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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 I could not recognize in below attempt that $0 \notin EV(T)$. Could you please have a check on it? I'm also happy to see other approaches.

We have $\dim E= \infty$ and $T \in \mathcal{K}(E)$, so $0\in \sigma(T)$. First, we prove that $0 \notin EV(T)$. Let $u, v\in E$ such that $Tu=Tv$.

  • Then $\int_0^x (u(t)-v(t)) \, d t = 0$ for a.e. $x\in (0, 1)$.
  • Then $\int_0^x (u(t)-v(t)) \, d t = 0$ for every $x$ in some dense subset of $(0, 1)$.
  • By dominated convergence theorem, $\int_0^x (u(t)-v(t)) \, d t = 0$ for every $x \in (0, 1)$.
  • Then $\int_I (u(t)-v(t)) \, d t = 0$ for every sub-interval $I$ of $(0, 1)$.
  • By dominated convergence theorem, $\int_B (u(t)-v(t)) \, d t = 0$ for every Borel subset $B$ of $(0, 1)$.
  • Then $u-v=0$ a.e.
  • Then $T$ is injective.
Akira
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    How do you go from integral over $I$ to integral over $B$? Use Lebesgue's differentiation theorem instead of DCT for this. – geetha290krm Jul 26 '23 at 09:28
  • @geetha290krm Because Lebesgue measure is regular, we can approximate $B$ be open set. On the other hand, we can approximate open set by open interval. – Akira Jul 26 '23 at 09:31
  • You can approximate $B$ by a finite union of intervals in a particular sense. Certainly, you cannot find sequence if intervals $I_n$ such that $1_{I_n}$ converges a.e. to $1_B$. So you have to explain how you are applying DCT. – geetha290krm Jul 26 '23 at 09:39
  • @geetha290krm I have posted a fix as below answer. Could you have a check on it? – Akira Jul 26 '23 at 10:25
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    It is correct, but I still think that Lebsgue's Theorem is the right approach, since it gives the result immediately. – geetha290krm Jul 26 '23 at 11:18
  • @geetha290krm Sure. It's also good that I come up with my own approach ;) – Akira Jul 26 '23 at 11:20

1 Answers1

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As @geetha290krm pointed out in a comment, the passing from intervals to Borel sets is not justified. Below is my fix.

Let $\mathcal C$ be the collection of all sub-intervals of $(0, 1)$. Then $\mathcal C$ is a $\pi$-system. Let $\mathcal D$ be the collection of all Borel subsets $B$ of $(0, 1)$ such that $\int_B (u(t)-v(t)) \, d t = 0$ . Then $\mathcal D$ is a $\lambda$-system. We have proved that $\mathcal C \subset \mathcal D$. By Dynkin's $\pi$-$\lambda$ theorem, $\sigma(\mathcal C) \subset \mathcal D$. Thus $\mathcal D$ indeed coincides with the Borel $\sigma$-algebra of $(0, 1)$.

Akira
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