# Spectrum of the operator on $L^2[0,1]$

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, $$f'(x) = -\frac{1}{\lambda}f(1-x)$$.

• 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
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 $$f(x)=\sin(\lambda^{-1}x).$$