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Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.24

  1. Prove that for every $\varepsilon>0$ there exists a constant $C_{\varepsilon}$ such that $$ |u(1)|^2 \leq \varepsilon\left\|u^{\prime}\right\|_{L^2}^2+C_{\varepsilon}\|u\|_{L^2}^2 \quad \forall u \in H^1(I) . $$
  2. Prove that if the constant $k>0$ is sufficiently large, then for every $f \in L^2(I)$ there exists a unique $u \in H^2(I)$ satisfying $$ (1) \quad \begin{cases} -u^{\prime \prime}+k u=f \quad \text{on} \quad I, \\ u^{\prime}(0)=0 \quad \text{and} \quad u^{\prime}(1)=u(1). \end{cases} $$ What is the weak formulation of problem (1)? What is the associated minimization problem?
  3. Assume that $k$ is sufficiently large. Let $T$ be the operator $T: f \mapsto u$, where $u$ is the solution of (1). Prove that $T$ is a self-adjoint compact operator in $L^2(I)$.
  4. Deduce that there exist a sequence $\left(\lambda_n\right)$ in $\mathbb{R}$ with $\left|\lambda_n\right| \rightarrow \infty$ and a sequence $\left(u_n\right)$ of functions in $C^2(\bar{I})$ such that $\left\|u_n\right\|_{L^2(I)}=1$ and $$ \left\{\begin{array}{l} -u_n^{\prime \prime}=\lambda_n u_n \quad \text {on} \quad I, \\ u_n^{\prime}(0)=0 \text { and } u_n^{\prime}(1)=u_n(1) . \end{array}\right. $$ Prove that $\lambda_n \rightarrow+\infty$.

There are possibly subtle mistakes that I could not recognize in my below attempt of (4.). Could you please have a check on it?


We need the following 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 $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.

Theorem 6.8 Assume $\dim E=\infty$ and $T$ be compact. Then $0 \in \sigma(T)$, $\sigma(T) \setminus \{0\}=E V(T) \setminus\{0\}$, 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 $M \in \sigma(T)$.

Theorem 6.11 Let $H$ be a separable Hilbert space and $T$ compact self-adjoint. Then there exists a Hilbert basis composed of eigenvectors of $T$.

If $u$ is a classical solution to $(1)$, then $$ \int_I [-u''v + kuv] = \int_I fv, \quad \forall v \in H^1 (I), $$ which (by integration by parts) implies $$ (2) \quad -u(1) v(1) + \int_I [u'v' + k uv] = \int_I fv, \quad \forall v \in H^1 (I). $$

We define a symmetric bilinear form $a$ on $H^1(I)$ by $$ a(u, v) := -u(1) v(1) + \int_I [u'v' + k uv]. $$

It follows from (1.) that $a$ is continuous and that if $k>0$ is sufficiently large then $a$ is coercive. By Lax-Milgram theorem, $Tf$ is uniquely characterized by $$ a(Tf, v) = \int_I fv, \quad \forall v \in H^1 (I). $$

In particular, $$ \langle Tf, f \rangle_{L^2} =\int_I f(Tf) = a(Tf, Tf) \ge 0. $$

By (3.), the map $T: L^2 (I) \to L^2 (I)$ is self-adjoint and compact. By Theorem 6.11, there is a Hilbert basis $(u_n)$ composed of eigenvectors of $T$. Then $0 \neq u_n \in N(T-\mu_n I)$ where $\mu_n$ is the eigenvalue associated to $u_n$ and $I : L^2(I) \to L^2(I)$ the identity map. By Proposition 6.9, $\mu_n \ge 0$. Because $N(T)=\{0\}$, we get $\mu_n \neq 0$. WLOG, we assume $\|u_n\|_{L^2}=1$. We have $Tu_n = \mu_n u_n$ and thus $-(\mu_n u_n)'' + k \mu_n u_n=u_n$ and thus $$ -u_n''=\frac{1-k\mu_n}{\mu_n} u_n. $$

Clearly, $u_n \in C(\bar I)$ implies $u_n \in C^2(\bar I)$. By Theorem 6.8, $\mu_n \to 0$. The claim then follows by taking $\lambda_n = \frac{1-k\mu_n}{\mu_n}$.

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  • $\begingroup$ which is the subtle point that you fear being subtly wrong? $\endgroup$
    – G. Gare
    Dec 4, 2023 at 16:18

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