Derivative of the limit of function sequence

Question

Suppose $(f_n)$ is a sequence of real functions that converges uniformly to $f:\mathbb R\to\mathbb R$. Is it possible to show that $$f'(x)=\lim_{n\to\infty}\frac{f_n(x+1/n)-f_n(x)}{1/n},$$ and in that case, what are the additional conditions that $(f_n)$ needs to satisfy?

Answer 1

$f$ does not have to be differentiable in general. There is a number of counterexamples here.

We need stronger assumptions.

Let $f_n : \mathbb{R} \to \mathbb{R}$ be a sequence of $C^1$-functions and $f : \mathbb{R} \to \mathbb{R}$ a $C^1$-function. Assume that $f_n \to f$ and $f_n' \to f'$ uniformly.

Then for every $c \in \mathbb{R}$ we have $$f'(c) = \lim_{n\to\infty} \frac{f_n\left(c+\frac1n\right) - f_n(c)}{\frac1n}$$

Proof.

Fix $c \in \mathbb{R}$.

We will show that $\phi_n(x) = \frac{f_n(x) - f_n(c)}{x-c}$ converges uniformly over $x \in \mathbb{R}$ to $\phi(x) = \frac{f(x) - f(c)}{x-c}$.

Let $\varepsilon > 0$ and pick $n_0 \in\mathbb{N}$ such that $m, n \ge n_0 \implies \|f_n' - f_m'\|_\infty < \frac\varepsilon2$. For $m, n \ge n_0$ and all $x \in \mathbb{R}$ the MVT on the function $f_n - f_m$ gives

$$|\phi_n(x) - \phi_m(x)| = \left|\frac{f_n(x) - f_n(c)}{x-c} - \frac{f_m(x) - f_m(c)}{x-c}\right| = \left|\frac{(f_n-f_m)(x) - (f_n-f_m)(c)}{x-c}\right| = |f_n'(\theta) - f_m'(\theta)|$$

for some $\theta$ between $x$ and $c$. Hence $|\phi_n(x) - \phi_m(x)| \le \|f_n' - f_m'\|_\infty < \frac\varepsilon2, \forall x \in \mathbb{R}$.

Clearly $\phi_n \to \phi$ pointwise so letting $m\to\infty$ we get $|\phi_n(x) - \phi(x)| \le \frac\varepsilon2 < \varepsilon, \forall x \in \mathbb{R}$. We conclude $\phi_n \to \phi$ uniformly.

Now we have

\begin{align} \left|\frac{f_n\left(c+\frac1n\right) - f_n(c)}{\frac1n} - f'(c)\right| &\le \left|\frac{f_n\left(c+\frac1n\right) - f_n(c)}{\frac1n} - \frac{f\left(c+\frac1n\right) - f(c)}{\frac1n}\right| + \left|\frac{f\left(c+\frac1n\right) - f(c)}{\frac1n} - f'(c)\right|\\ &= \left|\phi_n\left(c+\frac1n\right) - \phi\left(c+\frac1n\right)\right| + \left|\frac{f\left(c+\frac1n\right) - f(c)}{\frac1n} - f'(c)\right| \\ &\le \|\phi_n - \phi\|_\infty + \left|\frac{f\left(c+\frac1n\right) - f(c)}{\frac1n} - f'(c)\right|\\ &\xrightarrow{n\to\infty} 0 \end{align}

so we conclude

$$f'(c) = \lim_{n\to\infty} \frac{f_n\left(c+\frac1n\right) - f_n(c)}{\frac1n} $$

Answer 2

I will assume that each $f_n$ is a differentiable function. It doesn't follow from this that what you want to prove is true. Take, for instance, $f_n(x)=\sqrt{x^2+\frac1{n^2}}$. Then $(f_n)_{n\in\mathbb N}$ converges uniformly to the absolute value function, but this function is not differentiable at $0$ and so, in particular, we don't have$$f'(0)=\lim_{n\to\infty}\frac{f_n\left(\frac1n\right)-f_n(0)}{\frac1n}.$$

However, if all functions $f_n$ are of class $C^1$ and if $(f_n')_{n\in\mathbb N}$ converges uniformly to a function $g$, your equality holds.

Answer 3

This is a stronger version (uniformly integrability condition) of what you need:

If a sequence of absolutely continuous functions {$f_n$} converges pointwise to some $f$ and if the sequence of derivatives {$f_n’$} converges almost everywhere to some $g$ and if {$f_n’$} is uniformly integrable then $\lim\limits_{n\mapsto \infty} f_n’ = g= f’$ almost everywhere. Where the derivative of $f$ is $f’$. If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.

Proof : by FTC $f_n(x) – f_n(a) = \int_a^x f_n’ dx$

By Vitali convergence theorem : $\lim\limits_{n\mapsto \infty}\int_a^x f_n’ dx = \int_a^x g dx$

Therefore $\lim\limits_{n\mapsto \infty}( f_n(x) – f_n(a))= \int_a^x g dx$

$f(x)-f(a) = \int_a^x g dx$

$f(x)’=g$ almost everywhere

If the convergence is pointwise and $ g $ is continuous then $ f'$ = $ g $ everywhere.