Consider a linear operator $A\colon L^2[0,1]\rightarrow L^2[0,1]$ that acts as follows: $$Af(t) = \int_0^{t^2} f(s)ds$$

The problem is to compute its spectrum.

I know that the operator is compact (it is an integral operator with kernel $K(t,s)=\mathbb I\{s\leq t^2\}$), so its spectrum consists of 0 and eigenvalues.

Writing an equation for its eigenvalue $\lambda$, I got $$ \begin{cases} 2t\; f(t^2) = \lambda f'(t) \\ f(0) = 0 \end{cases} $$

I have no idea how to deal with this system or what I can do else.

Thank you for help.


Assuming: $$ A(f)(t)=\int_{0}^{t^2}f(s)\,ds = \int_{0}^{t}2u\, f(u^2)\,du=\lambda\cdot f(t) \tag{1}$$ we may look for first for analytic solutions in a neighbourhood of zero. If $$ f(z) = \sum_{n\geq 1} a_n\, z^{2n}\tag{2}$$ then: $$ A(f)(z) = \sum_{n\geq 1} \frac{a_n}{n+1} z^{2n+2} = \sum_{n\geq 2}\frac{a_{n-1}}{n} z^{2n} \tag{3}$$ so the only analytic eigenfunction is the null function. On the other hand, we may also assume $$ f(z) = \sum_{n\geq 0}c_n\cos(2\pi n z),\quad \sum_{n\geq 0}c_n=0 \tag{4}$$ since $(1)$ implies that $f$ is an even function in $L^2([0,1])$. $(4)$ implies: $$ f(z) = 2\sum_{n\geq 1}d_n \sin^2(\pi n z) \tag{5}$$ and we have: $$ A\left(\sin^2(\pi n z)\right) = \frac{2n\pi z^2-\sin(2n\pi z^2)}{4n\pi}\tag{6}$$ so we may get recursive formulas for the coefficients $d_n$ by integrating $(5)$ and $(6)$ against $z,z^3,z^5,\ldots$ on $[0,1]$. The spectrum of $A$ is bounded since the operator $A$ contracts the $L^2$ norm.

  • $\begingroup$ What does "$f$ is an even function in $L^2([0,1])$" mean? $\endgroup$ – quartermind Apr 27 '15 at 18:30
  • $\begingroup$ @quartermind: a square-integrable function and an even function. $\endgroup$ – Jack D'Aurizio Apr 27 '15 at 18:50

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