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Let $(f_n)_{n=1}^{\infty}$ be a sequence of real-valued functions defined on an open interval $I \subset \mathbb{R}$. Suppose that each $f_n$ is differentiable on $I$, and that:

  1. The sequence $(f_n)$ converges pointwise to a function $f$ on $I$.
  2. The sequence of derivatives $(f_n')$ converges pointwise on $I$ to a function $g$.

Question:
Under what additional assumptions (if any) can we conclude that:

  • $f$ is differentiable on $I$, and
  • $f'(x) = \lim_{n \to \infty} f_n'(x)$ for all $x \in I$?

I’m aware that uniform convergence of $(f_n')$ is sufficient (by the classical theorem on differentiating under the limit), but is pointwise convergence enough under some weaker conditions — perhaps boundedness, equicontinuity, or dominated convergence-style integrability?

Example of context:
Consider:

$$ f_n(x) = \begin{cases} x^n \sin\left( \frac{1}{x} \right) & x \in (0,1] \\ 0 & x = 0 \end{cases} $$

Then $f_n \to 0$ pointwise on $[0,1]$, and $f_n'$ converges pointwise as well — but the limit function $f$ is not differentiable at $0$.

So what are the minimal "reasonable" conditions to ensure that taking the derivative commutes with taking the limit?

Any counterexamples references are appreciated.

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  • $\begingroup$ The Lebesgue differentiation theorem (LDT) seems to be a strong source of counterexamples. Here we start with the sequence $(f_n')$ and integrate those to get the $(f_n)$, but we can't get stronger than $(f)' = f'$ a.e. without another hypothesis (perhaps one you list). Maybe see what you need to add to the proof assuming only pointwise convergence to get a valid proof... $\endgroup$ Commented Jun 23 at 9:06
  • $\begingroup$ Your first sentence talks about an open interval, but your example uses a closed interval. $\endgroup$ Commented Jun 23 at 9:28
  • $\begingroup$ See this 29 December 2006 sci.math post. $\endgroup$ Commented Jun 23 at 9:37
  • $\begingroup$ A collection of counter-examples: Let $g$ be any continuous function on $(0,1)$ such that $0\le g\le 1$ and $g(x_0)=1$ for a unique $x_0 \in(0,1)$. Let $f_n(x)=\int_0^{x} [g(t)]^{n}dt$. [To see that $f_n(x) \to 0$ pointwise, use Dominated Convergence Theorem]. $\endgroup$ Commented Jun 23 at 11:38
  • $\begingroup$ Why do you think that $f_n(1)\to 0$? $\endgroup$ Commented Jun 24 at 11:36

1 Answer 1

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This is a classical but subtle question in analysis. The key point is that pointwise convergence of both $f_n$ and $f_n'$ is generally not sufficient to conclude that $f$ is differentiable and that $\lim f_n' = f'$.

Counterexample (reproduced):

Let:

$$ f_n(x) = \begin{cases} x^n \sin\left( \frac{1}{x} \right), & x \in (0,1] \\ 0, & x = 0 \end{cases} $$

Each $f_n$ is differentiable, and $f_n(x) \to 0$ pointwise on $[0,1]$. However, $f$ is not differentiable at $x=0$, and $f_n'(x)$ does not converge uniformly.

Thus, pointwise convergence is too weak to guarantee differentiability of the limit.

Sufficient Conditions:

To ensure that $f$ is differentiable and that:

$$ f'(x) = \lim_{n \to \infty} f_n'(x), $$

one of the following is sufficient:

1. Uniform convergence of $f_n'$ on compact subsets of $I$

This is the classic theorem. If $f_n \to f$ pointwise and $f_n' \to g$ uniformly, then $f$ is differentiable and $f' = g$.

2. Equicontinuity + pointwise convergence of $f_n'$

If $(f_n')$ is equicontinuous and converges pointwise, then by Arzelà–Ascoli, the convergence is actually uniform on compact subsets, so we’re reduced to case 1.

3. Dominated convergence (in $L^1$ or $L^\infty$ sense)

If $|f_n'(x)| \leq g(x)$ for some integrable (or bounded) $g$, then we can often conclude convergence in $L^1$, and in some settings use this to show weak differentiability.

Summary:

  • Pointwise convergence of $f_n'$ is not enough.
  • Uniform convergence or domination by an integrable function is a standard sufficient condition.
  • The example you gave is a good demonstration of the failure of naive assumptions.

References:

  • Rudin, Principles of Mathematical Analysis, Ch. 7
  • Royden & Fitzpatrick, Real Analysis, Ch. 5–6
  • Folland, Real Analysis, for measure-theoretic treatment
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    $\begingroup$ Regarding the counter-example: "However, $f$ is not differentiable at $x=0$," If I understand the notation, function $f$ is the pointwise limit of $(f_n)$, therefore $f$ is the constant function $0$ on $[0, 1[$. So $f$ is differentiable at 0: its derivative is $0$. $\endgroup$ Commented Jun 24 at 11:33
  • $\begingroup$ Your counterexample is wrong as in OP. Moreover, your 3rd condition is not sufficient. Let $f_n(x)=\sqrt{x^2+\frac 1n}-\frac 1n$ for $x\in(0,1)$ and $f_n(x)=0$ for $x\in (-1,0]$. Then $f_n$ is differentiable and $|f'_n(x)|\le g(x)=1$ with $g$ bounded and integrable, with $f'_n\to 1$ for all $x>0$ and $f'_n(x)=0$ for all $x\le 0$. But $f$ is not differentiable in $0$. $\endgroup$ Commented Jun 24 at 12:01

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