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Let $f:\mathbb R \to \mathbb R$ be a $C^\infty$ function. Further, we have $C^\infty$ functions $f_n:\mathbb R \to \mathbb R$ for all $n \in \mathbb N$. Is it possible that $$\lim_{n \to \infty} f_n(x)=f(x)$$ for almost all $x$ but not for all?

Remark 1: A negative solution of this question would also solve my other question.

Remark 2: If the answer is negative, we could replace the first $C^\infty$ by $C^k$ and it would be interesting to know the least $k$ such that the answer still is negative.

Example: For $k=0$ the answer is positive:

  • Each periodic continuous function $f$ has a Fourier series which converges almost everywhere pointwise to $f$.
  • However, there are examples of continuous functions whose Fourier series diverges at certain points.
  • If we assume the continuous periodic function $f$ to be even $C^1$ the Fourier series converges everywhere to $f$. So the answer of remark 2 is either $k=1$ or we have to leave Fourier series for counter examples.
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Yes, this is possible. Set $f(x) = 0$ and $f_1(x) = (1+x^2)^{-1}$. Then set $$ f_n(x) = n f_1(nx) = \frac{n}{1 + n^2x^2} $$ Clearly $f_n(x) \to 0$ for $x \ne 0$ but $f_n(0)$ does not converge.

If you want the same phenomenon with a $C^k$ function $f$, just use $f(x) + f_n(x)$ instead of $f_n$.

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  • $\begingroup$ Thanks for you answer. It's a quite easy construction. Shame on my to not find it myself ;-). By the way, it was clear to me that if it fails for $C^\infty$ functions it would fail for all $C^k$ as well. $\endgroup$ Commented Nov 17, 2021 at 15:55

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