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Given $n \in \mathbb{N}$ we define the function $f_{i,n}: [0,1] \rightarrow \mathbb{R}$ for $i \in \{1,..., n\}$ by $f_{i,n} = 0$ on the interval $[0,(i-1)/n]$, $f_{i,n} = 1$ on $[i/n,1]$, and $f_{i,n}$ is linear on [(i-1)/n, i/n]. Let $A \in M_k(\mathbb{C})$ be a self-adjoint positive semi-definite matrix and $U$ a unitary matrix also of dimension $k$.

I want to know if there exists a constant $C$ independent of the matrix size $k$ such that the following inequality holds: $$\frac{1}{n}\sum_{i=1}^n ||f_{i,n}(U^*AU) - f_{i,n}(A)||_{1,tr} \leq C ||U^*AU -A||_{1,tr} $$ With $||.||_{1,tr}$ I mean the norm on $M_k(\mathbb{C})$ defined by $||A||_{1,u} = tr((A^*A)^{1/2})$, where $tr$ denotes the normalized matrix trace. For the problem I'm interested in it's also okay if the inequality holds when taking the $\limsup$ for $n \rightarrow \infty$ of the left hand side.

I have already used NumPy to check this numerically, and the results I get seem to indicate that such a $C$ might indeed exist. I did calculations for matrix sizes 5-30. For each matrix size, I generated 20 random matrices $A$ and $U$. Visually I saw that the values on the left hand side tended to stabilize when $n$ became large enough (certainly when approaching 60). For each matrix combination I calculated the average of the values of the left hand side for $n$ between 55 and 60 and then divided by $||U^*AU-A||_{1,tr}$. The following link shows a plot of my results:

X-axis show matrix size, y-axis shows the maximum and minimum of the 20 ratios I calculated for that matrix size. Repeatedly running this program always gives a graph showing very similar behavior. I'm not an expert in numerical simulations so I don't know if there is a better way to simulate this, but the plot seems to demonstrate that such a $C$ might exist, and that for example $C=2$ could already work.

I'm stuck when trying to prove this rigorously and was wondering if someone could help me with this question. I'm a $C^*$-algebraist myself and the reason I'm interested in this inequality is that if it holds, it would allow me to perform a functional calculus trick in certain $C^*$-algebras. In reality I want to obtain the above inequality for elements in more general classifiable $C^*$-algebras, but for this it is enough to solve the problem for matrices with the constant $C$ not depending on the matrix size.

For now the furthest I have gotten is showing that $$\frac{1}{n}\sum_{i=1}^n ||f_{i,n}(U^*AU) - f_{i,n}(A)||^2_{2,tr} \leq ||U^*AU -A||_{2,tr} $$ using 10.3.3 and ideas from 10.3.4 in https://www.math.ucla.edu/~popa/Books/IIun.pdf. Here $||A||_{2,tr} = \sqrt{tr(A^*A)}$. I seem to struggle with one-norm calculations because of the absolute value appearing in them.

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    $\begingroup$ Well, it seems Connes says it's not possible if $U^*AU$ is replaced with another positive semidefinite $B$. See Remark 1.2.7 in his Annals paper 1976. doi.org/10.2307/1971057 $\endgroup$ Commented Jul 9, 2021 at 6:13
  • $\begingroup$ Thanks. I realize now that I actually wanted the inequality to hold in this more general sense, but it seems from the reference to Connes' paper that it is impossible. I'm not too familiar with the $L^1(N,\tau)$ appearing in Remark 1.2.7 so I wanted to verify if I understood correctly why this remark shows it's impossible. Is the reasoning behind this that if a constant $K$ would exist such that the inequality holds for all finite rank operators in $\mathcal{B}(\mathcal{H})$, then the inequality would also hold for all of $L^1(\mathcal{B}(\mathcal{H}))$, contradicting remark 1.2.7? $\endgroup$ Commented Jul 9, 2021 at 16:45
  • $\begingroup$ Yes, that's my understanding. $\endgroup$ Commented Jul 12, 2021 at 0:32

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