Let $I$ be the open interval $(0, 1)$ and $H := L^2 (I)$ equipped with the usual inner product $\langle \cdot, \cdot \rangle$. Consider the linear map $T: H \to H$ defined by $$ (Tf) (x) = \int_0^x t f(t) dt + x \int_x^1 f(t)dt, \quad x \in I. $$
I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.28
Check that $T$ is a bounded linear operator.
Check that $T$ is a compact operator.
Check that $T$ is self-adjoint.
Show that $\langle Tf, f \rangle \ge 0$ for all $f \in H$, and that $\langle Tf, f \rangle = 0$ implies $f =0$.
Set $u=T f$. Prove that $u \in H^2(I)$ and compute $u''$. Check that $u(0)=u'(1)=0$
Determine the spectrum and the eigenvalues of $T$. Examine carefully the case $\lambda=0$.
In what follows, set $$ e_k(x) = \sqrt{2} \sin \left[\left(k+\frac{1}{2}\right) \pi x\right], \quad k=0,1,2, \ldots $$ Given $f \in H$ we denote by $\left(\alpha_k(f)\right)$ the components of $f$ in the basis $\left(e_k\right)$.
Check that $\left(e_k\right)$ is an orthonormal basis of $H$.
Deduce that the sequence $\left(\tilde{e}_k\right)$ defined by $$ \tilde{e}_k(x)=\sqrt{2} \cos \left[\left(k+\frac{1}{2}\right) \pi x\right], \quad k=0,1,2, \ldots, $$ is also an orthonormal basis of $H$.
Compute $\alpha_k(f)$ for the following functions: (a) $f_1(x)=\chi_{[a, b]}(x)= \begin{cases}1 & \text { if } x \in[a, b] \\ 0 & \text { if } x \notin[a, b]\end{cases}$ where $0 \leq a<b \leq 1$. (b) $f_2(x)=x$. (c) $f_3(x)=x^2$.
Finally, we propose to characterize the functions $f \in L^2 (I)$ that belong to $H^1 (I)$, using their components $\alpha_k(f)$.
Assume $f \in H^1 (I)$. Prove that there exists a constant $a \in \mathbb{R}$ (depending on $f)$ such that $(k \alpha_k(f)+a)_k \in \ell^2$, i.e., $$ (1) \quad \sum_{k=0}^{\infty} |k \alpha_k(f)+a|^2<\infty. $$
Conversely, assume that $f \in L^2 (I)$ and that $(1)$ holds for some $a \in \mathbb{R}$. Prove that $f \in H^1 (I)$.
In my below attempt of (11.), I am stuck at proving $\sum_{k=0}^{\infty} \left | \frac{1}{2} \alpha_k (f) - a \right |^2 < + \infty$. Could you provide some hints to finish the proof?
Fix $\varphi \in C^\infty_c (I)$. We have $$ \begin{align*} \int_I f \varphi' &= \int_I \sum_{k=0}^{\infty} \alpha_k (f) e_k \varphi' \\ &= \sum_{k=0}^{\infty} \alpha_k (f) \int_I e_k \varphi' \\ &= - \sum_{k=0}^{\infty} \alpha_k (f) \int_I e_k' \varphi \\ &= - \pi \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (f) \int_I \tilde e_k \varphi \\ &= - \int_I \left ( \pi \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (f) \tilde e_k \right ) \varphi. \end{align*} $$
It suffices to prove $$ v := \pi \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (f) \tilde e_k \in L^2 (I). $$
It remains to prove $$ C := \sum_{k=0}^{\infty} \left | \left(k+\frac{1}{2}\right) \alpha_k (f) \right |^2 < + \infty. $$
Indeed, $$ \begin{align*} C &= \sum_{k=0}^{\infty} \left | \left ( k \alpha_k (f) +a \right ) + \left ( \frac{1}{2} \alpha_k (f) - a \right ) \right |^2 \\ &\le \sum_{k=0}^{\infty} | k \alpha_k (f) +a |^2 + \sum_{k=0}^{\infty} \left | \frac{1}{2} \alpha_k (f) - a \right |^2. \end{align*} $$
Let $g := f + \frac{a \pi}{\sqrt 2}$. It suffices to prove that $g \in H^1 (I)$. We have $$ \begin{align*} \alpha_k (g) &= \int_I e_k g \\ &= \int_I e_k \left ( f + \frac{a \pi}{\sqrt 2} \right ) \\ &= \alpha_k (f) + \frac{a \pi}{\sqrt 2} \int_I e_k \\ &= \alpha_k (f) + \frac{2 a}{2k+1}, \end{align*} $$ which implies $$ \begin{align*} k \alpha_k (f) + a &= k \alpha_k (g) - \frac{2 k a}{2k+1} + a \\ &= k \alpha_k (g) + \frac{a}{2k+1}. \end{align*} $$
Notice that the sequence $k \mapsto \frac{a}{2k+1}$ belongs to $\ell^2$, so the sequence $k \mapsto k \alpha_k (g)$ is also in $\ell^2$. For $\varphi \in C^\infty_c (I)$, we have $$ \begin{align*} \int_I g \varphi' &= \int_I \sum_{k=0}^{\infty} \alpha_k (g) e_k \varphi' \\ &= \sum_{k=0}^{\infty} \alpha_k (g) \int_I e_k \varphi' \\ &= - \sum_{k=0}^{\infty} \alpha_k (g) \int_I e_k' \varphi \\ &= - \pi \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (g) \int_I \tilde e_k \varphi \\ &= - \int_I \left ( \pi \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (g) \tilde e_k \right ) \varphi. \end{align*} $$
It suffices to prove $v := \sum_{k=0}^{\infty} \left(k+\frac{1}{2}\right) \alpha_k (f) \tilde e_k \in L^2 (I)$. By (8.), it suffices to prove $\sum_{k=0}^{\infty} \left | \left(k+\frac{1}{2}\right) \alpha_k (g) \right |^2 < + \infty$. It follows from $f \in L^2 (I)$ that $g \in L^2 (I)$ and thus $\sum_{k=0}^{\infty} \left | \frac{1}{2} \alpha_k (g) \right |^2 < + \infty$. It remains to prove $\sum_{k=0}^{\infty} \left | k \alpha_k (g) \right |^2 < + \infty$, which is true.