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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 ...


12

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

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

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 ...


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 ...


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

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 ...


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

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

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

As another example: If we let $L(D)$ denote the space of holomorphic functions on a disk $D$, and if $D_1 \subset D_2$ is an inclusion of the disk of radius $r_1$ (around $0$, say) in the disk of radius $r_2$, with $r_1 < r_2$, then restriction $L(D_1) \to L(D_2)$ is compact with dense image. Indeed, suppose that $B_1 \to B_2$ is a compact operator ...


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

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

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

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

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

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

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

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 ...


4

As John noted below, the Volterra operator gives the desired counterexample. The original assertion, on the other hand, is true whe $K$ is normal: the difference being that a basis of eigenvectors can be obtained. So when $K$ is normal: All the nonzero elements of the spectrum of a compact operator are eigenvalues with finite-dimensional eigenspaces. ...


4

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 ...


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

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

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

If $M$ is infinite dimensional, then there exists $\{e_n\}_{n\in \mathbb{N}}\subset M$, which is an orthonormal set. Evidently $\{e_n\}_{n\in \mathbb{N}}\subset \overline{T(B(0,1))}$, but $\{e_n\}_{n\in \mathbb{N}}$ has no convergent subsequence, which contraditicts to the compactness of $T$.


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).



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