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 (1). Could you please have a check on it? I'm also happy to see other approaches.
Let $\| \cdot \|_p$ be the canonical norm of $L^p(0,1)$. Because the open interval $(0, 1)$ has a finite Lebesgue measure, we have $L^p (0,1) \subset L^1(0,1)$ and thus $Tu (x)$ is well-defined. Clearly, $T$ is linear. We have $$ \begin{align} \| T u\|^p_p &= \int_0^1 |Tu(x)|^p \, d x \\ &= \int_0^1 \left | \int_0^x u(t) \, d t \right |^p d x \\ &= \int_0^1 \int_0^x |u(t)|^p \, d t \, d x \\ &\le \int_0^1 \int_0^1 |u(t)|^p \, d t \, d x \\ &= \|u\|_p^p <\infty. \end{align} $$
Then $Tu$ indeed belongs to $E$. Also, $T$ is bounded with $\|T\|_{\mathcal L(E)} \le 1$.
1.
Let $(u_n)$ be a sequence in $E$ with $\|u_n\|_p \le 1$ for all $n \in \mathbb N^*$. Then $\|Tu_n\|_p \le 1$ for all $n \ge 1$. Let's prove that $(Tu_n)_n$ has a compact closure in $E$.
Let $\tau_h$ be the shift operator, i.e., $(\tau_h f) (x) =f(x+h)$ for all $x, h \in \mathbb R$. By Fréchet–Kolmogorov theorem, it suffices to prove that $$ \lim_{|h| \to 0} \sup_n \| \tau_h Tu_n-Tu_n\|^p_p =0. $$
For $h>0$, we have $$ \begin{align*} \| \tau_h Tu_n-Tu_n\|^p_p &= \int_0^1 \left | \int_0^{x+h} u_n (t) \, d t - \int_0^{x} u_n (t) \, d t \right |^p dx \\ &= \int_0^1 \left | \int_x^{x+h} u_n (t) \, d t \right |^p dx \\ &\le \int_0^1 \int_x^{x+h} |u_n (t)|^p \, d t \, dx \\ &= \int_0^1 \int_0^1 1_{[x, x+h]} (t) |u_n (t)|^p \, d t \, dx \\ &= \int_0^1 |u_n (t)|^p \left ( \int_{t-h}^t \, dx \right ) \, d t \\ &= h\int_0^1 |u_n (t)|^p\, d t \\ &= h\|u_n\|_p^p \\ &\le h. \end{align*} $$
The claim then follows.
