Consider the operator $T(f)(x): L^2[0,1] \rightarrow L^2[0,1]$ defined $$T(f)(x)=\int_{0}^{1-x} f(y) \ (1-y-x) \ dy.$$ (Assume $L^2[0,1]$ is the set of square integrable real valued functions over the interval $[0,1].$)
I know that this operator is self-adjoint, which I proved through showing $\langle T(f), g \rangle= \langle f, T(g) \rangle$ where the inner product is the integral inner product, but I have been unable to get the eigenfunctions/eigenvalues of $T$.
My attempt was this:
Suppose $$T(f)(x)=\lambda f(x)$$ for $\lambda \in \mathbb{R}.$ From the definition of $T$, we see that $f(1)=0.$
We then differentiate both sides of the equality to get
$$-\int_{0}^{1-x} f(y) \ dy= \lambda f'(x).$$ Here, we see $f'(1)=0.$
Differentiating both sides again, we get $$f(1-x) = \lambda f''(x),$$ which implies $f''(0)=0$ and through a change of variables, $$f(x)= \lambda f''(1-x).$$
I then differentiated $$f(1-x) = \lambda f''(x)$$ again to get $$-f'(1-x)=\lambda f'''(x)$$ which implies $f'''(0)=0.$
Differentiating one more time, gave me $$f''(1-x)=\lambda f''''(x)$$ with $f''''(1)=0.$
Combining this with $$f(x)= \lambda f''(1-x),$$ I got a 4th order differential equation $$f(x)=\lambda ^2 f''''(x)$$ with derived initial conditions: $f(1)=f'(1)=f''(0)=f'''(0)=f''''(1)=0.$
The general solution takes on form $$f(x)=C_1e^{\frac{x}{\sqrt{\lambda}}}+C_2e^{\frac{-x}{\sqrt{\lambda}}}+C_3\cos\left(\frac{x}{\sqrt{\lambda}} \right)+C_4\sin\left(\frac{x}{\sqrt{\lambda}} \right).$$ However, the problem is the initial conditions give the solution $f(x)=0,$ which I don't think is correct because I think Compact Self Adjoint Operators are supposed to have nonzero eigenfunctions. Is there some kind of error in my reasoning ?
To give some motivation, I am trying to mimic the ideas in https://www.maa.org/sites/default/files/pdf/news/Elkies.pdf (see the section about linear operators).
This is not an answer. It is just there for clarifying the first differentiation step (I thought, erroneously, at first, that the problem was there).
The action of the operator on a function $f$ can be decomposed in the following way (in order to get rid of the presence of "parameter" $x$ in an integrand) :
$$\tag{1}\int_{0}^{1-x} f(y) \ dy=\int_{0}^{1-x} f(y)(1-y) \ dy - x\int_{0}^{1-x} f(y) \ dy $$
Now, recall the derivation rule :
$$\tag{2}\psi(x) := \int_a^{g(x)} f(t)dt \ \Longrightarrow \ \psi '(x) = g'(x)f(g(x)) $$
See (Find the derivative of $\int_a^{g(x)} f(t)dt $ wrt $x$).
Then differentiate (1) by using (twice) property (2):
$$-f(1-x)(1-1+x)-1\int_0^{1-x}f(y) \ dy+x f(1-x)$$
finally giving:
$$-\int_0^{1-x}f(y) \ dy$$