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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 $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq ... 16 First, a general remark. A common way to prove that some operator T is not compact is to exhibit an infinite-dimensional subspace M on which T has a lower bound: that is, there exists c>0 such that$$\|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 ... 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 ... 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 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  ... 8 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 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 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 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.$$x_n cannot be 0, hence we may without loss of generality assume that \lVert x_n\rVert = 1. T is compact, hence T x_n has a ... 7 Suppose that g is not zero a.e. and let \epsilon>0 be such that E=\{x:|g(x)|>\epsilon\} has nonzero measure. Consider the orthogonal projection p=T_{\chi_{E}}. Assume that T_{g} is a compact operator. The operator T_{g} naturally induces a continuous linear operator of L^2(\mathbb{R}) into the Hilbert subspace pL^{2}(\mathbb{R})=\{\xi ... 6 Suppose, there is some \epsilon > 0 such that M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\} has positive measure \mu(M_\epsilon) > 0. Now pick a sequence of sets M_n \subset M_\epsilon with M_{n+1} \subset M_n and \mu(M_{n+1}) < \frac{1}{2}\mu(M_n) for all n \in \mathbb N. Note that 2. implies \mu(M_n) > 0 for all n\in ... 6 Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that \|T-T_k\| \to 0 for a suitable chosen sequence \{ T_k\} of finite rank operators. Try with$$ 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. $$Now let Tf = \lambda f, \lambda \ne 0. So we ... 6 No. The identity operator is not compact in any infinite dimensional Banach space. 6 There is nothing wrong with your argument here, assuming that \cal U_{\cal H} is the closed unit ball. (I imagine that "relative" is included in the definition of compactness since it's needed in the more general setting when defining compact operators between Banach spaces, or when defining compactness of T as "T(M) is relatively compact for any ... 6 Define$$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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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 ...

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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 ...

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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 ...

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The answer was found with the help of shwedka at dxdy forum. The result holds and there's an even more general formulation. Suppose that we have two measure spaces $(X,dm)$ and $(Y,dn)$; suppose also that $q\in[1,+\infty)$, $p\in (1,+\infty)$. The operators $K_i$ are given by $$K_i[f](y) = \int_{X}k_i(y,x)f(x)dm(x).$$ If $$|k_1(x,y)|\le k_2(x,y)\quad ... 5 A linear function from a finite dimensional Hausdorff topological vector space to itself is continuous, bounded, and compact. This is because a finite dimensional Hausdorff topological vector is isomorphic to \mathbb{R}^n and a linear function from \mathbb{R}^n to \mathbb{R}^n is continuous. Now closed bounded subsets of \mathbb{R}^n are compact and ... 5 A Banach space for which the finite rank operators are norm-dense in the compact operators is said to have the approximation property (AP). An explicit example of a Banach space without the AP is the space B(H) of bounded linear operators on an infinite-dimensional Hilbert space by deep work of Szankowski. Banach asked in his book of 1932 whether there ... 5 In the definition of Z, you probably want |v|>N instead of v>N. Also, in item 1, the definition of B, you have the Sobolev norm of x, so it's better to use norm notation for that. Let's also not use subscripts in superscripts... say, p<q and the embedding is H^q\to H^p. The H^q norm is given by$$\|f\|_{H^q}^2 = ...

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I think it is simpler to prove this without resorting to sequences. Key result: In a complete metric space, a set is relatively compact iff it is totally bounded. Let $B=B(0,1)$ be the open unit ball. To show that $TB$ is compact, it suffices to show that it is totally bounded. Choose $\epsilon>0$, and $n$ such that $\|T-T_n\| < \frac{1}{2} ... 5 No, it isn't. The image of any bounded sequence under a compact operator must have a norm-convergent subsequence. Consider the functions$f_n=n\chi_{[0,1/n]}$,$n=1,2,\ldots$. Each$f_n$has$L_1$-norm one, but the sequence$(Af_n)$has no subsequence which converges in$C[0,1]$. To see this, note$f_n$is the continuous function whose graph consists ... 5 Let$P_\epsilon$be the orthogonal projection onto$V_\epsilon$and $$T_n := T(\mathbb I - P_{1/n})$$$T_n$is a finite-rank operator and $$\lVert (T - T_n)v \rVert = \lVert T P_{1/n} v \rVert \leq \frac 1 n \lVert v \rVert$$ and so $$\lVert T - T_n \rVert \leq \frac 1 n \to 0$$ We can conclude$T$is compact because it is limit in the operator norm ... 5 Theorem, it is not true that$ \dim(\text{Range}(T)) = \dim(X) $implies that$ \dim(X) < \infty $. Another way of reasoning is as follows. Let$ T: X \rightarrow X $be a bijective compact operator. Then by the Bounded Inverse Theorem,$ T^{-1} $exists and is continuous. Hence,$ T $is also a homeomorphism. Let$ B_{X} $be the closed unit ball of$ X ...

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Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

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