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Let $\mathbb H$ be a Hilbert space, and let $\mathcal B(\mathbb H)$ be the space of bounded operators on $\mathbb H$, equipped with the operator-norm topology. Let $\mathbb R\ni t\mapsto A(t)\in \math …

Too long for a comment. Why don't use your third criterion: if $K_L$ is the kernel of the operator $L$, that gives you $$ K_A=K_B\circ K_C, $$ i.e. $ K_A(x,y)=\int K_B(x,z) K_C(z,y) dz. $

Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the Banach algebra of bounded operators on $\mathbb H$. Let $(A_k)_{k\ge 1}$ be a sequence in $\mathcal B(\mathbb H)$. $\bullet$ I …

Well, you have $$ [\frac{1}{idx},x]=1/i $$ although $\frac{1}{idx}$ is far from the Identity.

If $B$ is positive self-adjoint then $B=A^2$ with $A$ positive self-adjoint. If $X$ is bounded non-negative and commutes with $B$, it commutes as well with a function of $B$ such as $A=\sqrt B$. Then …

Take $f=g=1$, $\Delta$ the usual Laplace operator on $\mathbb S^1$, then $$ C_0=\partial_\theta e^{\partial_\theta^2}, $$ can be identified to the diagonal infinite matrix $ (ik e^{-k^2})_{k\in \mathb …

The following result is classical: let $\mathbb H$ be a Hilbert space, and let $A,B\in \mathcal B(\mathbb H)$, then $ [A,B]\not=I. $ In finite dimension, just take the trace, and if the dimension is i …

An operator is trace class whenever it is the product of Hilbert-Schmidt operators. There is a simple characterization of Hilbert-Schmidt operators pseudodifferential operators: a pseudodifferential o …

Let us start with the Campbell-Hausdorff formula for selfadjoint operators: let $H_j$ be bounded selfadjoint operators on a Hilbert space. Then $$ e^{i\tau H_1}e^{i\tau H_2}=e^{i\tau (H_1+H_2)-\frac{\ …

Let me first recall what is the so-called paving conjecture: for any $\epsilon >0$, there exists $r\in \mathbb N$ such that for any bounded operator $A$ on $\ell^2(\mathbb Z)$, there exists a partitio …

A good way to quantify the non-commutativity of $A,B$ is to compare their flows, e.g. to compare the solutions of $ \dot u=Au,\quad\dot v=Bv $ with same initial datum. A complete answer is given by th …