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., $$ (1) \quad \sum_{k=0}^{\infty} |k \alpha_k(f)+a|^2<\infty. $$
There are possibly subtle mistakes that I could not recognize in my below attempt of (10.). Could you please have a check on it?
We have $$ \begin{align*} \alpha_k(f) &= \int_I e_k f \\ &= \sqrt{2} \int_I \sin \left[\left(k+\frac{1}{2}\right) \pi x\right] f (x) d x \\ &= \frac{-\sqrt{2}}{\left( k+\frac{1}{2} \right) \pi} \int_I \left ( \frac{d}{dx} \cos \left [ \left( k+\frac{1}{2} \right) \pi x \right ] \right ) f (x) d x \\ &= \frac{-\sqrt{2}}{\left( k+\frac{1}{2} \right) \pi} \left \{ \cos \left [ \left( k+\frac{1}{2} \right) \pi x \right ] f(x) \bigg |_0^1 -\int_I \cos \left [ \left( k+\frac{1}{2} \right) \pi x \right ] f' (x) d x \right \} \\ &= \frac{-\sqrt{2}}{\left( k+\frac{1}{2} \right) \pi} \left \{ -f(0) -\int_I \cos \left [ \left( k+\frac{1}{2} \right) \pi x \right ] f' (x) d x \right \} \\ &= \frac{\sqrt{2} f(0)}{\left( k+\frac{1}{2} \right) \pi} + \frac{\tilde \alpha_k(f')}{\left( k+\frac{1}{2} \right) \pi}, \end{align*} $$ where $(\tilde \alpha_k (f'))_k$ is the components of $f'$ in the basis $(\tilde e_k)$. Then $$ \begin{align*} k \alpha_k(f) + a &= a + \frac{\sqrt{2} k f(0)}{\left( k+\frac{1}{2} \right) \pi} + \frac{k \tilde \alpha_k(f')}{\left( k+\frac{1}{2} \right) \pi}. \end{align*} $$
Notice that the sequence $k \mapsto \frac{k \tilde \alpha_k(f')}{\left( k+\frac{1}{2} \right) \pi}$ belongs to $\ell^2$. So we need to find $a$ such that the sequence $k \mapsto a + \frac{\sqrt{2} k f(0)}{\left( k+\frac{1}{2} \right) \pi}$ also belongs to $\ell^2$. A necessary condition is that $$ \lim_{k \to +\infty} \left ( a + \frac{\sqrt{2} k f(0)}{\left( k+\frac{1}{2} \right) \pi} \right )=0, $$ which implies $$ a = -\frac{\sqrt{2} f(0)}{\pi}. $$