6:58 AM
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Here is the paragraph that is not clear for me: Could anyone explain it for me please?

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The paragraph is given in the following pictures: If $A$ is an operator from $H_{1}$ to $H_{2}$, The paragraph is divided into 5 sentences. I could not understand the first, second, forth and fifth sentences. Could anyone explain them for me please? Thanks!

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Let $\mathfrak{F}(S)$ denotes the set of filters (including the improper filter) on a poset $S$, ordered reversely to set theoretic inclusion of filters. Let $Da$ for a lattice element $a$ denote its sublattice $\{ x \mid x \leq a \}$. Let $Z(X)$ denotes the set of complemented elements of the la...

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Let $T:l_{2} \rightarrow l_{2}$ be defined by $$T(x) = \left(x_{1},\frac{x_{2}}{2},\frac{x_{3}}{4},\ldots,\frac{x_{n}}{2^{n-1}},\ldots \right)$$ where $x = (x_{1},x_{2},.....).$ Find Im$T$. I know that the kernel of this operator is $\{0\}$ and I know that the sum of the dim Im$T$ + dim ker$... 1 Let$n \in \mathbb{N^{*}}$and define$T_{n}:l_{2} \rightarrow l_{2}$by $$T_{n}(x) = (\frac{x_{n}}{n},\frac{x_{n + 1}}{n + 1},... ).$$ where$x = (x_{1},x_{2},.....).$Then,$$T_{n}^{*}(x) = (0,...,0,\frac{x_{1}}{n},\frac{x_{2}}{n + 1},... ).$$ Where the zeros in the adjoint are$n-1$times. ... Two of them were created by Intuition. 4 hours later… 10:50 AM I've edited the two questions with the tag "image-of-operator", and essentially have killed the tag. @JohnMa Did you mean some other tag? I still see it here: 1 Let$n \in \mathbb{N^{*}}$and define$T_{n}:l_{2} \rightarrow l_{2}$by $$T_{n}(x) = (\frac{x_{n}}{n},\frac{x_{n + 1}}{n + 1},... ).$$ where$x = (x_{1},x_{2},.....).$Then,$$T_{n}^{*}(x) = (0,...,0,\frac{x_{1}}{n},\frac{x_{2}}{n + 1},... ).$$ Where the zeros in the adjoint are$n-1\$ times. ...

11:09 AM
oops, I mean "invertible-operators"
Um, the way the users "intuition" and "Idonotknow" ask their questions, I assume they are held by the same person. Their recent questions should be tagged "functional analysis" instead of the a little bit more advanced "operator theory".