# Can a sequence of trace class operators $\rho_{n} \in B(H)$ converge to a multiplication operator under trace norm?

Let $$H$$ be some infinite dimensional $$Hilbert$$ space. Now, let $$B(H)$$ be the set of all bounded linear operator over $$H$$ and let $$seq :=\{\rho_{n}\}_{n=1}^{\infty}$$ be a sequence of trace class operators in $$B(H)$$, the trace class operators form an ideal over $$B(H)$$.

-My first question is this. Under what conditions, if any, does the sequence $$seq$$ converge to a multiplication operator under the trace norm $$\| A\|_{1} : Tr(|A|)= Tr(\sqrt{A^{\dagger}A} )$$.

-My second question is a more specific version of the first. If the dynamics are generated by a contracting semigroup, i.e.

$$\rho_{n} = L_{n}\rho_{0}$$ where $$L_{n}$$ are contracting linear maps,

does the limit $$\lim_{n\rightarrow \infty}\|\rho_{n}\|_{1}$$ exists always? If so, what is the limiting operator and is it still trace class?

-My third question is essentially the second but for a particular case.

I have been working with the following operator. For $$\psi(x) \in H = L^{2}(\mathbb{R})$$ and $$\sigma_{t}\in B(H)$$ is defined as follows.

$$\sigma_{t}\psi(x) := \int_{\mathbb{R}} e^{-t(x-y)^{2}}K(x,y)\psi(y)dy$$.

Where $$K(x,y) \in L^{2}(\mathbb{R}^{2})$$ is a $$Hilbert-Schmidt$$ kernel. Note tha for $$t=0$$ this is a very tame integral transform. I am worried about the behviour as $$t\rightarrow \infty$$ in the trace norm sense. i.e. if the limit of $$\sigma_{t}$$ exists under $$\| \|_{1}$$, say $$\sigma_{\infty}$$, what is it? I am guessing that it should be some multiplication operator. $$\lim_{t\rightarrow \infty}\|\sigma_{t}\|_{1} = ? .$$

Thank you very much for your help.

• Your first question probably don't have a simple answer other than the defintion of convergence (and you forgot a square root in your definition of $|A|$). For the second one, when you say that $(\rho_n)$ is defined by a contracting semi-group, does it mean that for all $n,m$, $L_{n + m} = L_n \circ L_m$ (hence $L_n = L_1^n$) ? And for the third one, what is $\mathcal{R}$ ? Is it $\mathbb{R} ?$ Commented May 26, 2023 at 7:22
• @Cactus thank you for your interest. I fixed the error you pointed out and included the square root. Also, yes $\mathbb{R}$ is what I meant. Regarding your question about the properties of $L_{n}$. Indeed $L_{n}=L_{1}^{n}$. Commented May 26, 2023 at 10:05
• Sorry for my questions but I am still a bit confused about your operator in the question 3. Shouldn't you an absolute value or a square in the exponential ? Indeed, when $y \rightarrow -\infty$, you integral is not defined when $t > 0$ unless $K$ converges very fast toward zero when $y$ takes small values Commented May 26, 2023 at 11:25
• @Cactus exellent eye ! Indeed there should be a square there. Sorry for all of the typos. I should have been more diligent when writing the question. Commented May 26, 2023 at 12:30
• The $\|\cdot\|_1$ norm satisfies $\|AB\|_1\le \|A\|\,\|B\|_1.$ If $\rho_n$ is a contraction semigroup consisting of trace class operators then the sequence $\|\rho_n\|_1$ is nonincreasing. Indeed $\|\rho_{n+1}\|_1=\|\rho_1\rho_n\|_1\le \|\rho_1\|\,\|\rho_n\|_1\le \|\rho_n\|_1.$ Commented May 26, 2023 at 12:49

Question 1 I don't think there is a close answer other than the definition of convergence if there is no more assumption on $$(\rho_n)$$ for the trace norm, which is a Banach norm like an other.

Question 2 Not necessarily. Take for example $$H = \mathcal{l}^2(\mathbb{Z})$$, $$L$$ to be the shift operator, $$L(u)(k) = u(k - 1)$$ and $$\rho_0(u) = u(0)e_0$$ where for all $$i,j$$, $$e_i(j) = 1$$ if $$i = j$$, $$0$$ else. $$\rho_0$$ is a trace operator with $$\mathrm{Tr}(\rho_0) = 1$$ and $$L$$ is bounded with $$\|L\| = 1$$ thus for all $$n$$, $$\|L^n\| \leqslant 1$$ (and in this case, $$= 1$$).

However, we have for all $$u$$, $$(L^n\rho_0)(u)(k) = \rho_0(u)(k - n) = u(0)e_0(k - n)$$ so $$(L^n\rho_0)(u) = u(0)e_n$$ which diverges. You can prove it by noticing that for all $$n$$, $$\|L^n\rho_0\|_1 = 0$$ but it can not converge to something else than $$0$$.

However, if $$\rho_n \rightarrow \rho \in H$$ for the norme $$\|\cdot\|_1$$, then $$\rho$$ is a trace class operator (or else talking about the $$\|\rho_n - \rho\|$$ would be non-sens).

And in the case where $$\|L\|_1 = \lambda < 1$$, then $$\|L^n\rho_n\|_1 \leqslant \lambda^n\|\rho_0\|_1 \rightarrow 0$$ so $$L^n\rho_0 \rightarrow 0$$.

Question 3 You can prove that $$\sigma_t \rightarrow 0$$ for the operator norm using a series a Cauchy-Schwarz inequalities. However, are you sure that $$\sigma_t$$ has finite trace ? The product of two Hilbert-Schmidt operators has finite trace but Hilbert-Schmidt operators themselves aren't always I think.

If you set $$\rho_t = \sigma_t^*\sigma_t$$, we have $$\rho_t \in B_1(H)$$ and if $$(e_n)$$ is a Hilbert basis of $$H$$, $$\|\rho_t\|_1 = \sum_{n \geqslant 0} \left<\rho_te_n,e_n\right> = \sum_{n \geqslant 0} \|\sigma_te_n\| = \|\sigma_t\| \rightarrow 0.$$

The first question makes no sense because there's no such thing as a multiplication operator in the abstract setting of $$B(H)$$. 'Multiplication operator' has meaning when $$H$$ is concretely represented as $$L^2$$ of some measure space.

Second question - Let $$\|\cdot\|$$ be the operator norm and $$\|\cdot\|_1$$ the trace norm. $$B(H)$$ is a Banach algebra with the operator norm, so $$\|L_n\| = \|L_1^n\|\le \|L_1\|^n \le \varepsilon^n$$ for some $$0 < \varepsilon < 1$$ due to the assumption that $$L_1$$ is contracting. So for $$\rho_0$$ trace class $$\|\rho_n\|_1 = \|L_1^n \rho_0\|_1 \le \|L_1^n\| \|\rho_0\|_1 \le \varepsilon^n \|\rho_0\|_1 \rightarrow 0 \text{ as } n\rightarrow\infty$$ So $$\rho_n \rightarrow 0$$ in the $$\|\cdot\|_1$$ norm.

Third question - if this is a particular case of the second then $$\sigma_t \rightarrow 0$$ in the $$\|\cdot\|_1$$ norm as $$t\rightarrow \infty$$ :) I'm not sure why it's a particular case though.

These questions strangely mixed a couple of quite different concepts together, I'll give some examples or counterexamples I know of:

Q1: It is possible get a multiplication operator in the limit under proper settings. A typical example is Toeplitz matrice/operators. A bi-finite Toeplitz matrix is a multiplication operator on $$L^2(\mathbb T)$$, with its column being the Fourier series of the multiplication sign. If you choose the sign properly and pick finite submatrices from the bi-finite Toeplitz matrix, you will get such a convergent series. A famous example is be the following: simply take $$\rho_n$$ to be the $$n\times n$$ finite Toeplitz matrix below, and expand it consistently:
$$\begin{pmatrix} 0 & 1 & 1/2 & 1/3 & ...\\ -1 & 0 & 1 & 1/2 & \ldots \\ -1/2 & -1 & 0 & 1 & \ddots \\ -1/3 & -1/2 & -1 & 0 & \ddots \\ \vdots & \vdots & \ddots & \ddots & \ddots \end{pmatrix}$$ This sequence converges in trace norm to the multiplication operator with sign $$\pi-\theta$$, in fact you can also switch to the spectral norm by considering the adjoint.

Q2 That $$\{\|\rho_n\|_1\}_n$$ being a convergent sequence of numbers does not imply $$\{\rho_n\}_n$$ is convergent in $$\|\cdot\|_1$$ norm, e.g., you can take points on the unit sphere, then $$\{\|\rho_n\|_1\}_n$$ is a constant sequence, while $$\{\rho_n\}_n$$ need not converge since the unit ball is not compact in infinite dimension.

Q3 It seems that $$\sigma_t\to0$$ in various senses, you might want to multiply it by $$t$$ to avoid that, i.e., to redefine it as $$\sigma_t\psi(x)=t\int_{\mathbb R}e^{-t(x-y)^2}K(x,y)\psi(y)dy,$$ then it will look like an approximate of identity. I think as a way to verify your ideas, it may also be good to take $$K(x,y)=e^{-(x-y)^2}$$ and check what happens first.