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 $\| \cdot\|$ be the operator norm on $\mathcal L(E)$. 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 $E$ be a real Banach space and $T \in \mathcal L (E)$. Let $$ a_n := \log \|T^n\| \quad \forall n \in \mathbb N^*. $$
- Check that $a_{n+m} \le a_n + a_m$ for all $m, n \ge 1$.
- Deduce that $\lim_{n \to \infty} \frac{a_n}{n}$ exists and coincides with $\inf_{n \ge 1} \frac{a_n}{n}$.
- Conclude that $r(T) := \lim_n \|T^n\|^{1/n}$ exists and that $r(T) \le \|T\|$. Construct an example in $E=\mathbb{R}^2$ such that $r(T)=0$ and $\|T\|=1$. The number $r(T)$ is called spectral radius of $T$.
- Prove that $\sigma(T) \subset [-r(T), r(T)]$. Deduce that if $\sigma(T) \neq \emptyset$, then $$ \max_{\lambda \in \sigma (T)} |\lambda| \le r(T). $$
- Construct an example in $E=\mathbb{R}^3$ such that $\sigma(T)=\{0\}$, while $r(T)=1$.
In what follows we take $E=L^p(0,1)$ with $1 \leq p \leq \infty$. Consider the operator $T \in \mathcal{L}(E)$ defined by $$ T u(t)=\int_0^t u(s) \, d s . $$
- Prove by induction that for $n \geq 2$, $$ \left(T^n u\right)(t) = \frac{1}{(n-1) !} \int_0^t(t-s)^{n-1} u(s) \, d s. $$
- Deduce that $\|T^n\| \le \frac{1}{n!}$.
- Prove that the spectral radius of $T$ is $0$.
- Show that $\sigma(T) = \{0\}$.
There are possibly subtle mistakes that I could not recognize in below attempt of (7) and (8). Could you please have a check on it?
We extend $u \in L^p (0, 1)$ on the whole $\mathbb R$ by defining $u (s)=0$ for $s \in \mathbb R \setminus (0, 1)$. For $n \ge 1$, we define $$ K_n: \mathbb R \to \mathbb R, s \mapsto 1_{[0, 1]} (s) s^{n-1}. $$
Then $K_n \in L^1(0, 1)$ with $\|K_n\|_{L^1} = \frac{1}{n}$.
7.
By (6), $$ \begin{align*} (T^n u) (t) &= \frac{1}{(n-1) !} \int_{\mathbb R} 1_{[0, 1]} (t-s)(t-s)^{n-1} u(s) \, d s \\ &= \frac{1}{(n-1) !} (K_n * u) (t). \end{align*} $$
Then $T^n u= \frac{K_n * u}{(n-1) !}$. By Young's convolution inequality, $$ \|T^n u\|_{L^p} \le \frac{\|K_n\|_{L^1} \|u\|_{L^p}}{(n-1)!} = \frac{\|u\|_{L^p}}{n!}. $$
Then $\|T^n\| \le \frac{1}{n!}$.
8.
By (3) and (7), $$ \begin{align*} r(T) = \lim_n \|T^n\|^{1/n} \le \lim_n \frac{1}{\sqrt[n]{n!}}. \end{align*} $$
By Stirling's formula, $$ \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12 n+1}}<n !<\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12 n}}. $$
Then $\left(\frac{n}{e}\right)^n <n!$ and thus $\frac{n}{e} < \sqrt[n]{n!}$. Then $$ r(T) \le \lim_n \frac{2}{n} =0. $$
