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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 $e_1 = [1,0,\ldots]$. …

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 …

Consider $H = L^2[0,\infty)$ with $(,) = \langle,\rangle$, $A: H \to H$ the shift operator $A f(t) = f(t+1)$, so that $$ A^* f(t) = \cases{f(t-1) & if $t \ge 1$\cr 0 & otherwise\cr …

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 = …

What I would consider the obvious proof uses only the Banach space structure. If $\lambda$ is in the resolvent set, the graph $G(T)$ of $T$ maps in an obvious way to the graph of $(T-\lambda I)^{-1}$ …

No, and it's not true on Hilbert space either. For example, on $\mathbb C^2$ or $\mathbb R^2$ try $$ A = \pmatrix{0 & 0\cr 1 & 0\cr},\ x = \pmatrix{1\cr -1\cr},\ \lambda = 1$$ The spectrum is $\{0\}$, …

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

Yes, of course. By definition of the adjoint operator, $\{[-A^* y, y]: y \in \mathscr D(A^*)\}$ is the orthogonal complement in $H \oplus H$ of the graph $\{[x, Ax]: x \in \mathscr D(A)\}$ of $A$. …

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 …

You mean an infinite-dimensional separable Hilbert space. The answer is no. Suppose $p(z)$ has distinct roots $\alpha_1, \alpha_2$. Define a sequence $x_1, x_2, \ldots$ in the unit sphere of $H$ suc …

If the spectrum of $A$ is contained in a disk $\{z: |z - a| \le r\}$ where $|1-a| > r$, then the series $\sum_{n=0}^\infty (1-a)^{-1-n} (A - a I)^n$ converges to $(I-A)^{-1}$.

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.

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} …

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 …