Define the Frobenius norm of a matrix as $\left\Vert A \right\Vert_{\mathrm{F}}=\sqrt{\sum_{i,j} A_{ij}^2}$ and the operator norm as $\left\Vert A \right\Vert_{\mathrm{op}}=\sup_{x \not = 0} \frac{\left\Vert Ax\right\Vert_2}{\left\Vert x \right\Vert_2}$ where the the norm in the numerator and denominator are the standard Euclidean norm.

If $\mathbf{N}$ and $\mathbf{M}$ are normal matrices on a separable complex Hilbert space $H$, and $f$ is a Lipschitz function defined on the spectrum of both matrices $\Omega = \sigma(\mathbf{N}) \cup \sigma(\mathbf{M}) $ with Lipschitz constant $k$ then $\left\Vert f(\mathbf{N}) - f(\mathbf{M})\right\Vert_\mathrm{F} \leq k\left\Vert \mathbf{N} - \mathbf{M}\right\Vert_\mathrm{F}$. This is a result from Kittaneh (1985). Note that in this paper they use the notation $\left\Vert \cdot \right\Vert_2$ to be the Hilbert–Schmidt operator which I believe is the Frobenius norm in the finite dimensional case.

I found this recent survey paper on operator Lipschitz functions which states similar results. From what I can tell the norm isn't specified but I think the results are also for the Frobenius norm.

I would like to know

  1. Is the term "operator Lipschitz function" reserved for functions which have the property under the Frobenius norm, or is it a more general concept defined for any norms?
  2. Does the result from Kittaneh (1985) hold for operator norms and if so what is the reference? (The proof in this paper seems specific to the Frobenius norm). The review paper says for equation 3.1.2 "It follows easily from the spectral theory for pairs commuting normal operators", I'm not quite sure what this is and if it holds for norms generally (I haven't had much much formal training in functional analysis or measure theory)

Specifically for my research I have real symmetric matrices $\mathbf{N}$ and $\mathbf{M}$ with eigenvalues lying in the interval $[-1, 1]$. If I have a Lipschitz continuous function $f:[-1,1] \rightarrow \mathbb{R}$ (we can also add differentiability or infinitely differentiability as an assumption if it helps) does it hold that $\left\Vert f(\mathbf{N}) - f(\mathbf{M})\right\Vert_{\mathrm{op}} \leq k\left\Vert \mathbf{N} - \mathbf{M}\right\Vert_{\mathrm{op}}$ where $k$ is the Lipschitz constant of $f$?

I hope this question isn't too basic for MO, I asked on the mathematics SE but didn't get a response.

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    $\begingroup$ I’m pretty sure that $\lVert \cdot \rVert_F = \lVert \cdot \rVert_2$, with F standing for Frobenius. That said, I get the impression that “Frobenius inner product” is more frequently used to refer to the finite-dimensional special case of the Hilbert–Schmidt inner product than it is as a true synonym. $\endgroup$ – Branimir Ćaćić Oct 5 '20 at 12:41
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    $\begingroup$ In my question I'm using $\left\Vert A \right\Vert_F = \sqrt{\sum_{i,j}A_{i,j}^2}$ as the Frobenius norm and $\left\Vert A \right\Vert_2 = \sup_{x \not = 0} \frac{\left\Vert Ax \right\Vert_2}{\left\Vert x\right\Vert_2}$ where the norms for these vectors are the Euclidean norm. In Kittaneh they write $\left\Vert A \right\Vert_2$ as the Hilbert–Schmidt operator which I think is the Frobenius norm in the finite case (it also looks like they're using the Frobenius norm as I've defined it in the proof) $\endgroup$ – Henry Oct 5 '20 at 13:33
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    $\begingroup$ I think the inequality $\|f(M)-f(N)\|\le k \|M-N\|$ must definitely be true when enough assumptions on $f$ are made ($f\in C^2$ should work) because then calculus techniques for the map $M\mapsto f(M)$ become available. See Pedersen, Operator differentiable functions (2000) (I don't have the paper available right now, so can't check the details). $\endgroup$ – Christian Remling Oct 5 '20 at 13:57
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    $\begingroup$ The survey you mention is about Lipschitz estimates for the operator norm, at least most of it is. As Christian Remling mentions, for reasonably smooth functions there is some estimate, but it is usually more complicated than simply taking the Lipschitz constant of a function. $\endgroup$ – Mateusz Wasilewski Oct 5 '20 at 14:29
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    $\begingroup$ Henry, regarding the equations in the "review paper" you keep mentioning, there is a vital extra assumption that the operators commute with each other. Without that assumption the result you want fails to hold (indeed, if it didn't fail, then "operator Lipschitz" functions would just be "Lipschitz functions") $\endgroup$ – Yemon Choi Oct 5 '20 at 14:41

The term "operator Lipschitz function" is definitely not reserved to the Hilbert-Schmidt norm. On the opposite, I would say that it is mostly used for the operator norm (but not only, see for example https://arxiv.org/abs/0904.4095 ). In particular, the survey that you are citing is using the operator norm.

It is known that not every Lipschitz function is operator Lipschitz (for the operator norm). So the answer to you question 2) is negative. I think that this was a conjecture of Krein, and that the first countexample was produced by Farforovsakaya (1972). A very simple example (Davies and Kato) is given by the absolute value map, which is $1$-Lipschitz but only $\simeq \log(n)$-Lipschitz on $M_n(\mathbf{C})$ (for the operator norm). This is very closely related to the fact that the triangular truncation has norm $\simeq \log(n)$ on $M_n(\mathbf{C})$. In fact, any $1$-Lipschitz function is $O(\log n)$-Lipschitz on $M_n(\mathbf{C})$.

In another direction, sufficient conditions are known to ensure that a function is operator-Lipschitz (for the operator norm), in terms of Besov spaces (Peller, see the survey you cite). It is also known that Lipschitz functions are $O(\max(p,1/(p-1))$-Lipschitz on $S^p$, the Schatten $p$-class. This is the paper I cite above by Potapov and Sukochev.

Note: I am not sure, but I would expect that the fact that Lipschitz functions remain Lipschitz for the Hilbert-Schmidt norm was already known to Krein in the 1960's.

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    $\begingroup$ Since the estimate for the Hilbert-Schmidt norm follows from the basic theory of double operator integrals, you guess about Krein in the last paragraph seems quite reasonable. $\endgroup$ – MaoWao Oct 5 '20 at 14:46
  • $\begingroup$ Thank you for your answer Mikael. I'll look more in Besov spaces which may help me characterise functions for my purposes $\endgroup$ – Henry Oct 7 '20 at 9:40

There are several inaccuracies or mistaken impressions in the OP's original question – I say this not to be denigrating, since it is always tricky reading the literature in a different mathematical community from one's own. However it seems worth writing an answer rather than "sniping in comments".

First of all, Aleksandrov and Peller have written many papers on this theme, but in their setting they almost always mean the operator norm not the Frobenius norm. Hence, when you link to their paper https://arxiv.org/abs/1611.01593 you cannot use it to say things about the Kittaneh paper.

Regarding Q1. My impression is that among the infinite-dimensional functional analysis community, "operator Lipschitz" is interpreted with respect to the operator norm. This is for instance what is meant by the Aleksandrov—Peller paper that you linked to, contrary to your impression.

Q2. Note that the equation in A+P that you refer to is for commuting operators only! When two matrices commute and are normal in the technical sense of that word, they can be simultaneously diagonalized (over the complex numbers) and hence estimates of the form you seek can be obtained directly. The difficulty, in both the Kittaneh paper and the series of papers by A+P, lies in the fact that the given matrices might not commute with each other.

Finally, there is an example going back to Kato, and refined in work of Davies I think, which shows that the absolute value function is not "operator Lipschitz". So in full generality the answer to Q2 is no. (I don't have the references to hand but will try to update this later)

  • $\begingroup$ I think Mikael de la Salle's answer has said what I was trying to get at, but with more precise details $\endgroup$ – Yemon Choi Oct 5 '20 at 14:41
  • $\begingroup$ Yes, we more or less wrote the same answer... $\endgroup$ – Mikael de la Salle Oct 5 '20 at 14:43
  • $\begingroup$ Hi Yemon, thank you for your answer. I appreciate your patience with my lack of understanding in this field. It's very useful for me to know that the papers of Aleksandrov and Peller are about the operator norm. I understand that Lipschitz might not imply operator Lipschitz and even in the cases a function is both the constants may not be the same. I don't think I communicated it well but my question is what conditions can I find so that this is the case. Theorem 3.14.1 in the survey gives me a necessary condition for the constants to be equal, but I'm really after some sufficient conditions $\endgroup$ – Henry Oct 7 '20 at 9:36
  • $\begingroup$ @Henry No worries! Actually, I think you should probably "accept" Mikael's answer instead of mine since he actually provided more precise references. In particular, the Besov space approach that Aleksandrov and Peller looks like it might provide what you need. $\endgroup$ – Yemon Choi Oct 7 '20 at 20:18

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