Let $f \in L_2(0,\pi)$ have the Fourier expansion $f(x) = \sum_{n=2}^{\infty} f_n\sin(nx)$. Compute (formally) the boundardy value problem $$ u''(x) + u(x) = f(x) \qquad \mbox{ for } 0 < x < \pi $$ and $$ u(0) = u(\pi) = 0 $$ as a Fourier expansion and show that this expansion, as well as the term-wise differentiated series expansion converge uniformly. What properties of $u$ can you derive from this?
My attempt at an solution, I choose $u(x) = a_0/2 + \sum_{k=1}^{\infty} ( a_k \cos(k\omega x) + b_k \sin(k \omega x))$, then \begin{align*} & u''(x) + u(x) \\ & = \sum_{k=1}^{\infty} (-a_k(k\omega)^2 \cos(k\omega x) - b_k(k\omega)^2 \sin(k\omega x)) + a_0/2 + \sum_{k=1}^{\infty} ( a_k \cos(k\omega x) + b_k \sin(k \omega x)) \end{align*} Comparing coefficients yields $$ a_0 = 0, \quad $$ and $$ a_k(1 - k^2\omega^2) = 0, \quad k = 1,2,3,\ldots $$ and $$ b_k(1 - k^2\omega^2) = f_k, \quad k = 2,3,4,\ldots $$ and $b_1(1-\omega^2) = 0$, where $\omega = 2\pi/T = 2\pi/\pi = 2$. This implies $$ a_k = 0 $$ for all $k$, and $$ b_1 = 0, \quad b_k = \frac{f_k}{1 - 4k^2}, \quad k=2,3,4,\ldots $$ so that $$ u(x) = \sum_{k=2}^{\infty} \frac{f_k}{1-4k^2} \sin(2k x) $$ solves the boundary value problem.
But how to show uniform convergence of this series, and the termwise differentiated one, and what can I derive about $u$ from this?
Hint: use the following inequality to show the uniform convergence: $$ (\sum_{k=1}^na_kb_k)^2\le\sum_{k=1}^n|a_k|^2\sum_{k=1}^n|b_k|^2. $$