Let $ f(x) = \sum_{n=0}^{\infty} x^{2n} (n!)^{-2} $. Solve the equation $$ (f(x)-1) \: u'(x) = 0 $$ for $ u \in D' $. That is, what distributions u satisfy this equation?
I'm not quite sure about how to approach such a problem and would be grateful for any help!
It's obvious from the construction of $f$ that $f$ is convex and even, and therefore takes it minimum at $x=0$. We have $f(0)=1$, $f'(0)=0$ and $f''(0)=2$, so $f(x)-1$ has a double zero at $x=0$, and is non-zero everywhere else.
This means that the equation is equivalent to $x^2 \, u'(x) = 0$ which has solutions $u(x) = AH + B\delta + C,$ where $A,B,C$ are constants, $H$ is the Heaviside function, and $\delta$ of course is the Dirac delta distribution.