4

The question is : does there exist a sequence of derivable functions $f_n$ defined on $\mathbb{R}$, that converges uniformly to the zero function and such that the sequence of dérivatives converges pointwise to a function g that never vanishes?

Obviously, the sequence $(f_n ')$ cannot be bounded on every closed interval $[a,b]$. Otherwise, one can easily see by dominating convergence that $\int_a ^b g(t)dt=0$, and so $g$ must vanish somewhere on $[a,b]$, but I do not know what happens if we do not impose this condition on the sequence $(f_n ')$. I am bound to think this is impossible in the general cas, but I cannot see any counterexample. If you have any ideas, that will be helpful...

Marios Gretsas
  • 21,906
philippe
  • 131

1 Answers1

1

You may have seen user126154's answer. They demonstrate that if $f_n \to 0$ pointwise, $f_n : [0,1] \to \mathbb{R}$ is differentiable, and $f'_n$ is continuous on $[0,1]$, then $X = \{x \in [0,1] : g(x) = 0 \}$ is dense in $[0,1]$, where $f'_n \to g$ pointwise.

That means if a counterexample does exist, we are limited to discontinuous, unbounded sequences $(f'_n)$.

However, this could be impossible. To show that such a sequence cannot exist, here's an idea:

According to Theorem 2 in Haskell Curry's answer, the set $E_n$ of discontinuities of $f'_n$ is a meagre set.

Then, $E = \bigcup_{n=1}^\infty E_n$ is also a meagre set, with the property that every $f'_n$ is continuous on $D = \mathbb{R} - E$. By definition, $D$ is a comeagre set, which implies that int($D$) is dense in $\mathbb{R}$.

Maybe we could use the same argument in user126154's answer, by replacing $[0,1]$ with $D$? If so, it would prove that $X = \{x \in \text{int}(D) : g(x) = 0 \}$ is dense in $\text{int}(D)$, and hence, $X$ is dense in $\mathbb{R}$.

Consequently, not only does $g$ vanish somewhere, but $g$ vanishes on a dense subset of $\mathbb{R}$.

mathisawesome
  • 23