Hot answers tagged compact-operators
11
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 $v\in\ell^2$,
$v=(v_0,v_1,\ldots)$
$$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq ...
9
A very nice and short proof of Pitt's theorem (a bit more than a page and covering the case you're interested in as well) was recently given by Sylvain Delpech MR review here, online article here.
Added:
Let me adress your questions in the comments in a pedestrian way since I find this more illuminating than appealing to heavy artillery.
Lemma. Let ...
7
In order to distinguish the new definition of $ {\mu_{n}}(T) $ from the old one, let us call it $ {\mu^{\text{New}}_{n}}(T) $.
We shall assume throughout this discussion that $ \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty $.
By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have $ ...
6
a) Let
$$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..$$
Note that if $m<n$, then ...
6
Suppose, there is some $\epsilon > 0$ such that $M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\}$ has positive measure $\mu(M_\epsilon) > 0$. Now pick a sequence of sets $M_n \subset M_\epsilon$ with
$M_{n+1} \subset M_n$ and
$\mu(M_{n+1}) < \frac{1}{2}\mu(M_n)$ for all $n \in \mathbb N$.
Note that 2. implies $\mu(M_n) > 0$ for all $n\in ...
6
Suppose that $g$ is not zero a.e. and let $\epsilon>0$ be such that $E=\{x:|g(x)|>\epsilon\}$ has nonzero measure. Consider the orthogonal projection $p=T_{\chi_{E}}$. Assume that $T_{g}$ is a compact operator. The operator $T_{g}$ naturally induces a continuous linear operator of $L^2(\mathbb{R})$ into the Hilbert subspace $pL^{2}(\mathbb{R})=\{\xi ...
5
Define
$$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
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 ...
5
Hilbert-Schmidt operators are more than bounded, as in the case of separable Hilbert space there 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 ...
4
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 ...
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
...
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
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
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 ...
4
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 ...
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 ...
3
(a) You got the idea, but used it in the reversed direction. Define the linear continuous map $L_x\colon H\to \Bbb R$ by $L_x(f):=T(f)(x)$, and then use Riesz-Frechet theorem.
(b) Your hint will give the conclusion after a use of Bessel's equality. To see you hint works, notice that $\lVert g_x\rVert^2\leqslant \lVert T\rVert\cdot \lVert g_x\rVert$ by ...
3
Define $P_j\colon T_1\otimes T_2\to T_j$ by $P(x_1,x_2)=x_j$, $j\in \{1,2\}$. Then $T_j=P_jT$, and $P_j$ is linear and continuous.
Conversely, denote $B_j(0,1)$ the closed unit ball of $H_j$. If $H$ is endowed with the norm $\lVert (x_1,x_2)\rVert_H=\lVert x_1\rVert_{H_1}+\lVert x_2\rVert_{H_2}$, then $B_H(0,1)\subset B_1(0,1)\times B_2(0,1)$ hence
...
3
Here's one way to see (a): write $\langle Px, y \rangle = \langle Px, Py \rangle + \langle Px, y-Py \rangle$. Since $P$ is orthogonal projection, $y - Py$ is orthogonal to the range of $P$, so the second term vanishes. Thus $\langle Px, y \rangle = \langle Px, Py \rangle$. By the same argument, $\langle x, Py \rangle = \langle Px, Py \rangle$. Since ...
3
You also need $S$ to be positive semidefinite, i.e. $\langle x, S x \rangle \ge 0$ for all $x \in H$. Then you can take $A = \sqrt{S}$ using the continuous functional calculus. Note that any continuous function $f$ on $\sigma(S)$ with $f(0)=0$ is the uniform limit on $\sigma(S)$ of a sequence of polynomials $p_n$ with $p_n(0)=0$, and so $f(S)$ is the norm ...
3
$\newcommand{\R}{\mathbf R}$ $\newcommand{\geq}{\geqslant}$ Let us start with the remark that as $u$ is Schwartz, $u$ is in $L^2(\R^d \times \R^d)$. Hence, the operator $U$ given by
$$
\begin{align*}
U : L^2(\R^d) &\mapsto L^2(\R^d)\\
v &\mapsto \int_{\R^d} u(x, y) v(y) \, \mathrm{d}y.
\end{align*}
$$
is bounded, and self-adjoint by the condition ...
3
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 $T$.
3
The open mapping theorem is, as you say, about bijective linear operators between Banach spaces. As an operator $T : X \to X$, our $T$ may be injective, but this argument shows it cannot be surjective. If you think of it as an operator $T : X \to \operatorname{ran}(T)$, then it would be bijective, so this argument shows that $\operatorname{ran}(T)$ is not ...
3
If $u$ is not bounded below then for each $n$ there exists $z_n$ such that $\| u(z_n)\|<\frac{1}{n}\|z_n\|$. In particular $\|z_n\|\neq 0$. Now define $x_n=\frac{z_n}{\|z_n\|}$. Then $\{x_n\}$ is a sequence of unit vectors such that $\lim_{n\rightarrow\infty}u(x_n)=0$. The other implication is easy.
2
$\ell^2$ is Hilbert, thus $T$ is compact iff the image of any weakly convergent sequence under $T$ converges. You can show that $T$ is compact as follows:
Let $x^k\stackrel{\omega}{\rightarrow}x$, since $\ell^2$ is complete, it is enough to show that $\{Tx^k\}_k$ is Cauchy. Notice that $||Tx^k-Tx^l||_{\ell^2}^2=\sum_{n\geq1}2^{-2n}|x_n^k-x_n^l|^2$. We have ...
2
This is a simple consequence of Fredholm alternative: for a compact operator $T$, then a value $\lambda \neq 0$ is either an eigenvalue or is in the resolvent. In this case, you're interested in $\lambda = 1$. In fact the theorem is even stronger: it states that $\lambda I - T$ is injective iff it's surjective, which is exactly your question.
Since you ...
2
If the embedding $e: A \to B$ is compact, and there is a continuous linear map $S:B \to A$ such that
$T$ equals the composite $S \circ e$, then $T$ will be compact (since the composite of a compact operator and any other continuous linear map is always compact).
[This idea is used in elliptic PDE theory, to deduce compactness of the inverse
of elliptic ...
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