# Convergence of Fredholm determinants

Let $(X_N)_N$ be a sequence of trace class operators acting on, say, $L^2(\mathbb{R})$. What are the minimal assumptions in order to have the convergence of their Fredholm determinant $$\lim_N\det(I+X_N) ?$$ I know $X\mapsto \det(I+X_N)$ is continuous for the trace class norm topology (once restricted to trace class), ok. Let's say that $X_N$ converges weakly to $X$, it is enough to have the convergence of Fredholm determinants ? What conditions should we add ? I guess I just need a good reference on the topic. Any ideas ? Thanks in advance.

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The Fredholm determinant is not sequentially continuous in the strong topology. Take the sequence of 1-dimensional projectors $X_n=\langle e_n,\cdot\rangle e_n$ with the $e_n$ forming an ON basis. You then have $X_n \to 0$ strongly, but $0=\det(I-X_n)$ does not converge to $1=\det(I)$. Or what did you have in mind when saying that you would know the continuity for the strong topology (once restricted to trace class)?

If you, however, add the convergence $\|X_n\|_1\to\|X\|_1$, where $\|\cdot\|_1$ denotes trace class norm, then it is known that weak sequential convergence $X_n \to X$ implies convergence in trace class norm and, therefore, of the Fredholm determinant (see Thm. 2.21 and Addendum H of the book "Trace Ideals and Their Applications" by Barry Simon, 2nd ed., AMS 2005).

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Sorry, I meant "trace class norm"! I'm editing my post ... Thx ! –  Adrien Hardy Aug 12 '11 at 10:32
Folkmar, do you know other results if one restrict to operators which are symmetric positive integral kernel operators ? Anyway, thx for your answer –  Adrien Hardy Aug 12 '11 at 12:07
Weak operator convergence and convergence of the trace (which is then, in the case of symmetric positive integral trace class operators, just the trace class norm) will do. Have a look at Chapter 2 of Simon's book. –  Folkmar Bornemann Aug 12 '11 at 15:59