Compactness and spectrum of an intergral operator.

Question

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 08 '13 at 16:57
The spectrum is not ${0}$, it is $\varnothing$, i.e., the empty set. Unless $u=0$, that is. — , Jan 08 '13 at 17:36
but isnt ${0}$ always in the spectrum of an compact operator? — Johan, Jan 08 '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 08 '13 at 20:12

score 1 · Accepted Answer · answered Jan 08 '13 at 20:51

Assuming that your operator $T$ is indeed compact:

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.
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.

Compactness and spectrum of an intergral operator.

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