Brezis' exercise 6.20.2: prove that $\sigma (T) = \{0\}$

Question

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 $$ T u(x) = \int_0^x u(t) \, d t . $$

1. Prove that $T \in \mathcal{K}(E)$.
2. Determine $E V(T)$ and $\sigma(T)$.
3. Give an explicit formula for $(T-\lambda I)^{-1}$ when $\lambda \in \rho(T)$.
4. Determine $T^{\star}$.

There are possibly subtle mistakes that I could not recognize in below attempt of (2). Could you please have a check on it? I'm also happy to see other approaches.

---

I already proved that $0\in \sigma(T)$ and $0 \notin EV(T)$. Because $T \in \mathcal{K}(E)$, we have $\sigma (T) \setminus \{0\} = EV (T) \setminus \{0\}$ by Theorem 6.8 in the same book. Next, we will prove that $\sigma(T) = \{0\}$. Let $\lambda \in \mathbb R \setminus \{0\}$ and $u\in E$ such that $$ (T-\lambda I)u=0. \tag{1} $$

Then $$ \int_0^x u(t) \, dt = \lambda u(x) \quad \text{for a.e.} \quad x \in (0, 1). \tag{2} $$

It suffices to prove that equation (1) has uniqueness. Let $v \in L^p(0, 1)$ be another solution of (1), i.e., $$ \int_0^x v (t) \, dt = \lambda v (x) \quad \text{for a.e.} \quad x \in (0, 1). \tag{3} $$

We can modify $u,v$ on a null set such that equation (2, 3) holds for every $x \in (0, 1)$. By dominated convergence theorem, $u$ and $v$ are continuous on $(0, 1)$. Let $L := \frac{1}{|\lambda|}>0$ and $\varphi (x) := |u(x)-v(x)|$ for $x \in (0, 1)$. Then $\varphi$ is continuous and $$ \varphi (x) \le L \int_0^x \varphi(t) \, dt. \quad \forall x \in (0, 1). $$

Let $$ \psi(x) := e^{-L x} \int_0^x \varphi(t) \, dt \quad x \in [0, 1). $$

For every $x \in (0, 1)$, $$ \begin{align*} \psi' (x) &= -Le^{-Lx} \int_0^x \varphi(t) \, dt + e^{-Lx} \varphi (x) \\ &= -L \psi (x) + e^{-Lx} \varphi (x) \\ &\le -L \psi (x) + Le^{-Lx} \int_0^x \varphi (t) \, dt \\ &= -L \psi (x) + L \psi (x) \\ &= 0. \end{align*} $$

We have $\psi$ is non-negative and differentiable on $(0, 1)$ such that $\psi(0)=0$ and $\psi' \le 0$. Then $\psi \equiv 0$ and thus $\varphi = 0$ a.e. Because $\varphi$ is continuous, we get $\varphi \equiv 0$ everywhere. This completes the proof.

Answer 1

It can be shown by induction that $$(T^nf)(x)={1\over (n-1)!}\int\limits_0^x(x-t)^{n-1}f(t)\,dt$$ Hence $$|(T^nf)(x)|\le {1\over (n-1)!}\int\limits_0^1|f(t)|\,dt\le {1\over (n-1)!}\|f\|_p$$ Therefore $$\|T^n\|^{1/n}\le {1\over ((n-1)!)^{1/n}}\to 0$$ Hence $\sigma(T)=\{0\}.$ Concerning the formula for $(\lambda I-T)^{-1}, $ $\lambda\neq 0,$ the simplest way is by applying the ODE. An alternative way is $$(\lambda I-T)^{-1}=\sum_{n=0}^\infty \lambda^{-n-1}T^n\\ =\lambda^{-1}I+\sum_{n=1}^\infty \lambda^{-n-1}T^n$$ By applying the formula for $T^n$ and $$\sum_{n=1}^\infty {1\over (n-1)!}(x-t)^{n-1}=e^{x-t}$$ we get $$[(\lambda I-T)^{-1}f](x)\\ =\lambda^{-1}f(x) + e^x\int\limits_0^xe^{-t}f(t)\,dt$$