Tag Info

13

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 $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq ... 10 A very nice and short proof of Pitt's theorem (a bit more than a page and covering the case you're interested in as well) was recently given by Sylvain Delpech MR review here, online article here. Added: Let me adress your questions in the comments in a pedestrian way since I find this more illuminating than appealing to heavy artillery. Lemma. Let ... 8 In order to distinguish the new definition of  {\mu_{n}}(T)  from the old one, let us call it  {\mu^{\text{New}}_{n}}(T) . We shall assume throughout this discussion that  \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty . By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have  ... 8 Hilbert-Schmidt operators are more than bounded, as in the case of separable Hilbert space they are compact. Indeed, if \{e_n\} is an orthonormal basis, and P_N the projection over \operatorname{Span}\{e_j,1\leqslant j\leqslant N\}, then \{TP_N\} converges in norm to T and TP_N is finite ranked. The Hilbert-Schmidt operators form an ideal of ... 8 Yes, K is compact. Note first that we can assume that K is selfadjoint; indeed, if K satisfies the hypothesis, so do K^* and K+K^*, and if we prove that the real and the imaginary parts of K are compact, then we have that K is compact. Using the Weyl-von Neumann-Berg Theorem (II.4.1 or II.4.2 in Davidson's C^*-algebras by example) we can ... 6 Suppose it weren't so. Then there'd be an \varepsilon > 0 such that for every n \in \mathbb{N} there is an x_n \in E with$$\lVert T x_n\rVert > \varepsilon \lVert x_n\rVert + n\cdot \lVert ST x_n\rVert.$$x_n cannot be 0, hence we may without loss of generality assume that \lVert x_n\rVert = 1. T is compact, hence T x_n has a ... 6 a) Let$$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..$$Note that if m<n, then ... 6 Suppose, there is some \epsilon > 0 such that M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\} has positive measure \mu(M_\epsilon) > 0. Now pick a sequence of sets M_n \subset M_\epsilon with M_{n+1} \subset M_n and \mu(M_{n+1}) < \frac{1}{2}\mu(M_n) for all n \in \mathbb N. Note that 2. implies \mu(M_n) > 0 for all n\in ... 6 Suppose that g is not zero a.e. and let \epsilon>0 be such that E=\{x:|g(x)|>\epsilon\} has nonzero measure. Consider the orthogonal projection p=T_{\chi_{E}}. Assume that T_{g} is a compact operator. The operator T_{g} naturally induces a continuous linear operator of L^2(\mathbb{R}) into the Hilbert subspace pL^{2}(\mathbb{R})=\{\xi ... 5 This is how I would do it: First assume that \{a_n\} does not converge to zero. This means that there exists \varepsilon>0 and a subsequence \{a_{n_k}\}_k with |a_{n_k}|\geqslant\varepsilon. Now consider the sequence of vectors \{e_k\}, where e_k has a 1 in the n_k position, and zero elsewhere. Then Te_k is the sequence with a_{n_k} in ... 5 I think the statement is true as long as \mathcal{Y} is a Banach space (\mathcal{X} can be any normed space). Is that correct? Yes, that is correct. In a complete metric space - such as a Banach space - a subset is relatively compact if and only if it is totally bounded. Showing that T(B_\mathcal{X}) is totally bounded if T is the norm-limit of ... 5 Define$$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 ...

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

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 ...

5

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 ...

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 $X ... 5 A linear function from a finite dimensional Hausdorff topological vector space to itself is continuous, bounded, and compact. This is because a finite dimensional Hausdorff topological vector is isomorphic to$\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)$of bounded linear operators on an infinite-dimensional Hilbert space by deep work of Szankowski. Banach asked in his book of 1932 whether there ... 5 This is indeed only true in infinite dimensions. Suppose your compact operator$A$has finite spectrum. Then there are only finitely many eigenvectors with nonzero eigenvalue. Let$F$be the subspace spanned by those eigenvectors and let$E$be its orthogonal complement. Since$F$is finite dimensional,$E$is infinite dimensional, and in particular$E ...

4

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.

4

It is "well-known" that $\ell^\infty$ is isomorphic to $L^\infty[0,1]$. The standard embedding of $L^\infty[0,1]$ into $L^2[0,1]$ is not compact (e.g. the image of the unit ball contains an orthonormal basis).

4

In my opinion, the proof strategy is straightforward. Let $(x_n)$ be a sequence in $B_{c_0}$, the unit ball of $c_0$. As $((Tx_n)(1))_n$ is a bounded sequence of scalars, there is a bounded subsequence converging to some scalar y_1. Now, inductively, you have a sub-sub-... sequence, call it for simplicity $(x_m)$ of $(x_n)$, and an element $y = (y_i)_i$ ...

4

More is true, in fact. If $H$ is a Hilbert space, and $T\colon H \to H$ a compact normal operator, then $H$ is the closure of the direct sum of the eigenspaces of $T$. For $\lambda \in \sigma_P(T)$, let $E_\lambda$ be the eigenspace of $T$ corresponding to the eigenvalue $\lambda$, and let $S = \bigoplus\limits_{\lambda\in\sigma_P(T)} E_\lambda$. Then $S$ ...

4

I suppose the theorem you want to prove is this one: Let $(X,\Vert \cdot \Vert)$ be a Banach space. There exists a linear continuous operator $T \colon X \to X$ compact if and only if $\dim X <+\infty$. One way (if) is clear: indeed, if $\dim X<+\infty$ then every operator $T \colon X \to X$ is compact (since its range is finite dimensional: this ...

4

Theorem. Let $\Omega$ be an open set in a Banach space $X$ and let $F \in C(\Omega,X)$. If the Fréchet derivative $F'(x_0)$ exists for some $x_0\in\Omega$, then $F'(x_0)$ is a (linear) compact operator. Proof. Assume that $F'(x_0)$ is not compact. Then one can find $\epsilon_0>0$ and a sequence $\{y_n\}_n$ such that $\|y_n\|\leq 1$ and $$\|F'(x_0)y_k - ... 4 No: let e^{(n)}_k:=\delta_{nk}. Then the sequence \{e^{(n)}\} is bounded in \ell^p, for 1\leqslant p\leqslant \infty. If n_1\neq n_2, then$$\lVert e^{n_1}-e^{n_2}\rVert_q=\begin{cases} 2^{1/q},&\mbox{if }1\leqslant q<\infty\\\ 1&\mbox{if }q=+\infty, \end{cases}$$which proves that there is no convergent subsequence in \ell^q. So we ... 4 The following proof is adapted from Bruce Barnes, Majorization, range inclusion, and factorization for bounded linear operators. One maybe able to simplify the proof somewhat. Lemma Let T,S\in B(X,Y) and R(T)\subseteq R(S), then \exists M > 0 such that for all \alpha\in Y^*,$$ \|T^*\alpha\| \leq M\|S^*\alpha\| \tag{1}$$where * denotes the ... 4 Here's a different proof. Assume first that A is finite-rank. Then \text{Tr}(A^*A)<\infty, and so$$ 0\leq\text{ Tr}(A^*A)=\sum_{n=1}^\infty\langle A^*Ae_n,e_n\rangle=\sum_{n=1}^\infty\|Ae_n\|^2<\infty,  and so $\|Ae_n\|\to0$. If $A$ is any compact operator, there exists a sequence of finite-rank operators $\{A_m\}$ with $\|A_m- A\|\to0.$ Then ...

Only top voted, non community-wiki answers of a minimum length are eligible