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.


1 Answer 1


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 $x<y$, \begin{align} |f(y)-f(x)|&=\frac1{|\lambda|}\,\left|\int_x^yf(t)dt\right|\leq\frac1{|\lambda|}\,\int_x^y|f(t)|dt=\frac1{|\lambda|}\,\int_0^1|f(t)|\,1_{[x,y]}(t)\,dt\\ &\leq\frac{\|f\|_2}{|\lambda|}\,\left(\int_0^1(1_{[x,y]})^2\,dt\right)^{1/2}=\frac{\|f\|_2}{|\lambda|}\,\sqrt{y-x}. \end{align} So $f$ is continuous. Then looking at (1) again we get that $f$ is differentiable; thus, after differentiation, (1) is the equation $f=\lambda f'$. This implies that $f(t)=c\,e^{ t/\lambda}$. But $f(0)=0$, so the only solution for (1) is $f=0$, and $\lambda$ cannot be an eigenvalue.

The case $\lambda =0$ is trivial: if $Vf=0$, then $f=0$.

  • $\begingroup$ Excellent answer. +1. $\endgroup$
    – Potato
    Sep 1, 2013 at 1:42

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