# Brezis' exercise 8.24.4: there exists a sequence $\left(\lambda_n\right)$ in $\mathbb{R}$ with $\left|\lambda_n\right| \rightarrow \infty$

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 $$\lambda_n = \frac{1-k\mu_n}{\mu_n}$$.

• which is the subtle point that you fear being subtly wrong? Dec 4, 2023 at 16:18