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Hopelessly false. Consider the one-dimensional case (so the $T_j$ are just numbers). Any function $f$ for which $\sum_j T_j$ convergent implies $\sum_j f(T_j)$ convergent is linear in a neighbourhoo …

You might look at Barry Simon's book "Trace Ideals and Their Applications", especially for applications in mathematical physics.

I doubt that you'll find a very general condition that requires multiplicity $1$, other than those that are essentially restatements of that fact. There will be conditions related to particular forms …

We can define the Fredholm determinant on operators on Hilbert space which differ from the identity by a trace-class operator. This satisfies $$\det(\exp(T)) = \exp(\text{Tr}(T))$$ for trace-class o …

By "$I$ is well-defined", I presume you mean you have a continuous injection $\iota$ of $X$ into $Y$. If $X$ and $Y$ are not isomorphic it will not be a bijection, because there is no continuous line …

Perhaps not what you're looking for, but you may be interested in the following result. Let $\cal H$ be a separable infinite-dimensional Hilbert space, and $H$ any self-adjoint unbounded linear opera …

The most important thing about residual spectrum, I think, is that some classes of operators, e.g. normal operators on Hilbert space, don't have any.

Let's assume the matrix $A$ has all its row sums equal to $\lambda$, the largest eigenvalue. We can rescale the rows and columns of any other nonnegative irreducible matrix by a similarity transformat …

Consider the right shift $R([x_1, x_2, \ldots]) = [0, x_1, x_2, \ldots]$ on $\ell^2$. I claim the open ball of radius $1/2$ about $R$ contains no diagonalizable operators. Let $e_1 = [1,0,\ldots]$. …

Apply the SNAG (Stone-Naimark-Godement-Ambrose) theorem to the unitary group generated by these operators.

Even one positive definite operator on an infinite-dimensional Hilbert space need not have any eigenvectors at all: it might have continuous spectrum. The more general statement is the Spectral Theor …

What you take as the domain is to some extent a matter of choice. You do want to be able to define $Su$, for $u$ in the domain, as a member of $L^2(\mathbb R)$. The real questions, I think, are whet …

Here's a counterexample (subject perhaps to what you consider "natural"). Take a separable Hilbert space with orthonormal basis $\{u_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by $$ A u_n = …

Yes, using Runge's theorem. I'll assume wlog your $D$ is the unit disk $\{z: |z| \le 1\}$. Given positive integer $n$, take $$ K_n = \{z: (|z| \le 1 \ \text{or}\ 1 + 1/n \le |z| \le n)\ \text{and} …

That closed form is $c^{1/2} (2n-1)$ when $c$ is a positive real. These eigenvalues must be analytic as functions of $c$ as long as they don't collide or go off to $\infty$, and they don't as long a …