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These conditions are not equivalent. Take a non-zero operator $x$. Then certainly $x\neq -x$. However, $$|(-xh, k)| = |-(xh,h)| = |(xh, k)|$$ for any $h,k\in H$. Separation in the context of locally convex spaces means the following: Let $\mathscr{P}$ be a family of seminorms. Then $\mathscr{P}$ is (by definition) separating if for each $x\neq 0$ ...
Suppose $f$ is as stated with exponential bound $Ce^{-\delta|x|}$, and suppose that $\overline{g} \in L^{2}(\mathbb{R})$ is orthogonal to the functions $\{ x^{n}f(x) \}_{n=0}^{\infty}$ in the $L^{2}$ inner-product. Define $$H(\lambda) = \int_{-\infty}^{\infty}g(x)f(x)e^{-i\lambda x}\,dx.$$ Suppose that $|\Im\lambda| \le \delta' < ... 2 No, they do mean that$X$is a countable [countably infinite] set, and the measure is the counting measure on$X$, $$\mu(A) = \operatorname{card} A$$ for any subset$A\subset X$. In particular, the$\sigma$-algebra of$\mu$-measurable sets is$\mathfrak{P}(X)$, the entire power set of$X$, and the only null set is$\varnothing$. Thus for$f\colon X \to ...