# Function sequence converging almost everywhere to smooth function implies convergence everywhere?

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.

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 $$f_n$$.

• 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. Commented Nov 17, 2021 at 15:55