Consider the operator T on $L^2[0,1]$, given by $T(f(x)) = \int_{1-x}^1 f(y)dy$. I want to find the spectrum of this operator. I know the only possible candidates are 0 and non-zero Eigen values of T, since it is a compact operator.

Since it is infinite dimensional $0$ has to be in spectrum.

Now I've to check only non-zero eigen values $\lambda$ of T, but $T(f(x))= \lambda f(x)$ gives the following ODE, $f'(x) = -\frac{1}{\lambda}f(1-x)$.

Please help me how to proceed from here.

  • $\begingroup$ I'm not 100% positive, but since an eigenvalue would have to make that true for all $f$ in $L^2$, if it's not true for a single $L^2$ function it isn't an eigenvalue. Pretty sure for NO $\lambda$ will that be true for ALL $L^2$ functions, just find a single one it doesn't work for. I'd try simple ones like polynomials $\endgroup$
    – Alan
    May 16 at 16:50
  • $\begingroup$ @Alan Not really. Eigenvalues are always associated to an eigenfunction (just like in the finite dimensional case). So it is enough to find a single (nonzero) function that solves the eigenvalue equation. $\endgroup$ May 16 at 16:54

1 Answer 1


An eigenfunction $f$ corresponding to $\lambda\neq 0$ satisfies $f(0)=0.$ Moreover $$f'(x)=-{1\over \lambda}\,f(1-x)$$ Hence $f'(1) =0.$ Applying next derivative gives $$f''(x)=-{1\over \lambda^2}\,f(x)$$ Thus $$ f(x)=a\,\cos (\lambda^{-1}x)+b\,\sin(\lambda^{-1}x)$$ Substituting $x=0$ gives $a=0.$ Next $$0=f'(1)=b\lambda^{-1}\cos(\lambda^{-1}) $$ Therefore $\cos(\lambda^{-1})=0,$ hence $\lambda^{-1}=\pi/2 +n\pi,$ $n\in \mathbb{Z}.$ The corresponding eigenfunction is equal $f(x)=\sin(\lambda^{-1}x).$


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