Uniform Convergence and Differentiation

Quick Recap: The Derivative

We say $f : A \to \mathbb{R}$ is differentiable at $c \in A$ if the limit

$f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$

exists. When $f'(c)$ exists, it equals the slope of the tangent line to the graph of $f$ at $c$.

Two key facts we will use:

If $f$ is differentiable at $c$, then $f$ is continuous at $c$.
(Mean Value Theorem) If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $\alpha \in (a,b)$ with $f'(\alpha) = \frac{f(b)-f(a)}{b-a}$.

The Question

Background: We proved that if $f_n \to f$ uniformly and each $f_n$ is continuous, then $f$ is continuous. The natural next question is:

If $f_n \to f$ uniformly and each $f_n$ is differentiable, must $f$ be differentiable?
And if so, does $f_n' \to f'$?

A Shocking Example

Example: $h_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$

(a) Show that $h_n \to 0$ uniformly on $\mathbb{R}$.

(b) Compute $h_n'(x)$ for each $n$.

(c) Does $(h_n')$ converge for any $x$?

Conclusion: $h_n \to 0$ uniformly, yet $(h_n')$ diverges everywhere.

What Is Really Being Asked

Differentiating the limit function means computing:

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

Saying $f' = \lim_n f_n'$ is asking to interchange the following limits:

$\lim_{x \to c} \lim_{n \to \infty} \frac{f_n(x)-f_n(c)}{x-c} \overset{?}{=} \lim_{n \to \infty} \lim_{x \to c} \frac{f_n(x)-f_n(c)}{x-c}$

This is even more delicate than the continuity interchange. The fix: require the derivatives $(f_n')$ to converge uniformly.

The Differentiable Limit Theorem

Theorem 1

Let $f_n \to f$ pointwise on $[a,b]$, and assume each $f_n$ is differentiable on $[a,b]$. If $(f_n')$ converges uniformly on $[a,b]$ to a function $g$, then $f$ is differentiable and $f' = g$. That is,

$\left(\lim_{n\to\infty} f_n\right)' = \lim_{n\to\infty} f_n'$

What the theorem says in plain language: To differentiate through a limit, you do not need $(f_n)$ to converge uniformly — you need $(f_n')$ to converge uniformly. If the derivatives converge uniformly to some $g$, then the limit function $f$ is differentiable and its derivative is exactly $g$.

Proof Sketch

Fix $c \in [a,b]$ and $\varepsilon > 0$. We want to show $f'(c) = g(c)$, i.e., there exists $\delta>0$ such that whenever $|x-c|<\delta$:

By the triangle inequality, for any $n \in \mathbb{N}$ and $x \neq c$:

$\left|\frac{f(x)-f(c)}{x-c} - g(c)\right| \leq$ + +

Term (I), Term (II), Term (III)

Step 1. Use uniform convergence of $(f_n')$ to choose $N \in \mathbb{N}$ so that terms (I) and (III) are each $< \varepsilon/3$.

Step 2. How is the MVT used to control term (I)?

Step 3. With $N$ fixed, use differentiability of $f_N$ at $c$ to choose $\delta > 0$ so term (II) $< \varepsilon/3$ whenever $|x-c| < \delta$.

Conclude: for $0 < |x-c| < \delta$,

$\left|\frac{f(x)-f(c)}{x-c} - g(c)\right| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$ $\square$

Key insights about the proof:

Three-tool argument. Each tool handles exactly one term — drop any one hypothesis and the proof collapses: Term (I) uses MVT; Term (II) uses differentiability of $f_N$; Term (III) uses uniform convergence of $(f_n')$.
Order of quantifiers matters. Pick $N$ first. Then pick $\delta$ from the differentiability of the single function $f_N$. Reversing the order breaks the argument.

Examples

Example 1: $f_n(x) = \frac{x^2}{n} + x$ on $[-2,2]$

(a) Find the pointwise limit $f(x) = \lim_{n\to\infty} f_n(x)$.

(b) Compute $f_n'(x)$ and find $g(x) = \lim_{n\to\infty} f_n'(x)$ pointwise.

(c) Show $(f_n')$ converges uniformly to $g$ on $[-2,2]$.

(d) Conclude by Theorem 1 that $f' = g$. Verify directly that $f'(x) = g(x)$.

Example 2: $g_n(x) = \frac{x^n}{n}$ on $[0,1]$

(a) Show $(g_n)$ converges uniformly on $[0,1]$ and find $g = \lim_{n \to \infty} g_n$.

(b) Compute $g_n'(x)$ for each $n$. Show $(g_n')$ converges pointwise on $[0,1]$. Is the convergence uniform?

(c) Compute $\lim_{n\to\infty} g_n'(x)$ and compare it with $g'(x)$. Are they equal?

Note: This example shows that even when $(f_n)$ converges uniformly, the sequence $(f_n')$ may converge non-uniformly.

Hypothesis	Conclusion	Example
$f_n \to f$ uniformly, each $f_n$ differentiable	No conclusion about $f'$	$h_n(x) = \frac{\sin(nx)}{\sqrt{n}}$
$f_n \to f$ pointwise, $f_n' \to g$ uniformly	$f$ is differentiable and $f' = g$	Example 1

Activity

Problem 1: $f_n(x) = \frac{nx + x^2}{2n}$ on $\mathbb{R}$

(a) Compute $f = \lim_{n\to\infty} f_n$ pointwise. Identify $f$.

(b) Compute $f_n'(x)$ and find $\lim_{n\to\infty} f_n'(x)$ pointwise.

(c) Show $(f_n')$ converges uniformly on every bounded interval $[-M, M]$.

(d) Apply Theorem 1 to conclude $f' = \lim_{n \to \infty} f_n'$. Verify directly.

Problem 2. True/False — justify each.

(a) If $f_n \to f$ uniformly and each $f_n$ is differentiable, then $f$ is differentiable.

(b) If $f_n \to f$ pointwise, each $f_n$ is differentiable, and $f_n' \to g$ pointwise, then $f' = g$.

(c) If $f_n \to f$ pointwise, each $f_n$ is differentiable, and $f_n' \to g$ uniformly, then $f' = g$.