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This question has the same flavor of this and this questions, but asks for something stronger.

Assume that

  • $A$ is a symmetric $n \times n$ matrix,
  • $H$ is a $n \times n$ perturbation matrix.

Moreover let

  • $A = U \Lambda U^\top$ and $A+H = V \Sigma V^\top$, and
  • $f: \mathbb R \rightarrow \mathbb R$ be a $L$-Lipschitz function, extended to diagonal matrices so that $f(D) = \operatorname{Diag}(f(D_{11}) \ldots f(D_{nn}))$.

I would like to prove that there exists a universal constant $C$ independent of $n$ such that $$ \|U f(\Lambda) U^\top - V f(\Sigma) V^\top\|_2 \leq C L \|H\|_2 $$ where $\|\cdot\|_2$ is the Euclidean operator norm.

Thanks to Weyl's inequality we have that $$ \|\Lambda - \Sigma\|_2 \leq \|H\|_2 $$ and thus $$ \|f(\Lambda) - f(\Sigma)\|_2 \leq L \|H\|_2. $$ In other words, we have Lipschitz continuity of eigenvalues. On the other hand, continuity of eigenprojections is established in

Kato, T.Perturbation theory for linear operators, Grundlehren der mathematischen Wissenschaften, 132. Berlin-Heidelberg-New York: Springer-Verlag, pp. XXI+619, 1976, MR0407617, Zbl 0342.47009.

However, I was not able to find a quantitative result on convergence speed of eigenprojections. I believe that a quantitative result on convergence speed of eigenprojections would help me to prove the statement above, I explain why in detail below.

Does anyone know about a quantitative result of this kind? Do you think there is an easier way to prove the result above?

Proof Sketch. Here is why I believe that understanding convergence speed of eigenprojections should lead to prove the result above.

The intuition that suggests me that the above results hold is the following. If $\lambda_i$ is not well-separated from $\lambda_{i+1} \leq \lambda_i$ (say $|\lambda_i - \lambda_{i+1}| < \epsilon$), then the eigenprojection relative to $\sigma_i$ cannot converge to the eigenprojection relative to $\lambda_i$ at scale $\|H\|_2 \approx \epsilon$ because we cannot distinguish between the eigenspaces relative to $\lambda_i$ and $\lambda_{i+1}$ with so much noise. However, the continuity of eigenprojections ensures that the eigenvectors relative to a simple eigenvalue eventually converge. Thus, it seems plausible to conjecture that the convergence speed should be dictated by the spectral gaps $|\lambda_i - \lambda_{i+1}|$ for $i = 1 \dots n - 1$.

Moreover, I conjecture that the only phenomenon preventing convergence is the one above. More specifically, I conjecture that at scale $\|H\|_2 \approx \epsilon$ the eigenvector relative to $\sigma_i$ is, up to error $C \cdot \epsilon$, contained in the span of eigenvectors relative to $\lambda_j$ such that $|\lambda_j - \lambda_i| \leq C \cdot \epsilon$.

Moreover, whenever $|\lambda_i - \lambda_{i+1}| \approx \epsilon$ then $|f(\lambda_i) - f(\lambda_{i+1})| \approx L \epsilon$, thus applying a Lipschitz function makes up for not-yet converged eigenvectors. So my idea of proof goes like this: ''if the eigenvalues are well-separated, eigenprojections converge. If they are not then f is almost constant''.

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I don't know about the fine details of the actual question, but can provide some (perhaps unhelpful) context: a related general notion is that of operator differentiable functions, defined as those $f\in C(I)$ for which $T\mapsto f(T)$ is Frechet differentiable on bounded self-adjoint operators.

The review article by Pedersen in Publ. RIMS, Kyoto Univ. 36 (2000), 139-157, discusses the topic in some depth. As one would expect, nice functions are operator differentiable while nasty ones aren't, but the details are somewhat subtle: for example, $f\in C^2(I)$ is sufficient and $f\in C^1(I)$ is necessary for $f$ to be operator differentiable.

This means that you obtain the desired estimate if you make the stronger assumption $f\in C^2$, but that is probably cracking a nut with a sledge-hammer, and it's not so clear either if the subtleties of the theory are even relevant here since you are dealing with matrices, not self-adjoint operators on an infinite-dimensional Hilbert space.

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