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
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- 4$\begingroup$ $\hat{f(\xi)}=0$ only for $\xi\neq 0$, else you omit $\delta$ whose Fourier transform is $1$ (leading to constants). $\endgroup$zwim– zwim2019-04-02 20:19:54 +00:00Commented Apr 2, 2019 at 20:19
- $\begingroup$ Also to avoid the Fourier transform : $f'=0$ as distributions means for any $\phi \in C^\infty_c, f \ast \phi \in C^\infty$ and $ 0 = f' \ast \phi = (f \ast \phi)'$ thus $f \ast \phi$ is constant. With $\int \phi=1$ and $\phi_n(x) =n \phi(nx)$ you obtain $f = \lim_{n \to \infty} f \ast \phi_n$ is constant. $\endgroup$reuns– reuns2019-04-02 20:38:42 +00:00Commented Apr 2, 2019 at 20:38
- $\begingroup$ @zwim thanks. the support of a distribution is $0$ gives that the function is a combination of $delta$ function and its derivative. $\endgroup$89085731– 890857312019-04-02 20:40:25 +00:00Commented Apr 2, 2019 at 20:40
- $\begingroup$ If $T$ is a distribution of order $m$ supported on $K$ then $|<T,\phi>| \le C\sum_{n\le m} \sup_{x \in K} |\phi^{(n)}(x)|$. If $K= \{0\}$ then $|<T,\phi>| \le C\sum_{n\le m} \sup_{x \in K} |\phi^{(n)}(0)|$ so that $<T,\phi - \sum_{n=0}^m \frac{\phi^{(n)}(0)}{n!} x^n> = 0$ and $<T,\phi> = \sum_{n=0}^m \frac{\phi^{(n)}(0)}{n!} <T,x^n>$ ie. $T = \sum_{n=0}^m a_n\delta^{(n)}$ with $a_n= (-1)^n\frac{<T,x^n>}{n!}$ $\endgroup$reuns– reuns2019-04-02 22:05:07 +00:00Commented Apr 2, 2019 at 22:05
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1 Answer
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0 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 $\rho$).
