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Assume that

$f_n(x)\to f(x)$ pointwise at any $x\in\mathbb{R}$

$f_n,f$ are all differentiable at any $x\in\mathbb{R}$

$f_n'(x)\to g(x)$ pointwise at any $x\in\mathbb{R}$ and $g$ is differentiable at any $x\in\mathbb{R}$

Does it imply $g(x)=f'(x)$ for all $x$ ?

Without $g$ being differentiable, it was in my previous question pointwise convergence in relation to derivatives and turned out to be easy and quite standard. This one is probably harder. I have searched 3 books and over 100 posts on convergence, but without finding a suitable counterexample.

tong_nor
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  • Someone closed this topic as a duplicate of https://math.stackexchange.com/questions/4261920/point-wise-convergence-and-differentiation-counter-examples, but it's not the same, or I'm horribly overlooking something. In the quoted topic, $f_n'=g_n$ and this sequence was pointwise convergent to a function $g$ such that $g(0)=1$ and $g(x)=0$ otherwise, such $g$ is obviously nondifferentiable, while in my question $g$ should be differentiable. – tong_nor Feb 11 '22 at 20:33

1 Answers1

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Define the triangle function $T$ by $T(x):=\min\{x,2-x\}$ for $x \in[0,2]$, and $T(x):=0$ for $x \notin [0,2]$. Let $$g_n(x):=\sum_{k=-\infty}^{\infty} n \, T(n^2(x-k/n))\,,$$ and $$f_n(x):=\int_0^x g_n(t) \, dt \,, $$ for all real $x$, so $f'_n =g_n$ identically. Note the graph of $g_n$ consists of equally spaced triangles, each shifted by $1/n$ from the preceding one, with base $2/n^2$, height $n$, and area $1/n$ each. Then $f'_n=g_n$ tends pointwise to $g=0$, yet for each $n$, the nondecreasing function $f_n$ satisfies $f_n(k/n)=k/n$ for all integer $k$, so $f_n(x)$ tends uniformly to $f(x)=x$.

Remarks

  1. Actually, replacing $T(\cdot)$ by a $C^\infty$ bump function of integral 1 supported in $[0,2]$, one can make all $f_n$ be $C^\infty$ functions. https://en.wikipedia.org/wiki/Bump_function

  2. If $\sup_n |f_n'|$ is bounded (or integrable on compact sets) then $g=f'$.

Yuval Peres
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