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Let our Banach space be $\ell^1$, which is the set of all $a \in \mathbb{R}^{\mathbb{N}}$ such that $\sum_n |a_n|<\infty$. The norm is the $L^1$ norm. For $i \in \mathbb{N}$, let $e_i \in \ell^1$ denote the $i^{th}$ standard basis vector, i.e, the sequence whose $i^{th}$ index is $1$, but all other indices are $0$. Consider the linear map $T: \ell^1 \to ... 1 The other implication is true as well, at least if you assume that$Y$and$Z$are closed in$X$. In this case, the projection maps are bounded (by the closed graph theorem) which is needed in the proof. Start with a bounded sequence$(x_n)=(y_n\oplus z_n)$in$X=Y\oplus Z$. You are looking for a subsequence$(x_{n_k})$such that the sequence ... 2 Using the spectral theorem: A selfadjoint operator$T \ne 0$on a Hilbert space is compact iff $$T = \lambda_1 E_{1} + \lambda_2 E_{2} + \cdots,$$ where$\{ E_{j} \}$is a finite or countably infinite set of disjoint orthogonal projections onto finite-dimensional subspaces, and the sequence$\{ \lambda_{j} \}$, if infinite, converges to ... 2 Presumably you are asking about the absolute value of the diagonal terms, rather than the diagonal terms. In this case the answer is: Yes, since otherwise the diagonal elements would converge to zero. This would mean that$I$is a norm limit of the finite-rank operators $$I_n = \sum_{k \le n} \lambda_k P_k,$$ where$\{P_k\}$are the orthogonal projectors ... 3 The idea of compact came from sequences. You can see how one might come up with the name 'compact' to describe a set where you can't have an infinite set of points that are a minimum fixed positive distance from each other; that would not be a 'compact' set. Finding a cluster point gives you way to find a limit, and that's the importance of a 'compact' set. ... 1 Every operator is in particular a mapping (in the set theoretical sense) and each mapping$X \to Y$is (also in set theoretical sense) some subset of the product$X \times Y$(identified naturally with it's graph). If we are in the functional analysis setting$X$and$Y$are (at least) topological vector spaces. When we deal with linear operators then the ... 0 Suppose$X$and$Y$are Banach spaces and$U$is the open unit ball in$X$. A linear map$T: X\to Y$is said to be compact if the closure of$T(U)$is compact in$Y$. It is clear that$T$is then bounded. Thus$T\in \mathcal B (X, Y)$. Since$Y$is a complete metric space, the subsets of$Y$whose closure is compact are precisely the totally bounded ... 2 Edit: previous answer was wrong. The assertion$\|KT_n\|\to0$does not hold in general. Let$H=\ell^2(\mathbb N)$, and $$T_n(a_1,a_2,\ldots,)=(a_n,a_{n+1},\ldots).$$ Then$T_n\to0$in the strong operator topology. Consider the rank-one operator$P$given by$P(a_1,a_2,\ldots)=(a_1,0,0,\ldots)$. Then $$PT_n(a_1,a_2,\ldots)=(a_n,0,0,\ldots).$$ If ... 1 The set of compact operators (in a Hilbert space) is exactly the set of norm limits of finite rank operators. This is perhaps a more natural definition than the one you indicate. Many of the nice properties of finite rank operators have analogues for compact operators. You can view compactness has a slight generalization of being finite rank that preserves ... 2 The motivation to study compact operators is that a lot of the operators that we are interested in are compact. The reason the property of being compact is so special, as you say, that it deserved to be given a name is that it has many useful and interesting consequences, of course. In fact, you can generalize this: for almost all X and Y, the motivation ... 0 Please note that if$T$were surjective, then its inverse must be continuous (if$T$is compact, it is closed, hence so must be$T^{-1}$if it exists. By the closed graph theorem,$T^{-1}$is then continuous). As T.A.E. explained,$\{1/2^n :n=1,2,3,\dots\}\subseteq \sigma_p(T)$. It is easy to see that no other numbers belong to the point spectrum of$T$. ... 0 If$S^2$is compact, then$|S|=(S^2)^{1/2}$is compact. By looking at the polar decomposition$S=U|S|$, we deduce that$S$is compact. For non-normal, the statement is not true. Consider the shift-like operator$S$that, for a fixed orthonormal basis$\{e_n\}$, sends$e_{2n}\longmapsto e_{2n+1}$,$e_{2n+1}\longmapsto0$. The$S$is not compact, but ... 2 Let$\{ e_{n} \}_{n=1}^{\infty}$be the standard basis for$\ell^{2}$. Specifically, $$e_1 = \{ 1,0,0,0,\cdots \} \\ e_2 = \{ 0,1,0,0,\cdots \} \\ e_3 = \{ 0,0,1,0,\cdots \} \\ \vdots$$ You can write $$Tx = \sum_{n=1}^{\infty}\frac{1}{2^{n}}(x,e_n)e_n$$ If you truncate this ... 1 I think there is rather a simple way to solve this problem. Let us assume that$T$is compact and let$T=S^2$where$S$is a self-adjoint operator. We need to show that$S$is compact. Observe that (Take this as an exercise) an operator$A$is compact if and only if$AA^*$and$A^*A$both are compact. Now since$T$is compact it follows that$S^2 = SS^* = ...