Let $T : C[0,1] \to C[0,1]$ where

$Tf(x) := \frac{1}{x}\int_{0}^{x}f(t)dt , x > 0$

$(Tf)(0) = f(0)$

I'm trying to solve for the spectrum $\sigma(T)$

What I have achieved so far is to show that $T$ is bounded, specifically that $||T|| = 1$. This means that $\lambda \in \sigma(T) \implies |\lambda| \leq 1$ (Correct?)

Also, I attempted to solve $Tf = \lambda f$. Setting $g(x) =\int_{0}^{x}f(t)dt$, we get:

$g(x) = x\lambda g'(x)$, since $g' = f$

Then, since the D.E. is separable, we get:

$\int\frac{dg}{g} = \int\frac{dx}{\lambda x} \implies ln(g) = \frac{ln(x)}{\lambda} + C_0 \implies g(x) = Cx^{\frac{1}{\lambda}}$ for some $C > 0$.

This gives me that $f(x) = g'(x) = \frac{C}{\lambda}x^{\frac{1}{\lambda}-1}$

How can I proceed from here ?

  • 1
    $\begingroup$ When $f$ is a constant, $Tf=f$, so $1$ is in the point spectrum. As you proceeded, the solution of $Tf=\lambda f$ when $-1 \le \lambda <1$ can not extend to a function continuous on the whole $[0,1]$, so what leaves to you is to decide whether $\lambda$ belongs to the continuous spectrum. $\endgroup$
    – Zerox
    Nov 9, 2021 at 16:38
  • $\begingroup$ You may also want to look at the case $\lambda = 0$ separately. $\endgroup$
    – podiki
    Nov 9, 2021 at 16:54

2 Answers 2


Firstly, yes, since $\|T\|=1$, we have $Tf=\lambda f \implies |Tf|=|\lambda||f|\leq |f| \implies |\lambda|\leq 1$ for eigenvalues.

From $f(x)=\frac{C}{\lambda}x^{\frac{1}{\lambda}-1}$, with $\lambda\neq 0$. We need $Re\left(\frac{1-\lambda}{\lambda}\right)\geq 0$ for continuity, which is equivalent to $$Re(\bar{\lambda}-|\lambda|^2)\geq 0$$ or $Re(\lambda)\geq |\lambda|^2$.

  • $\begingroup$ So all values satisfying above constitute the whole spectrum? The continuous spectrum is empty? $\endgroup$
    – JustANoob
    Nov 9, 2021 at 18:30

The resolvent equation for $T$ is $$ \left(\lambda f(x) -\frac{1}{x}\int_0^x f(t)dt\right)= g(x) \\ \lambda x f(x)-\int_0^x f(t)dt = xg(x) \\ \lambda x \frac{d}{dx}\int_0^x f(t)dt-\int_0^x f(t)dt = x g(x) \\ -\frac{d}{dx}\left(e^{-\lambda x^2/2}\int_0^x f(t)dt\right)=-xe^{-\lambda x^2/2}g(x) \\ -e^{-\lambda x^2/2}\int_0^x f(t)dt =-\int_0^xte^{-\lambda t^2/2}g(t)dt \\ \int_0^x f(t)dt = e^{\lambda x^2/2}\int_0^x te^{-\lambda t^2/2}g(t)dt \\ f(x)=\lambda xe^{\lambda x^2/2}\int_0^x te^{-\lambda^2 t^2/2}g(t)dt+e^{\lambda x^2/2}xe^{-\lambda x^2/2}g(x) \\ f(x)=\lambda xe^{\lambda x^2/2}\int_0^xte^{-\lambda^2t^2/2}g(t)dt+xg(x) $$ Therefore, the resolvent of $T$ is given by $$ R(\lambda)g = \lambda x e^{\lambda x^2/2}\int_0^x t e^{-\lambda^2 t^2/2}g(t)dt+xg(x). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.