Brezis' exercise 8.31.2: if $\int_I f=0$ then $\|u\|_{L^2(I)} \leq \frac{1}{(1+\pi^2)} \|f\|_{L^2(I)}$

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

Exercise 8.31 Consider the Sturm-Liouville operator $$A u=-u^{\prime \prime}+u$$ on $$I$$ with Neumann boundary condition $$u' (0) = u' (1)=0$$.

1. Compute the eigenvalues of $$A$$ and the corresponding eigenfunctions.
2. Given $$f \in L^2 (I)$$ with $$\int_I f=0$$, let $$u$$ be the weak solution of $$(1) \quad \begin{cases} -u'' + u = f \quad \text {on} \quad I, \\ u'(0)=u'(1)=0. \end{cases}$$ Prove that $$\|u\|_{L^2(I)} \leq \frac{1}{(1+\pi^2)} \|f\|_{L^2(I)} .$$

I am trying to solve question (2.). In below attempt, I get the inequality with a bigger constant and have not utilized $$\int_I f=0$$. Could you elaborate on how to obtain the desired constant $$\frac{1}{\left(1+\pi^2\right)}$$?

It follows from (1) that $$\int_I [ -u''v + uv ] = \int_I f v, \quad \forall v \in H^2(I),$$ which (by integration by parts) implies $$\int_I [ u'v' + uv ] = \int_I f v, \quad \forall v \in H^2(I),$$

Substituting $$v=u$$, we get $$\int_I |u|^2 \le \int_I fu,$$ which (by Cauchy-Schwarz inequality) implies the desired inequality.

• The problem gave a hint no? [Apply Question 6 in Problem 49]. Jan 5 at 21:31
• @CheeHan I would like to see if there is a more direct approach without appealing to Question 6 of Problem 49. Jan 5 at 21:33

Clearly the corresponding eigen-problem $$\begin{cases} -u'' + u = \lambda u\quad \text {on} \quad I, \\ u'(0)=u'(1)=0, \end{cases} \tag{*}$$ has eigenvalues $$\lambda_n=(n\pi)^2+1$$ with the corresponding eigen-functions $$u_n=\cos(n\pi t)$$, $$n=0,1,2,\cdots$$. Observe $$\int_Iu_n=\int_Iu_mu_n=0 \text{ for }m\neq n,\int_Iu_n^2=\frac12, m,n=1,2,3,\cdots.$$ Integrating (1) and using $$\int_If=0$$, one has $$\int_Iu=0$$ and hence one has to rule out the eigenvalue $$1$$. Let $$u=\sum_{n=1}^\infty u_n\cos(n\pi t), f=\sum_{n=1}^\infty f_n\cos(n\pi t)$$ in (1). Then $$\sum_{n=1}^\infty((n\pi)^2+1) u_n\cos(n\pi t)=\sum_{n=1}^\infty f_n\cos(n\pi t)$$ and hence $$((n\pi)^2+1) u_n=f_n\text{ or }u_n=\frac{f_n}{(n\pi)^2+1}.$$ So $$\|u\|^2_{L^2}= \frac12\sum_{n=1}^\infty|u_n|^2=\frac12\sum_{n=1}^\infty\frac1{((n\pi)^2+1)^2}|f_n|^2\le \frac12\frac1{(\pi^2+1)^2}\sum_{n=1}^\infty|f_n|^2=\frac1{(\pi^2+1)^2}\|f\|_{L^2}^2$$ which gives the desired the estimate.