# Showing that spectrum of operator is not compact

• Leitmotif
In summary, the second derivative operator T with domain of all functions in X such that x(0)=x(pi)=0 is not a compact operator, as it is unbounded and has an unbounded set of eigenvalues.
Leitmotif

## Homework Statement

Let $$X=C[0,\pi]$$.
Define $$T:\mathcal{D}(T) \to X$$, Tx = x" where
$$\mathcal{D}(T) = \{ x \in X | x(0)=x(\pi)=0 \}$$.

Show that $$\sigma(T)$$ is not compact.

None.

## The Attempt at a Solution

Well, functions sin(Ax) and sin(-Ax), for A=0,1,2,... are in the domain, and eigenvalues are all integers, which are not bounded, aka. not compact. But I am not sure how to show this in general case. Any help in the right direction is appreciated.

Last edited:
Yikes, the preview looked fine, but not sure what happened with that LaTeX code there. The operator is second derivative, Tx=x", and the domain is all functions in X so that x(0)=x(Pi)=0.

What do you mean 'general case'? I think you've shown that the second derivative operator T is unbounded, so it can't be compact. What else is there to show?

Well, I guess I came up with an example of a function that's in the domain, and whose set of eigenvalues are not bounded, and eigenvalues are in the spectrum. I just wasn't really sure if that is enough to show that it's true for ALL functions in the domain.

Also, I noticed I made a mistake. If f(x)=sin(Ax), then $$f''(x)=-A^2sin(Ax)$$, so for any integer A, $$-A^2$$ is an eigenvalue, which is still an unbounded set.

That what is true for ALL functions in the domain? Compactness or boundedness are properties of the operator T, not of specific functions. Your example shows T is not a bounded operator, I think it's just fine.

Point taken. Thanks for the input :)

## 1. What is a compact operator?

A compact operator is a type of linear operator in functional analysis that maps elements from one Banach space to another Banach space. It is known for its important role in the study of functional analysis and operator theory.

## 2. How can you show that the spectrum of an operator is not compact?

To show that the spectrum of an operator is not compact, one can use the Fredholm alternative theorem, which states that the spectrum of a compact operator is either finite or consists of 0 as its only accumulation point. If the spectrum of an operator has an infinite number of eigenvalues or accumulation points, then it is not compact.

## 3. What are some common examples of non-compact operators?

Some common examples of non-compact operators include the identity operator, which has infinite spectrum, and the differentiation operator, which has a continuous spectrum that is unbounded.

## 4. What are the applications of studying compact operators?

The study of compact operators has various applications in fields such as functional analysis, operator theory, and partial differential equations. It is also used in engineering and physics to model and analyze various physical systems.

## 5. How do non-compact operators differ from compact operators?

Non-compact operators differ from compact operators in that they do not have finite or zero accumulation points in their spectrum. Non-compact operators also do not have a finite-dimensional range, unlike compact operators which have finite-dimensional ranges.

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