Argue, why given Operators are compact or not.

In summary, the operators (i) and (ii) are related to e^t and there are various ideas for showing their compactness, such as the image of the closed unit ball being relatively compact, the operator being bounded, and using equicontinuity and Arzela-Ascoli's theorem. Some progress has been made in showing the compactness of (i) by showing the boundedness of T_nx and the existence of a convergent subsequence, but for (ii) it can be shown that T_nx is not cauchy using x(t)=t^n.
  • #1
jonathan krill
2
0
Which of the operators [itex]T:C[0,1]\rightarrow C[0,1][/itex] are compact?

$$(i)\qquad Tx(t)=\sum^\infty_{k=1}x\left(\frac{1}{k}\right)\frac{t^k}{k!}$$ and
$$(ii)\qquad Tx(t)=\sum^\infty_{k=0}\frac{x(t^k)}{k!}$$

ideas for compactness of the operator:

- the image of the closed unit ball is relatively compact under T.
- for any sequence [itex](x_n)\subset B_1(0)[/itex], the sequence [itex](T_nx)[/itex] contains a cauchy sequence.
- show that the operator is not bounded, which is equivalent of it not being continuous which is necessary to be a compact operator
- show [itex]T(C[0,1])[/itex] is equicontinuous, and then argue via arzela-ascoli to get compactness of the operator
- obviously they are somehow related to [itex]e^t[/itex]

sorry read the sticky to late, pls some moderator move this to HW forum.
 
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  • #2
Moved to homework. @jonathan krill have you made any attempts at a solution? You give some ideas for showing compactness, but have you tried any of them? If so, please post what you obtained. (Note that in homework threads you are supposed to fill out the homework template; part of that is showing your attempts at a solution.)
 
  • #3
made progress showing (i) is compact the [itex] T_nx[/itex] are polynomials of finite rank, so for sequences in [itex]C[0,1][/itex] [itex] T_nx[/itex] is bounded hence [itex] T_nx[/itex] contains a convergent subsequence. hence [itex] T[/itex] is compact.

for (ii) One probably can show that [itex] T_nx[/itex] is not cauchy using [itex] x(t)=t^n[/itex]
 

FAQ: Argue, why given Operators are compact or not.

What does it mean for an operator to be compact?

An operator is said to be compact if it maps a bounded set of inputs to a set of outputs that is relatively compact in the output space. This means that the operator takes a set of inputs and produces a set of outputs that is contained within a finite region of the output space.

How do you determine if an operator is compact?

One way to determine if an operator is compact is by using the Heine-Borel theorem, which states that a set in a metric space is compact if and only if it is closed and bounded. This means that to show an operator is compact, we need to show that it maps bounded sets to closed sets.

Can an operator be compact on one space but not on another?

Yes, an operator can be compact on one space but not on another. This is because the definition of compactness depends on the underlying space and the properties of the operator.

What are some examples of compact operators?

Some examples of compact operators include integral operators, compact linear operators on normed vector spaces, and compact self-adjoint operators on Hilbert spaces.

What are the applications of compact operators in science?

Compact operators have various applications in science, such as in quantum mechanics, functional analysis, and differential equations. They are also used in signal processing, image processing, and machine learning for data compression and feature extraction.

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