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 my below attempt of (6.), I have proved that if $\lambda$ is an eigenvalue of $T$, then $\cos \left ( \frac{1}{\sqrt \lambda} \right ) =0$. To finish (6.), I have to prove the reverse. It is suffices to prove the system $(2)$ implies the system $(1)$. Could you elaborate on how to prove this?
- We have $$ \begin{align*} u (x) &= x \left ( \int_0^1 f(t)dt - \int_0^x f(t)dt \right ) + \int_0^x t f(t) dt \\ &= x \int_0^1 f(t)dt - x \int_0^x f(t)dt + \int_0^x t f(t) dt. \end{align*} $$
Because $f$ and $tf(t)$ are integrable, we get $u$ is absolutely continuous and thus differentiable a.e. on $I$. By chain rule, $$ u' (x) = \int_0^1 f(t)dt - \int_0^x f(t)dt. $$
Clearly, $u'$ is absolutely continuous with $u'' (x) = -f(x)$.
- We need auxiliary results (from the same book):
Let $H$ be a real Banach space and $T:H \to H$ a bounded linear operator. 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 $N(T-\lambda \operatorname{id})$ is called the eigenspace corresponding to $\lambda$.
Theorem 6.8 Assume $\dim E=\infty$ and $T$ be compact. Then $0 \in \sigma(T)$ and $\sigma(T) \setminus \{0\}=E V(T) \setminus\{0\}$ and one of the following cases holds:
- $\sigma(T)=\{0\}$,
- $\sigma(T) \backslash\{0\}$ is a finite set,
- $\sigma(T) \backslash\{0\}$ is a sequence converging to 0.
Proposition 6.9 Assume $H$ is a Hilbert space and $T$ self-adjoint. Let $$ m=\inf _{\substack{u \in H \\|u|=1}}(T u, u) \quad \text { and } \quad M=\sup _{\substack{u \in H \\|u|=1}}(T u, u). $$ Then $\sigma(T) \subset[m, M], m \in \sigma(T)$, and $M \in \sigma(T)$.
It follows from (2.) and Proposition 6.8 that $0 \in \sigma(T)$ and $\sigma(T) \setminus \{0\}=E V(T) \setminus\{0\}$. It follows from (3.) and Proposition 6.9 that $0 \in \sigma (T) \subset [0, +\infty)$.
To check whether $0$ is an eigenvalue of $T$ or not, we consider the equation $Tf = 0$. It follows from (5.) that $(Tf)''=-f$ and thus $f=0$. Then $T$ is injective and thus $0 \notin EV(T)$.
Let $\lambda >0$ and we consider the equation $$ (1) \quad \begin{cases} Tf = \lambda f, \\ f \neq 0. \end{cases} $$
From (5.), we have $(Tf)''=-f$ and $(Tf) (0) = (Tf)' (1)=0$. Then $(1)$ implies $$ (2) \quad \begin{cases} \lambda f''+f=0 \quad \text {on} \quad I, \\ f (0) = f'(1)=0, \\ f \neq 0. \end{cases} $$
From the first equation of $(2)$, we get $f$ is of the form $$ f (x) = A \cos \left ( \frac{x}{\sqrt \lambda} \right ) + B \sin \left ( \frac{x}{\sqrt \lambda} \right ), $$ which implies $$ f'(x) = \frac{B}{\sqrt \lambda} \cos \left ( \frac{x}{\sqrt \lambda} \right ) - \frac{A}{\sqrt \lambda} \sin \left ( \frac{x}{\sqrt \lambda} \right ). $$
We have $f(0)=0$ implies $A=0$. We have $f \neq 0$ and $f'(1)=0$ implies $B \neq 0$ and $\cos \left ( \frac{1}{\sqrt \lambda} \right ) =0$. It follows that $\lambda \in EV(T)$ implies $\cos \left ( \frac{1}{\sqrt \lambda} \right ) =0$.
