Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. Let $I:E \to E$ be the identity map. For $T \in \mathcal L(E)$,
- we denote by $N(T)$ its kernel and by $R(T)$ its range.
- we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
- for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.
I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E=L^p(0,1)$ with $1 \leq p<\infty$. Given $u \in E$, set $$ T u(x) = \int_0^x u(t) \, d t . $$
- Prove that $T \in \mathcal{K}(E)$.
- Determine $E V(T)$ and $\sigma(T)$.
- Give an explicit formula for $(T-\lambda I)^{-1}$ when $\lambda \in \rho(T)$.
- Determine $T^{\star}$.
There are possibly subtle mistakes that I could not recognize in below attempt of (3). Could you please have a check on it? I'm also happy to see other approaches.
Let $v \in E$. We look for $u\in E$ such that $(T-\lambda I)u=v$. Let $y:=Tu$. By Lebesgue differentiation theorem, $y$ is differentiable a.e. and $y'=u$ a.e. Let $\mu := -\frac{1}{\lambda}$. Then $y' + \mu y = \mu v$ a.e. By theory of ODE, we assume $y$ has a form $y (t) = K(t) e^{-\mu t}$ for some function $K$ we are looking for. Then $y' (t) = K'(t) e^{-\mu t} - \mu K(t) e^{-\mu t}$. Then $y' + \mu y = \mu v$ implies $K'(t) e^{-\mu t} = \mu v(t)$. Then $K'(t) = \mu v(t) e^{\mu t}$. Intuitively, we define $$ K(x) := \mu \int_0^x v(t) e^{\mu t} \, dt. $$
Then $$ \begin{align*} u (x) &= y'(x) \\ &= K'(x) e^{-\mu x} - \mu K(x) e^{-\mu x} \\ &= \mu v(x) - \mu^2 e^{-\mu x} \int_0^x v(t) e^{\mu t} \, dt \\ &= \mu v(x) - \mu^2 \int_0^x v(t) e^{\mu (t-x)} \, dt. \end{align*} $$
