Brezis' exercise 8.29.5: if $f \geq 0$ a.e. on $I$, then $T f \geq 0$ on $I$

Question

Let $I$ be the open interval $(0, 1)$. Consider the symmetric bilinear form $a$ defined on $H^1 (I)$ by $$ a(u, v) = \int_I [ u'v' + uv ] + [u(1) - u(0)] [v(1) - v(0)]. $$

I am trying to solve a problem in Brezis' Functional Analysis

Exercise 8.29

1. Check that $a$ is a continuous coercive bilinear form on $H^1 (I)$.
2. Deduce that for every $f \in L^2 (I)$, there exists a unique $u \in H^2 (I)$ satisfying $$ (1) \quad a(u, v)=\int_I f v, \quad \forall v \in H^1 (I) . $$ 3.Check that $u$ satisfies $$ (2) \quad \begin{cases} -u''+u=f \quad \text {on} \quad I, \\ u'(0)=u(0)-u(1), \\ u'(1)=u(0)-u(1). \end{cases} $$ Show that any solution $u \in H^2 (I)$ of $(2)$ satisfies $(1)$.

Let $T: L^2 (I) \to L^2 (I)$ be the operator defined by $T f=u$.

4. Check that $T$ is self-adjoint and compact.
5. Show that if $f \geq 0$ a.e. on $I$, then $u=T f \geq 0$ on $I$.
6. Check that $\langle T f, f \rangle_{L^2} \ge 0$ for all $f \in L^2 (I)$.
7. Determine the set $E V(T)$ of eigenvalues of $T$.

There are possibly subtle mistakes that I could not recognize in my below attempt of (5.). Could you please have a check on it?

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We fix $G \in C^1 (\mathbb R)$ such that

• $G$ is strictly increasing on $(0, + \infty)$,
• $G(t) = 0$ for all $t \in (- \infty, 0]$.

Let $K := \operatorname{ess-sup}_{I} f$. We will prove that $u \le K$. WLOG, we assume $K < + \infty$ and thus $K \in \mathbb R$. Let $v = G(u-K)$. By Corollary 8.11 (in the same book), $v \in H^1 (I)$ with $v' = u' G' (u-K)$. We plug $v$ into $(1)$ and get $$ \int_I [ |u'|^2 G' (u-K) + u G (u-K) ] + [u(1) - u(0)] [ G (u(1)-K) - G (u(0)-K)] = \int_I f G(u-K), $$ which implies $$ \int_I [ |u'|^2 G' (u-K) + (u-K) G (u-K) ] + [u(1) - u(0)] [ G (u(1)-K) - G (u(0)-K)] = \int_I (f-K) G(u-K). $$

We have $(f-K) G(u-K) \le 0$ a.e. on $I$ and thus $\int_I (f-K) G(u-K) \le 0$. We have $G$ is increasing, so $[u(1) - u(0)] [ G (u(1)-K) - G (u(0)-K)] \ge 0$. We have $G' \ge 0$, so $\int_I |u'|^2 G' (u-K) \ge 0$. As such, $$ \int_I (u-K) G (u-K) \le 0. $$

On the other hand, $tG(t) \ge 0$ for all $t \in \mathbb R$. Then $(u-K) G (u-K)=0$ a.e. on $I$. Then $u-K \le 0$ and thus $u \le K$ a.e. on $I$. The lower bound for $u$ is obtained by applying this upper bound to $-u$.