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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 ...


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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 ...


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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 ...


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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 ...


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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^* = ...



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