2
votes
2answers
36 views

Fredholm Index: Finite Corank $\Rightarrow$Closed Range [duplicate]

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus ...
4
votes
1answer
79 views

What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
0
votes
0answers
13 views

inequality for compact operator

Let $K(x)$, $x\ge0$ be a nonnegative-valued continuous function with support $(0,\infty)$ and such that $\int_0^\infty K(x)\,dx=1$. Let $\mathcal{K}$ be an integral operator given by $$ ...
5
votes
1answer
36 views

Is the Neumann series a compact operator?

Let $X$ be an infinite dimensional Banach space and $A:X\to X$ be a compact operator with the operator norm $\|A\|<1$. Then $I-A$ is invertible and the Neumann series $$ S_N = \sum_{k=0}^N A^k $$ ...
3
votes
2answers
82 views

Subspaces in the image of compact operator

Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem ...
0
votes
1answer
32 views

Sufficient condition for compact embedding between Banach spaces

Let $(X,\|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be two Banach spaces and $X\subset Y$. Then $X$ is compactly embedded in $Y$, if the unit ball $B_X(0,1)$ in $X$ is a relatively compact subset in $Y$. ...
2
votes
1answer
50 views

Show that the span of eigenvalues of a compact operator is a closed

Let $(X, ||\cdot||)$ be a real Banach space and $T: X \to X$ a compact operator (so $\{x_n\}_{n=1}^\infty$ bounded implies that $\{Tx_n\}_{n=1}^\infty$ has a convergent subsequence). Let $x_1, \dots, ...
1
vote
1answer
35 views

Compact embedding

Prove that the embedding $j\colon (C^1[0,1],\|\cdot\|)\to(L^1[0,1],\|\cdot\|_{L^1})$ where $\|f\|=\max\{\|f\|_\infty,\|f'\|_\infty\}$ and $\|f\|_\infty$ denotes the supremum norm, ...
1
vote
1answer
54 views

The trace class operators are the dual of the compact operators

I know that the map from the trace class operators $L_1(H)$ to the dual of the compact operators $K'(H)$ given by $A \mapsto tr( \cdot A)$ is an isometric isomorphism. Linearity is obvious by the ...
1
vote
0answers
43 views

properties of integral operator $x^{-1}\int_0^xf(x,y)v(y)dy $

here we have two cases to study $(1)$ let us fix any $f \in C^{1}[ [0,1] \times [0,1]]$ ($k \neq 0$). Set $$[T(v)](x) := x^{-1}\int_0^xf(x,y)v(y)dy $$ for any $x \neq 0$ otherwise $[T(v)](0) := ...
1
vote
1answer
47 views

Is the canonical injection between $ C^{k+1}[0, 1] $ and $ C^k [0, 1] $ compact?

If $ k=0 $ by the Ascoli - Arzelá theorem the answer is yes, but i don't know how to proceed in the general case ($ k> 0, k \in \mathbb{N} $). I tried to build a counter example using Riesz lemma ...
4
votes
0answers
119 views

Non linear compact map

Suppose to have two Banach spaces $E$ and $F$, with $E$ reflexive. Suppose to have a continuous map $T:E \to F$ which maps bounded subsets into precompact subsets. $T$ is not assumed to be linear. ...
0
votes
1answer
80 views

Compact Operators and Complete Metrics Spaces

I have a couple of questions about compact operators and compactness in complete metric spaces: 1.I have the following implications: Let $Y$ be a metric space with $A$ a subset of $Y$. $A$ is ...
1
vote
1answer
37 views

Determine when operator is compact

Let $B$ be the Banach space of bounded complex functions on $[0,1]$ with sup-norm. For $q \in B$, define the (multiplication) operator $T_q : B\rightarrow B$ by $(M_q f)(t) = q(t)f(t)$. Which $q$ ...
1
vote
1answer
48 views

a compact operator on $l^2$ defined by an infinite matrix

Let $A$ be an infinite matrix such that $\displaystyle \sum_{i,j}|a_{i,j}|^2<\infty$. Then $A$ defined a compact operator on $l^2$.
0
votes
2answers
69 views

Multiplication operator with a function non-vanishing on the cantor set

Let $M_f$ be the multiplication operator, which acts on bounded functions $g$ on the unit interval as $g\mapsto fg$, with $f:[0,1]\rightarrow \mathbb{C}$ such that $f$ is nonzero only on the Cantor ...
2
votes
1answer
258 views

Closure of the range of a compact operator

Let $X$ be an infinite-dimensional Banach space, and let $Y$ be a banach. Let $T$ be a compact operator from $X$ to $Y$, ie. if $(x_n)$ is a sequence in $X$ then there is a subsequence s.t. ...
3
votes
1answer
70 views

$L(\ell_{p})$ contains only one proper closed ideal

I am trying to solve the following problem: Show that if $1<p<\infty$ and $T:\ell_{p}\rightarrow\ell_{p}$ is not compact then there is a complemented infinite dimensional subspace $E$ of ...
0
votes
1answer
65 views

Why is the set of compact operators closed in the space of all bounded operators between Banach spaces?

Let $X$ and $Y$ be Banach space. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. Also, $K(X,Y)$ is the set of all compact operators from $X$ to $Y$. Why is $K(X ,Y)$ ...
3
votes
1answer
159 views

Frechet derivative of compact operator is compact

... this seems to be a well known fact as mentioned in this and in this manuscript. However, I was not able to find a proof or to prove it by myself. So my question is: How to prove this? Any hint ...
2
votes
1answer
76 views

Is there a noncompact operator from $\ell^\infty$ to a reflexive space?

It is well known that bounded operators from $c_{0}$ to a reflexive Banach space $X$ are in fact all compact. Indeed, since it can be shown that an operator is compact iff for any weakly convergent ...
5
votes
1answer
97 views

Pitt's theorem and reflexivity

Does it follow from Pitt's theorem that the space of bounded operators from $\ell_2$ to $\ell_p$ ($p<2$) is actually reflexive? We have $$ \mathcal{B}(\ell_2, \ell_p) = \mathcal{K}(\ell_2, \ell_p) ...
0
votes
1answer
448 views

There are compact operators that are not norm-limits of finite-rank operators

Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators. Tanks for answer
4
votes
1answer
370 views

Is the inclusion $C^1[0,1]\subset C[0,1]$ compact?

I am working on this problem but i couldn't succeed . Consider the space $C^1[0,1]$ with the norm $$\|f\|=\max \{\|f\|_{C[0,1]}, \|f'\|_{C[0,1]}\},$$ I don't know if the inclusion map is compact, ...
4
votes
0answers
79 views

The control of norm in quotient algebra

Let $B_1,B_2$ be two Banach spaces and $L(B_i,B_j),K(B_i,B_j)(i,j=1,2)$ spaces of bounded and compact linear operator between them respectively. If $T \in L(B_1,B_1)$, we have a $S \in K(B_1,B_2)$ and ...
3
votes
2answers
460 views

Show that a finite-dimensional Banach space has a bijective compact operator

It is clear that if $ T: X \rightarrow X $ is a bijective compact operator, where $ X $ is a Banach space, then $ \dim(\text{Range}(T)) = \dim(X) $, which implies that $ \dim(X) $ must be $ < ...
2
votes
3answers
267 views

Compact integral and multiplication operator in Banach spaces

Let $ A\colon C[0,1] \to C[0,1] $ $$ A(x)(t) = f(t)x(t) + \int_0^t x(s)ds,\quad f \in C[0,1]: f(1) \neq 0, \forall t \in [0,1] $$ Is $A$ a compact operator or not?
1
vote
2answers
210 views

Proof of the Pitt's theorem

I'm reading the book Topics in Banach Space Theory by Albiac F. Kalton N. J. I got stuck at the proof of the Pitt's theorem. In the second paragraphs authors tries to prove ad absurdum that for ...
5
votes
2answers
1k views

Compact operators and completely continuous operators

A compact operator between Banach spaces is an operator that maps bounded sets into relatively compact sets, while a completely continuous operator maps all weakly convergent sequences into convergent ...
2
votes
1answer
541 views

A compact operator is completely continuous.

I have a question. If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps ...
4
votes
1answer
269 views

Criterion for a bounded linear operator to be compact

Let $X,Y$ be Banach spaces. $T\colon X\to Y$ be a bounded linear operator. How can I prove that $T$ is compact if and only if there is $\lbrace x_n^*\rbrace\subset X^*$ such that $\|x_n^*\|\to 0$ and ...
13
votes
3answers
854 views

Compactness of a bounded operator $T\colon c_0 \to \ell^1$

Pitt Theorem says that any bounded linear operator $T\colon \ell^r \to \ell^p$, $1 \leq p < r < \infty$, or $T\colon c_0 \to \ell^p$ is compact. I know how to prove this in case $\ell^r \to ...