# Spectrum of $Tf(x) := \frac{1}{x}\int_{0}^{x}f(t)dt$

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 ?

• 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. Nov 9, 2021 at 16:38
• You may also want to look at the case $\lambda = 0$ separately. Nov 9, 2021 at 16:54

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 $$Re(\lambda)\geq |\lambda|^2$$.
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 $$R(\lambda)g = \lambda x e^{\lambda x^2/2}\int_0^x t e^{-\lambda^2 t^2/2}g(t)dt+xg(x).$$