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First, a general remark. A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the ...

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Let $T_n$ such that $T_n(e_k)=\begin{cases}\lambda_ke_k&\mbox{ if }k\leq n\\\ 0&\mbox{ if }k>n \end{cases}$. Then $T_n$ is finite ranked hence compact and for $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq ... 10 A very nice and short proof of Pitt's theorem (a bit more than a page and covering the case you're interested in as well) was recently given by Sylvain Delpech MR review here, online article here. Added: Let me adress your questions in the comments in a pedestrian way since I find this more illuminating than appealing to heavy artillery. Lemma. Let ... 10 In order to distinguish the new definition of  {\mu_{n}}(T)  from the old one, let us call it  {\mu^{\text{New}}_{n}}(T) . We shall assume throughout this discussion that  \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty . By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have  ... 10 Here is the second installment that answers the converse in the affirmative. The result is not easy to establish. One method of proof uses the spectral calculus for self-adjoint operators, but this is like cracking a nut with a sledgehammer. I provide a softer approach below, which exploits the geometric properties of Hilbert spaces. Lemma 1 Every bounded ... 8 Suppose that g is not zero a.e. and let \epsilon>0 be such that E=\{x:|g(x)|>\epsilon\} has nonzero measure. Consider the orthogonal projection p=T_{\chi_{E}}. Assume that T_{g} is a compact operator. The operator T_{g} naturally induces a continuous linear operator of L^2(\mathbb{R}) into the Hilbert subspace pL^{2}(\mathbb{R})=\{\xi ... 8 Hilbert-Schmidt operators are more than bounded, as in the case of separable Hilbert space they are compact. Indeed, if \{e_n\} is an orthonormal basis, and P_N the projection over \operatorname{Span}\{e_j,1\leqslant j\leqslant N\}, then \{TP_N\} converges in norm to T and TP_N is finite ranked. The Hilbert-Schmidt operators form an ideal of ... 8 Yes, K is compact. Note first that we can assume that K is selfadjoint; indeed, if K satisfies the hypothesis, so do K^* and K+K^*, and if we prove that the real and the imaginary parts of K are compact, then we have that K is compact. Using the Weyl-von Neumann-Berg Theorem (II.4.1 or II.4.2 in Davidson's C^*-algebras by example) we can ... 7 This is how I would do it: First assume that \{a_n\} does not converge to zero. This means that there exists \varepsilon>0 and a subsequence \{a_{n_k}\}_k with |a_{n_k}|\geqslant\varepsilon. Now consider the sequence of vectors \{e_k\}, where e_k has a 1 in the n_k position, and zero elsewhere. Then Te_k is the sequence with a_{n_k} in ... 7 The zero representation of an algebra on a Hilbert space H is the map that sends every element of the algebra to the zero operator on H. Murphy's book gives the following definitions: If A is a C*-subalgebra of B(H), it is said to be irreducible, or to act irreducibly on H, if the only closed vector subspaces of H that are invariant for ... 7 Define$$T_j(x):=\left(x_1,\frac{x_2}2,\dots,\frac{x_j}j,0,\dots,0\right).$$It's a compact operator (because it's finite ranked) and$$T(x)-T_j(x)=\left(0,\dots,0,\frac{x_{j+1}}{j+1},\dots\right),$$hence$$\lVert T(x)-T_j(x)\rVert_{\ell^1}=\sum_{k=j+1}^{+\infty}\frac{|x_k|}{k}\leq \frac 1{j+1}\sum_{k=j+1}^{+\infty}|x_k|\leq \frac 1{j+1}\lVert ...

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Suppose it weren't so. Then there'd be an $\varepsilon > 0$ such that for every $n \in \mathbb{N}$ there is an $x_n \in E$ with $$\lVert T x_n\rVert > \varepsilon \lVert x_n\rVert + n\cdot \lVert ST x_n\rVert.$$ $x_n$ cannot be $0$, hence we may without loss of generality assume that $\lVert x_n\rVert = 1$. $T$ is compact, hence $T x_n$ has a ...

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I take it you are assuming that $\mu(X)<\infty$; I don't think it works otherwise. In this case, compose $T$ with point evaluation at $x\in X$ to obtain the bounded linear functional on $L^p$ that sends $f\mapsto (Tf)(x)$. This shows that there exists $K(x,\cdot)\in L^q$, $1/p+1/q=1$, with $\|K(x,\cdot)\|_q\le \|T\|$, such that (Tf)(x)= \int ... 6 No. The identity operator is not compact in any infinite dimensional Banach space. 6 By using the polar decomposition, we can write T=V|T|. So |T|=V^*T\in J, and then J contains a positive non-compact operator. On a side note, this argument also shows that J contains all adjoints of its operators, since now T^*=|T|V^*\in J. So from now on we assume T\geq0, non-compact, T\in J. This means that there is \lambda>0 with ... 6 For the boundedness, observe that \begin{align*} \| Tf \|_2^2 &= \int_{0}^{\infty} |Tf(x)|^2 \, dx \\ &\leq \int_{0}^{\infty} \frac{1}{x^2} \int_{0}^{x} \int_{0}^{x} |f(u)||f(v)| \, dudvdx \\ &= \int_{0}^{\infty}\int_{0}^{\infty} |f(u)||f(v)| \left( \int_{u \vee v}^{\infty} \frac{dx}{x^2} \right) \, dudv \\ &= ... 6 I thought I would just offer an alternate solution using Minkowski's integral inequality applied to: \| Tf \|_2 = \left\{ \int_0^{\infty} \left( \int_0^x x^{-1} f(t)\,dt \right)^{2}\,dx \right\}^{1/2} $$But first we do a variable substitution t \rightarrow xt so the inner integral is integrating over a fixed space:$$ \| Tf \|_2 = \left\{ ...

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Suppose, there is some $\epsilon > 0$ such that $M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\}$ has positive measure $\mu(M_\epsilon) > 0$. Now pick a sequence of sets $M_n \subset M_\epsilon$ with $M_{n+1} \subset M_n$ and $\mu(M_{n+1}) < \frac{1}{2}\mu(M_n)$ for all $n \in \mathbb N$. Note that 2. implies $\mu(M_n) > 0$ for all $n\in ... 6 I think the statement is true as long as$\mathcal{Y}$is a Banach space ($\mathcal{X}$can be any normed space). Is that correct? Yes, that is correct. In a complete metric space - such as a Banach space - a subset is relatively compact if and only if it is totally bounded. Showing that$T(B_\mathcal{X})$is totally bounded if$T$is the norm-limit of ... 6 Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that$\|T-T_k\| \to 0$for a suitable chosen sequence$\{ T_k\}$of finite rank operators. Try with $$T_k: (x_1, x_2, \ldots)\mapsto (x_1, \frac{x_2}{2}, \ldots, \frac{x_k}{k}, 0, 0, \ldots ).$$ It is obvious that$rk(T_k)=k$, i.e., it is finite rank ... 6 here's my solution: The function$\min \{ x, y \}$can be written as follows: $$\min\{x, y \}= \begin{cases} & y, \mbox{ if } 0 \le y \le x \\ & x, \mbox{ if } x \le y \le 1 \end{cases}$$ so we found the form for$T$: $$Tf(x) = \int_0^x yf(y) \, dy + x \int_x^1 f(y) \, dy.$$ Now let$Tf = \lambda f, \lambda \ne 0$. So we ... 6 There is nothing wrong with your argument here, assuming that$\cal U_{\cal H}$is the closed unit ball. (I imagine that "relative" is included in the definition of compactness since it's needed in the more general setting when defining compact operators between Banach spaces, or when defining compactness of$T$as "$T(M)$is relatively compact for any ... 6 Differential operators are badly discontinuous in general, and not defined for all functions. This was recognized as a problem early in the study of PDEs of classical Math/Physics. However, it was found that the inverse problems written as Fredholm integral equations gave rise to operators that are very continuous, and, in modern terms, often compact. This ... 6 Inverse operators are a bit more abstract than what one may first think. The notation and ease of defining inverses from linear algebra is misleading. Really, the inverse operator (when it is well-defined) takes$w\in W$and gives$v\in V$, with the added condition that$Kv = w$. This is the definition of the inverse operator. What the operator looks like ... 6 Same approach as previous answers, using limited operators: If$T\colon X \to X$is a compact operator and$S \colon X \to X$is a bounded operator, then both$TS$and$ST$are compact. Let$S: C^1[0,1] \to C^1[0,1]$such that$u(t) \mapsto u(\sqrt{t})$. Readily we see that$\|S\| = 1$, so$S$is bounded. But$TS = Id$. If$T$is compact so is$Id$, but ... 6 I do not know off the top of my head a characterisation of when the completely continuous operators coincide with the compact operators, but certainly such a space need not be reflexive; for example, consider the James space. In particular it is shown in Proposition 4.9 of Niels Laustsen's paper Maximal ideals in the algebra of operators on certain Banach ... 6 a) Let $$f_n(x)=\left\{\begin{array}{cl} 2^nx, & x\in[0,2^{-n}]\\1\ ,& x\in[1-2^{-n},1]\end{array}\right..$$ Then$f_n\in C([0,1])$and$\|f_n\|_{C([0,1])}=1$. Denote$g_n=Hf_n$. Then $$g_n(x)=\left\{\begin{array}{cl} 2^{n-1}x, & x\in[0,2^{-n}]\\1-2^{-n-1}x^{-1},& x\in[1-2^{-n},1]\end{array}\right..$$ Note that if$m<n$, then ... 6 A Banach space for which the finite rank operators are norm-dense in the compact operators is said to have the approximation property (AP). An explicit example of a Banach space without the AP is the space$B(H)$of bounded linear operators on an infinite-dimensional Hilbert space by deep work of Szankowski. Banach asked in his book of 1932 whether there ... 6 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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