Assume the Dirichlet problem for the Poisson equation in $I=(-1,1)$ \begin{equation*} \tag{$\dagger$} \begin{cases} u''= -f & \text{in} \quad I\\%\equiv (-1,1)\\ u = 0 & \text{on} \quad \partial I \equiv \{-1,1\} \label{eq:bvp} \end{cases} \end{equation*} and let $f\equiv c$ in $I$, $c\in\mathbb{R}$, and $f\equiv0$ in $\mathbb{R}\setminus I$. Then, the solution can be recovered in closed form as \begin{gather} u = \frac{c}{2}\big(1-x^2\big) \quad \text{in} \quad \overline{I} \end{gather} Now, take the Fourier expansion of $u,f$ \begin{gather} u(x) = \sum_{-\infty}^{\infty}\hat{u}(k)e^{i\pi kx} \quad,\quad f(x) = \sum_{-\infty}^{\infty}\hat{f}(k)e^{i\pi kx} \end{gather} The BVP \ref{eq:bvp} in frequency domain reads \begin{equation*} \tag{$\star$} \begin{cases} \hat{u}= \dfrac{\hat{f}}{\pi^2k^2} & \text{for}\ k \in \mathbb{Z}\setminus \{0\}\\ \hat{u} =\ ? & \text{for}\ k=0 \label{eq:bvp_freq} \end{cases} \end{equation*} Since $f$ is constant we have that $\hat{f}$ is zero almost everywhere except at $\hat{f}(k=0) = c$. Thus, we obtain $\hat{u} \equiv 0$ which implies $u \equiv 0 \ \text{in} \ \overline{I}$.
Question a. What went wrong? I guess that we must have $\hat{f}(k=0) =0$, but why?
Question b. If $\hat{f}(k=0) =0$ is a strict requirement, does that mean that we cannot solve \eqref{eq:bvp} in frequency domain? If yes, how could we reformulate it?
Cheers!
Update 1. After the comment of @Rahul, I think a possible way to go through could be to extend $f$ to take the value of zero at the boundary $\partial I$. Then, proceed with mollifying the extended function, taking the Fourier transform of this smooth approximation and passing to the limit. I haven't tried that yet. But.. isn't there any simpler way!?