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Let $V$ be the vector space of all sequences which are eventually zero. Let $L$ be the backwards shift-- this is obviously "locally nilpotent". Given $V$ the norm $$ \| (x_n) \| = \sum_n a_n x_n, $$ …

Let $T:\ell^2\rightarrow\ell^2$ be the backwards shift operator, $T(a_n) = (a_2,a_3,\cdots)$. This is a contraction. For any $\lambda\in\mathbb C$, consider the sequence given by $a_n = \lambda^n$. …

You can use essentially the same definition. If $T: E_1 \supseteq D(T)\rightarrow E_2$ is a linear map between Banach spaces, then we define $x^*\in D(T^*)$ with $T^*(x^*)=y^*$ to mean that $y^*(x) = …

No. Consider $P:\ell^2(\mathbb Z)\supset D(P)\rightarrow\ell^2(\mathbb Z)$ defined by $P(e_n) = e^n e_{n} $ with $$D(P)=\{ (\xi_n)\in\ell^2(\mathbb Z) : \sum_{n=-\infty}^\infty e^{2n} |\xi_n|^2 < \in …

This is a complete, but rather abstract characterisation. $T$ has closed range if and only if there is $H_0\subseteq H$ closed and a bounded self-adjoint injective map $R:H_0\rightarrow H_0$ with …

I'm not sure what you can expect here. Notice that if $T$ is antilinear then defining the "graph" as $$ G(T) = \{ (T\xi, \xi) : \xi\in D(T) \} $$ does not give a subspace: it's not closed under (comp …