# Eigenvalues and Eigenfunctions of a Particular Self Adjoint Operator.

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 $T$.

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).

• Looks correct to me. The condition $f''''(1)=0$ is superfluous, since it corresponds to $f(1)=1$. Then you find a condition where the vector $(C_1,C_2,C_3,C_4)$ must be in the kernel of a $4\times 4$ matrix $M_\lambda$. This is a condition on $\lambda$, i.e. you have to pick $\lambda$ such that $M_\lambda$ is not injective. – Jonas Dahlbæk May 16 '17 at 16:43
• $f(1)=0$ not $1.$ – Vivek Kaushik May 16 '17 at 17:48
• Indeed, that was a typo. – Jonas Dahlbæk May 16 '17 at 17:53

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 :

$$\tag{2}\psi(x) := \int_a^{g(x)} f(t)dt \ \Longrightarrow \ \psi '(x) = g'(x)f(g(x))$$

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:

$$-\int_0^{1-x}f(y) \ dy$$

• Ok, based on the decomposition, I am getting the derivative of $\int_{0}^{1-x} f(y) (1-y) dy=- x f(1-x)$ and using the product rule on the other term, I am getting derivative of $-x \int_{0}^{1-x}f(y) dy$ to be $xf(1-x)- \int_{0}^{1-x} f(y) dy.$ Summing those two together yield the same answer as before. – Vivek Kaushik Oct 4 '16 at 6:27
• Ok, you are right. I transform my answer. I prefer this way, because, it can be of some help to readers because this first differentiation step (for people not accustomed to this kind of technique) is not evident at all. Moreover, it may help others to find where is the true problem. – Jean Marie Oct 4 '16 at 6:40
• I think that your problem is in the fact that you decompose onto a basis of 4 solutions, and thereby assume that you have a vector space of solutions which is 4-dimensional. But there is no reason for that... Had you stopped further, taking for example a fifth and sixth derivative, would, continuing the reasoning, yield a 6-dimensional space of solutions... – Jean Marie Oct 4 '16 at 7:08
• $e^{\frac{x}{\sqrt{\lambda}}}$ and $e^{-\frac{x}{\sqrt{\lambda}}}$ are not solutions because of condition $f(x)=0$. – Jean Marie Oct 4 '16 at 7:22
• ... nor $f(x)=\cos\left(\frac{x}{\sqrt{\lambda}} \right)$ or $f(x)=\sin \left(\frac{x}{\sqrt{\lambda}}\right)$ because of the restrictive conditions $f(1)=f ' (1)=0$ or because they don't verify $f(1-x)=\lambda f '' (x)$. I spent some time again trying to find other solutions, but without success. On your side, what is your conclusion ? Could you say what is the origin of the problem ? – Jean Marie Oct 4 '16 at 21:49