just as in the linear case. This is the definition used, for example, in "Control of Uncertain Systems" by Munther A. Dahleh and Ignacio J. Diaz-Bobillo.

I can't comment on whether such a definition is "correct", but it is certainly useful; for example in control theory, where you are interested in knowing whether your system - considered as a nonlinear operator between inputs and outputs - is stable, i.e., produces bounded outputs for bounded inputs, i.e., has bounded norm.

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    $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – Max Horn
    Commented Aug 17, 2023 at 21:23
  • $\begingroup$ Corrected to not be a question. $\endgroup$
    – Ben Francis
    Commented Aug 17, 2023 at 21:31
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    $\begingroup$ If $A$ is nonlinear, why should this supremum be finite in general? You hint it might be useful to do this "in some cases". Could you elaborate which cases you have in mind? $\endgroup$
    – Max Horn
    Commented Aug 17, 2023 at 21:38
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    $\begingroup$ @MaxHorn: And, if that quantity is finite, why is it a norm? $\endgroup$
    – Alex M.
    Commented Aug 17, 2023 at 22:02
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    $\begingroup$ There is no reason why the supremum need be finite, even for linear $A$; that's precisely the concept of an unbounded operator: an operator that does not have a finite norm. $\endgroup$
    – Ben Francis
    Commented Aug 22, 2023 at 22:08

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