Volterra operator is defined as operator $V:L^2[0,1]\rightarrow L^2[0,1]$ by \begin{eqnarray} (V)(f(x))=\int_0^xf(y)dy \end{eqnarray} Would you help me to prove that this operator is compact but has no eigenvalues.
Volterra Operator is compact but has no eigenvalue
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2I noticed that you used `eqnarray` for your math. [Don't do that](http://tex.stackexchange.com/questions/196/eqnarray-vs-align)! – 2012-10-24
1 Answers
Note that $ Vf(x)=\int_0^1f(t)k(x,t)\,dt, $ where $k(x,t)=1_{[0,x]}(t)$. It is a general fact that such an operator is Hilbert-Schmidt (and in particular compact) if and only if $k\in L^2([0,1]^2)$. Or one can show that the measurable function $k$ is a uniform limit of simple functions, and these simple functions can be used as kernels to define operators that approximate $V$. As these operators are finite-rank, $V$ is compact.
As for the eigenvalues, if $\lambda\ne0$ and $Vf=\lambda f$, then we get $\tag{1} f(x)=\frac1\lambda\,\int_0^xf(t)\,dt. $ Using that $f$ is in $L^2$ we have, for $x
The case $\lambda =0$ is trivial: if $Vf=0$, then $f=0$.
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0Excellent answer. +1. – 2013-09-01