Solving $f'=0$ in distribution sense

Question

When we solve $f'=0$, we can solve it by using Fourier transform in the space of tempered distribution to get $$i\xi\hat{f}(\xi)=0,$$ which gives us the distribution $\hat{f}(\xi)=0$, thus $f=0$. However, the solution is $f=const$. I want to know what is the problem

Answer 1

The distributional solutions to $i\xi \, \hat{f}(\xi) = 0$ are $\hat{f}(\xi) = C \, \delta(\xi),$ where $C$ is a constant.

This is in accordance with the solutions to $f'(x) = 0,$ which are $f(x) = C \, 1(x),$ transforming to $\hat{f}(\xi) = C \, 2\pi\,\delta(\xi).$

---

I will now show that the distributions satisfying $x \, u(x) = 0$ are $u(x) = C \, \delta(x)$ for constant $C$.

Let $u$ be a tempered distribution such that $x \, u(x) = 0$ and fix a test function $\rho$ such that $\rho(0) = 1.$ Set $C = \langle u, \rho \rangle.$

Given an arbitrary test function $\phi$ set $\hat{\phi} = \phi - \phi(0)\,\rho.$ Then $\hat{\phi}$ is a test function and $\hat{\phi}(0) = 0.$ There then exists a test function $\psi$ such that $\hat{\phi}(x) = x \, \psi(x).$

Now we have $$\begin{align} \langle u, \phi \rangle &= \langle u, \hat{\phi} + \phi(0) \, \rho \rangle \\ &= \langle u, \hat{\phi} \rangle + \phi(0) \, \langle u, \rho \rangle \\ &= \langle u, x \, \psi \rangle + \phi(0) \, C \\ &= \langle x \, u, \psi \rangle + C \langle \delta, \phi \rangle \\ &= \langle 0, \phi \rangle + \langle C \, \delta, \phi \rangle \\ &= \langle C \, \delta, \phi \rangle. \end{align}$$ Since this is true for any test function $\phi$ we have $u = C \, \delta,$ where $C$ is a constant (which here depends on the choice of $\rho$).