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31 votes
3 answers
5k views

When is an integral transform trace class?

Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator $$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$ the operator $K$ is Hilbert ...
Marc Palm's user avatar
  • 11.2k
18 votes
1 answer
564 views

Is the space of Hankel operators complemented in B(H)?

Let $H$ be $\ell^2({\mathbb N})$ and let $S:H\to H$ be the unilateral forward shift, so that $S^*S=I\neq SS^*$. Then a bounded operator $T:H\to H$ is Hankel if and only if it satisfies $TS=S^*T$. Let ...
Yemon Choi's user avatar
  • 25.8k
10 votes
1 answer
598 views

What happens if we rotate the kernel of an integral operator?

Given an integral operator $K$ on $L^2(\mathbb R)$ with kernel $k(x, y)$, consider the integral operator $L$ on $L^2(\mathbb R)$, whose kernel has the form $k(\alpha x+\beta y, \gamma x+\delta y)$, ...
limanac's user avatar
  • 452
6 votes
3 answers
852 views

Are nuclear operators closed under extensions?

Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram $\require{AMScd}$ \begin{CD} 0 @>>> X_1 @>f_1>> X_2 ...
santker heboln's user avatar
4 votes
1 answer
279 views

Reference request: Baire's theorem for operator ranges

Let $F$ be a Banach space. A vector subspace $U \subseteq F$ is called an operator range if there exists a Banach space $E$ and a bounded linear mapping $T: E \to F$ such that $TE=U$. By a quotient ...
Jochen Glueck's user avatar