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.

  • 1
    $\begingroup$ 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 $\mathbb{R} ?$ $\endgroup$
    – Cactus
    Commented May 26, 2023 at 7:22
  • $\begingroup$ @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 $L_{n}=L_{1}^{n}$. $\endgroup$
    – Hldngpk
    Commented May 26, 2023 at 10:05
  • $\begingroup$ 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 $\endgroup$
    – Cactus
    Commented May 26, 2023 at 11:25
  • $\begingroup$ @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. $\endgroup$
    – Hldngpk
    Commented May 26, 2023 at 12:30
  • $\begingroup$ 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 $\|\rho_{n+1}\|_1=\|\rho_1\rho_n\|_1\le \|\rho_1\|\,\|\rho_n\|_1\le \|\rho_n\|_1.$ $\endgroup$ Commented May 26, 2023 at 12:49

3 Answers 3


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.


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