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''.