i can't find any counterexample

[muted swan]

If f is C∞ on an interval (−a,a) and can be expressed as an infinite sum of functions f(x)=∑fn(x) does it follow that each fn is C∞, and can we differentiate the series term by term?

I want to prove that this is false but i can't find any satisfying function

[eternal acorn]

Can't you take:

• f constant at 1 over [0,2]
• f1 constant at 0 for [0,1] then constant at 1
• f2 constant at 1 over [0,1] then constant at 0
• for all n>2 fn=0

?

[muted swan]

ok bien vu ca a l'air de marcher, mais est ce que tu penses que si on rajoute l'hypothese fn est C∞ pour tout n alors on peut quand meme trouver un contre exemple ?

For those who dont speak french : If we suppose that each fn is Cinfinite, can we still find a counter example ?

[polar matrix]

then you have uniform convergence so you can differentiate term by term

so no counterexamples

i think

but the general case doesn't seem to be true

[polar matrix]

actually yes very simple counterexample

take $$
f_0 = \begin{cases}
x^2,\quad &x\geq0\
0,\quad &x<0
\end{cases}
$$
$$
f_1 = \begin{cases}
0,\quad &x\geq0\
x^2,\quad &x<0
\end{cases}
$$
$$
f_n = 0
$$
for $n\geq 2$.
then clearly $f = \sum_{n=0}^\infty f_n = x^2 \in C^\infty$ but $f_0$ and $f_1$ are not in $C^\infty$

[proper plume]

artemetra

[polar matrix]

@muted swan if you have any more constraints, let me know but letting fn be any real functions is a veeery weak restraint