Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $u \in L^\infty ([0,1])$ be fixed and define the operator $T$ on $C([0,1])$ by $$ Tf(x) = \int_0^x u(t) f(t) dt $$ Show that T is compact and into itself and determine its spectrum.

My try: The compactness follows from Arzela-ascoli.

The point spectrum: after derivation we get $$u(x)f(x) = \lambda f'(x) \Longleftrightarrow \left(f(x) e ^{-\frac{1}{\lambda} \int_0^x u(t)dt} \right)' = 0 \Longleftrightarrow f(x) = C e ^{\frac{1}{\lambda} \int_0^x u(t)dt} $$

but $Tf(0) = 0 \Longrightarrow C = 0$. Does this mean that $\sigma(T) = {0}$ or what have I missed?

share|improve this question
    
What $u$ is doesn't matter too much. The $u\equiv 1$ case is done here, and with minor modifications you can generalise. –  Willie Wong Jan 8 '13 at 16:57
1  
The spectrum is not $\{0\}$, it is $\varnothing$, i.e., the empty set. Unless $u=0$, that is. –  user53153 Jan 8 '13 at 17:36
    
but isnt ${0}$ always in the spectrum of an compact operator? –  Johan Jan 8 '13 at 19:17
    
It is, but it is often not an eigenvalue but only the unique accumulation point of the set of eigenvalues. –  Branimir Ćaćić Jan 8 '13 at 20:12

1 Answer 1

up vote 1 down vote accepted

Assuming that your operator $T$ is indeed compact:

  1. Since $T$ is compact, and since $C([0,1])$ is infinite dimensional, it follows that $\sigma(T)$ is a non-empty, countable set containing $0$, and that all non-zero elements of $\sigma(T)$ must be eigenvalues.
  2. Your computation shows that for $\lambda \neq 0$, $Tf(x) = \lambda f(x)$ if and only if $f(x) = 0$, so that $T$ admits no non-zero eigenvalues. On the other hand, it also shows that $Tf(x) = 0$ if and only if $u(x)f(x) = 0$ for almost every $x$.

Putting these two facts together, it would seem that $\sigma(T) = \{0\}$, with $\ker T$ consisting of all $f \in C([0,1])$ such that $u(x)f(x) = 0$ almost everywhere.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.