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 $T^*$.
There are possibly subtle mistakes that I could not recognize in below attempt of (4). Could you please have a check on it? I'm also happy to see other approaches.
Let $q \in (1, \infty]$ be the Hölder conjugate of $p$, i.e., $\frac{1}{p} + \frac{1}{q}=1$. Then $E^*=L^q(0, 1)$. For $u \in E$ and $v \in E^*$, we have $$ \begin{align*} \langle Tu, v \rangle_{E, E^*} &= \int_0^1 Tu(x) v(x) \, dx \\ &= \int_0^1 \int_0^x u(t) \, d t \, v(x) \, dx \\ &= \int_0^1 \int_0^1 1_{[0, x]} (t) u(t) \, d t \, v(x) \, dx \\ &= \int_0^1 \int_0^1 1_{[t, 1]} (x) u(t) \, d t \, v(x) \, dx \\ &= \int_0^1 \left ( \int_0^1 1_{[t, 1]} (x) v(x) \, dx \right ) u(t) \, d t \\ &= \int_0^1 \left ( \int_t^1 v(x) \, dx \right ) u(t) \, d t. \end{align*} $$
Then $$ T^* v(t) = \int_t^1 v(x) \, dx. $$
