# Volterra Operator is compact but has no eigenvalue

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.

• – Tomás Oct 23 '12 at 22:55
• I noticed that you used eqnarray for your math. Don't do that! – kahen Oct 24 '12 at 7:53

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 then we would have $$ce^{x/\lambda}=\frac1\lambda\,\int_0^x\,c\,e^{t/\lambda}\,dt=ce^{x/\lambda}-c,$$ which implies that $c=0$. 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 $f=0$.