I'm trying to solve an exercise in Brezis' Functional Analysis
Let $(e_n)_{n \ge 1}$ be an orthonormal sequence in $H := L^2 (0, 1)$ with the real inner product $\langle \cdot, \cdot \rangle$. Fix $p \in H$.
- Prove that $$ \sum_{n=1}^\infty \left | \int_0^t p(s) e_n (s) \, \mathrm d s \right |^2 \le \int_0^t |p(s)|^2 \, \mathrm d s \quad \forall t \in (0, 1). \tag{1} $$
- Deduce that $$ \sum_{n=1}^\infty \int_0^1 \left | \int_0^t p(s) e_n (s) \, \mathrm d s \right |^2 \mathrm d t \le \int_0^1 (1-t) |p(t)|^2 \, \mathrm d t. \tag{2} $$
- Let $E := \{e_1, e_2, \ldots\}$. Assume here that $\operatorname{span} E$ is dense in $H$. Prove that (1) and (2) become equalities.
- Conversely, assume here that equality holds in (2) and that $p \neq 0$ a.e. Prove that $\operatorname{span} E$ is dense in $H$.
- Could you have a check on my attempt on (4.)?
- Is there an approach that sidesteps the separability of $H$?
Integrating both sides of (1) w.r.t $t$, we get (2). It follows from equality holding in (2) that equality holds in (1) for a.e. $t \in (0,1)$. By continuity, equality holds in (1) for all $t \in (0,1)$. Assume the contrary that $E$ is not dense in $H$. Because $H$ is separable, there is an orthonormal sequence $(e'_n)_{n \ge 1}$ such that $\langle e_n, e'_m \rangle=0$ for all $n,m \ge 1$ and that $\operatorname{span} (E \cup E')$ is dense in $H$. Here $E' := \{e'_1, e'_2, \ldots\}$. We have $$ \sum_{n=1}^\infty \left | \int_0^t p(s) e_n (s) \, \mathrm d s \right |^2 + \sum_{n=1}^\infty \left | \int_0^t p(s) e'_n (s) \, \mathrm d s \right |^2 = \int_0^t |p(s)|^2 \, \mathrm d s \quad \forall t \in (0, 1). $$
Then $$ \int_0^t p(s) e'_n (s) \, \mathrm d s = 0 \quad \forall t \in (0, 1), \forall n \ge 1. $$
Then $$ pe'_n = 0 \quad \text{a.e.} \quad \forall n \ge 1. $$
Because $p \neq 0$ a.e., $e'_n =0$ a.e. for all $n \ge 1$, which is a contradiction! This completes the proof.