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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 $x\neq 0$ ...


2

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' < ...


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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 ...



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