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