difference of operators defined through 2 different functional calculi.

Question

Lets say, we have a fixed function $f$ and two normal/self-adjoint operators $\mathcal{L} ,\tilde{\mathcal{L}} $ with discrete spectrum. Are there some conditions (on $f$ or the normal operators , such as lipschitz continuity of f) that ensure: \begin{equation} \label{ineq: pertOfLaplCont} \| f(\mathcal{L}) - f(\tilde{\mathcal{L}}) \| \leq C \cdot \| \mathcal{L} - \tilde{\mathcal{L}} \| \end{equation} Here $f(\mathcal{L}) $ shall describe the functional calculus defined through $\mathcal{L} $ and so on.

Thanks for answers!

Answer 1

This an interesting question, which has been studied quite a lot by the Russian operator theory school (Birman, Solomyak, Peller ...).

A function $f\colon I\to \mathbb{C}$ with the property that there exists $C>0$ such that $\|f(S)-f(T)\|\leq C\|S-T\|$ for all normal operators with spectrum contained in $I$ is called operator Lipschitz. It is well-known that the set of operator Lipschitz function (say, on $\mathbb{R}$) is strictly smaller than the set of Lipschitz functions. For example, the absolute value fails to be operator Lipschitz. In fact, every operator Lipschitz function on $\mathbb{R}$ is differentiable.

A sufficient condition for a function $f$ on $\mathbb{R}$ to be operator Lipschitz is that it belongs to the Besov class $B^1_{\infty,1}(\mathbb{R})$, a slightly more elementary one is that the derivative of $f$ is the Fourier transform of a complex Borel measure on $\mathbb{R}$.

In contrast, operator Lipschitz functions on the complex plane (i.e. those for which all normal operators are allowed as arguments) are quite boring - they are all of the form $f(z)=az+b$.

If you want to learn more, I can recommend the article Aleksandrov, Peller. Operator Lipschitz functions, arXiv:1611.01593. There you can find proofs for the results I stated.

Answer 2

Since OP requests this in the comment section:

Let $A$ be a unital $C^*$-algebra and $f:\mathbb{R}\to\mathbb{C}$ a continuous function. Show that the map $A_{sa}\to A$ with $a\mapsto f(a)$ is continuous.

Proof:

First note that if $f$ is any polynomial the result is immediate, since scaling, adding and multiplying are all continuous operations. Now suppose that $f$ is an arbitrary continuous function and let $(a_n)\subset A_{sa}$ with $a_n\to a$ (hence $a\in A_{sa}$). Let $\varepsilon>0$ and set $$E=\bigcup_{\lambda\in\sigma(a)}(\lambda-\varepsilon,\lambda+\varepsilon).$$ Due to convergence, there exists $n_0$ such that for all $n\geq n_0$ it is $\|a_n-a\|<\varepsilon$. For such $n$, if $\lambda\in\sigma(a_n)$, we have that $\lambda\in E$: if $\lambda\not\in E$, then $dist(\lambda,\sigma(a))\geq\varepsilon$. Remember that if $u\in\text{inv}(A)$, then $B(u,1/\|u^{-1}\|)\subset\text{inv}(A)$ (see the proof that $\text{inv}(A)$ is open in a unital Banach algebra). Now since $dist(\lambda,\sigma(a))\geq\varepsilon$, we have that $a-\lambda\in\text{inv}(A)$ and $\|(a-\lambda)^{-1}\|=1/dist(\lambda,\sigma(a))$. Hence $B(a-\lambda,\varepsilon)\subset B(a-\lambda, dist(\lambda,\sigma(a))\subset\text{inv}(A)$. But $\|(a_n-\lambda)-(a-\lambda)\|<\varepsilon$, a contradiction. So we have that for all $n\geq n_0$ it is $\sigma(a_n)\subset E\subset\overline{E}$ and $\overline{E}$ is obviously a compact subset of the real line.

Keep in mind that $f(b)$ makes sense as $f\vert_{\sigma(b)}(b)$ (if our notation is extremely strict). Now since $\overline{E}$ is compact, $f\vert_{\overline{E}}$ can be approximated uniformly on $\overline{E}$ by a polynomial $p$, i.e. $\|f\vert_{\overline{E}}-p\|_\infty<\varepsilon$. Also, by the continuous functional calculus (that is isometric), we have $\|p(c)-f(c)\|<\varepsilon$ for all $c$ that have $\sigma(c)\subset\overline{E}$. Moreover, for this polynomial we have $p(a_n)\to p(a)$, so pick $n_1$ such that for all $n\geq n_1$ it is $\|p(a_n)-p(a)\|<\varepsilon$. Then we have for $n\geq\max\{n_0,n_1\}$: $$\|f(a_n)-f(a)\|\leq\|f(a_n)-p(a_n)\|+\|p(a_n)-p(a)\|+\|p(a)-f(a)\|=3\varepsilon$$ and we are done.

Index: $\sigma$ denotes the spectrum, dist is the distance of a number and a set, we use B for open balls and inv is the set of invertibles.

I have only seen this as an exercise years ago, this is my own solution so I'm afraid I don't have any references for this..