# 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.

• Oct 23, 2012 at 22:55
• I noticed that you used eqnarray for your math. Don't do that! Oct 24, 2012 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 $$x, \begin{align} |f(y)-f(x)|&=\frac1{|\lambda|}\,\left|\int_x^yf(t)dt\right|\leq\frac1{|\lambda|}\,\int_x^y|f(t)|dt=\frac1{|\lambda|}\,\int_0^1|f(t)|\,1_{[x,y]}(t)\,dt\\ &\leq\frac{\|f\|_2}{|\lambda|}\,\left(\int_0^1(1_{[x,y]})^2\,dt\right)^{1/2}=\frac{\|f\|_2}{|\lambda|}\,\sqrt{y-x}. \end{align} So $$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$$.