# Compactness and spectrum of an intergral operator.

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?

-
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
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

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.