Suppose that $F:\mathbb R\to\mathbb R$ is bounded and continuously differentiable everywhere. Suppose also that the sequence of bounded (not necessarily even continuous) functions $(F_n)_{n\in\mathbb N}$ converges to $F$ uniformly. If $(x_n)_{n\in\mathbb N}$ and $(\varepsilon_n)_{n\in\mathbb N}$ are real sequences such that $x_n\to x\in\mathbb R$ and $\varepsilon_n\to0$, is it always true that $$\frac{F_n(x_n)-F_n(x_n-\varepsilon_n)}{\varepsilon_n}\to F'(x)?$$ Any hints (or counterexamples) would be appreciated.


Consider the sequence

$$f_n(x) = \frac{x}{1 + nx^2}.$$

Note that

$$\sup f_n(x) = f_n(1/\sqrt{n}) = \frac{1}{2 \sqrt{n}}.$$

Then $f_n(x) \to f(x) = 0$ uniformly and $f'(0) = 0$.

By the mean value theorem, we have for some $\xi_n \in(x_n - \epsilon_n,x_n)$

$$\frac{f_n(x_n) - f_n(x_n - \epsilon_n)}{\epsilon_n} = f'_n(\xi_n) = \frac{1 - n\xi_n^2}{(1 + n\xi_n^2)^2}.$$

If $x_n = 1/n^2$ and $\epsilon_n = 1/(2n^2)$ then $f'(\xi_n) \to 1 \neq f'(0).$

  • $\begingroup$ Great, thank you very much! $\endgroup$ – triple_sec Apr 26 '16 at 4:25
  • $\begingroup$ @triple_sec: You're welcome. $\endgroup$ – RRL Apr 26 '16 at 4:26
  • $\begingroup$ Note that the functions $f_n$ are even differentiable. If you only assume continuous or less, you can come up with even easier counter examples in the same spirit as the above. $\endgroup$ – user113529 Apr 26 '16 at 4:33

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