(on a bounded domain). I believe that such a function could not exist since every $C^\infty$ function can be approximated by a sequence of polynomials and every polynomial has a finite number of roots so it would not be possible for something that vanishes countably often to converge to something that vanishes uncountably often.
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ How about a compactly supported function ? $\endgroup$– fwdCommented Mar 17, 2023 at 18:59
-
$\begingroup$ Your polynomial approximation idea works the same way if the function is only assumed to be continuous, and such a function can be zero on $(-1,0)$ and $=x$ on $[0,1)$ $\endgroup$– reunsCommented Mar 17, 2023 at 19:00
-
$\begingroup$ And so with $g(x)=e^{-1/(x(1-x))})1_{x\in [0,1]}$ you can also construct a smooth function $\phi\in C^\infty(0,1),\phi(x)=\sum_{n=0}^\infty \frac1{n!}\sum_{a\in \{0,2\}^n} g(3^{n+1} (x-3^{-n-1}-\sum_{m=1}^n a_m 3^{-m})$ vanishing only at the Cantor set, which is not a countable union of points and closed intervals. $\endgroup$– reunsCommented Mar 17, 2023 at 19:38
-
$\begingroup$ Your claim that a sequence of functions with only countably many zeros cannot converge to a sequence with uncountably many zeros is also wrong: Take the sequence of constant functions $f_n(x) = 1/n$. $\endgroup$– Eike SchulteCommented Mar 18, 2023 at 10:19
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The following function is $C^\infty$ on ${\bf R}$ and is zero outside the unit interval. $$ f(x) = e^{-{1\over 1-x^2}}{\bf 1}_{[-1,1]}(x) $$ Such function is called a bump function.
$\begingroup$
$\endgroup$
2
$x \sin (1/x) $
This function has a sequence of roots tending to zero.
-
1$\begingroup$ The question asked for an uncountable set of roots. $\endgroup$ Commented Mar 18, 2023 at 12:43
-