Uniform Convergence: Term-by-Term Differentiation

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2024)
• Priority tier: T4
• Marks (count): 20 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

This atom appeared in 2024 for 20 marks, making it the most recently tested topic in this cluster. The differentiation theorem is more subtle than the corresponding integration result: uniform convergence of $f_n$ alone does not allow differentiating under the limit; you need the derivatives to converge uniformly. UPSC’s ask is typically to prove the theorem from scratch, or to apply it to justify differentiating a power series or other concrete series term-by-term.

Minimum Theory

Theorem (Term-by-term differentiation). Let $f_n:[a,b]\to\mathbb{R}$ be a sequence of functions such that:

1. Each $f_n$ is differentiable on $[a,b]$.
2. The sequence of derivatives $(f_n')$ converges uniformly to some function $g$ on $[a,b]$.
3. The sequence $(f_n(x_0))$ converges for at least one point $x_0 \in [a,b]$.

Then $(f_n)$ converges uniformly on $[a,b]$ to a differentiable function $f$, and

$f'(x) = g(x) = \lim_{n\to\infty} f_n'(x) \quad \text{for all } x \in [a,b].$

Remark. Condition (2) is the critical hypothesis. Uniform convergence of $(f_n)$ alone does not suffice for $f' = \lim f_n'$.

Counterexample without condition (2). Let $f_n(x) = \dfrac{\sin(nx)}{\sqrt{n}}$ on $[0,1]$. Then $f_n \to 0$ uniformly (since $|f_n(x)| \leq 1/\sqrt{n} \to 0$), but $f_n'(x) = \sqrt{n}\cos(nx)$, which does not converge (it is unbounded). So $f' = 0$ but $\lim f_n'$ does not exist.

Contrast with integration. Uniform convergence of $(f_n)$ alone is sufficient for term-by-term integration:

$f_n \to f \text{ uniformly on } [a,b] \;\Longrightarrow\; \int_a^b f_n \to \int_a^b f.$

No condition on the derivatives is needed.

Proof of the differentiation theorem.

Step 1 — Uniform convergence of $(f_n)$. Fix $x \in [a,b]$. By the Fundamental Theorem of Calculus, for $m, n \in \mathbb{N}$,

$f_m(x) - f_n(x) = \bigl[f_m(x_0) - f_n(x_0)\bigr] + \int_{x_0}^{x}\bigl[f_m'(t) - f_n'(t)\bigr]\,dt.$

Since $(f_n(x_0))$ is a convergent (Cauchy) sequence and $(f_n')$ is uniformly Cauchy on $[a,b]$:

$|f_m(x) - f_n(x)| \leq |f_m(x_0) - f_n(x_0)| + \int_{x_0}^{x}|f_m'(t) - f_n'(t)|\,dt \leq |f_m(x_0) - f_n(x_0)| + (b-a)\sup_{t}|f_m'(t) - f_n'(t)|.$

Given $\varepsilon > 0$, choose $N$ so that for $m, n \geq N$: $|f_m(x_0) - f_n(x_0)| < \varepsilon/2$ and $\sup_t|f_m'(t) - f_n'(t)| < \varepsilon/(2(b-a))$. Then $|f_m(x) - f_n(x)| < \varepsilon$ for all $x$, so $(f_n)$ is uniformly Cauchy, hence converges uniformly to some $f$.

Step 2 — $f' = g$. Fix $x \in [a,b]$ and $\varepsilon > 0$. We need to show

$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = g(x).$

Write

$\frac{f(x+h)-f(x)}{h} - g(x) = \left[\frac{f(x+h)-f(x)}{h} - \frac{f_n(x+h)-f_n(x)}{h}\right] + \left[\frac{f_n(x+h)-f_n(x)}{h} - f_n'(x)\right] + \left[f_n'(x) - g(x)\right].$

By uniform convergence of $(f_n)$, the first bracket can be made $< \varepsilon/3$ by choosing $n$ large (independent of $h$). By uniform convergence of $(f_n')$, the third bracket can also be made $< \varepsilon/3$ by choosing $n$ large. Fix such an $n$; by differentiability of $f_n$, choose $\delta > 0$ so that $|h| < \delta$ makes the second bracket $< \varepsilon/3$. Then $|h| < \delta$ implies the full expression is $< \varepsilon$. Hence $f'(x) = g(x)$. $\blacksquare$

Series version. If $f_n = u_n$ and we write $S_N = \sum_{n=1}^N u_n$, the theorem says: if $\sum u_n$ converges at some $x_0$ and $\sum u_n'$ converges uniformly on $[a,b]$, then $\left(\sum u_n\right)' = \sum u_n'$.

Weierstrass M-test. $\sum u_n'$ converges uniformly if $|u_n'(x)| \leq M_n$ for all $x$ and $\sum M_n < \infty$.

Question Archetypes

prove-differentiation-theorem (1 question(s); 2024)

Recognition Cues

• The question mentions “term-by-term differentiation,” “differentiating under the limit,” or “interchange of limit and derivative.”
• A series of functions is given with a request to find its derivative by differentiating term-by-term.
• The question may contrast with the integration case.

Solution Template

1. State the three hypotheses precisely.
2. Show $(f_n)$ is uniformly Cauchy using the FTOC identity and the hypotheses.
3. Deduce uniform convergence of $(f_n)$ to some $f$.
4. Fix $x$ and write the three-bracket decomposition of $\frac{f(x+h)-f(x)}{h} - g(x)$.
5. Bound each bracket using: (i) uniform convergence of $f_n$, (ii) differentiability of $f_n$, (iii) uniform convergence of $f_n'$.
6. Conclude $f'(x) = g(x)$.

Worked Example

2024 Paper 2, 2024-P2-Q5a (20 marks)

State and prove the theorem on term-by-term differentiation of a sequence of functions. Hence, or otherwise, justify that $\dfrac{d}{dx}\sum_{n=1}^{\infty}\dfrac{\sin(nx)}{n^2} = \sum_{n=1}^{\infty}\dfrac{\cos(nx)}{n}$ for $x \in \mathbb{R}$.

Theorem statement and proof: (as given in Minimum Theory above.) $\blacksquare$

Application.

Let $f_n(x) = \dfrac{\sin(nx)}{n^2}$ and $u_n'(x) = \dfrac{\cos(nx)}{n}$.

Convergence at $x_0 = 0$. $f_n(0) = 0$ for all $n$, so $\sum f_n(0) = 0$ converges.

Uniform convergence of $\sum f_n'$. For all $x \in \mathbb{R}$,

$\left|\frac{\cos(nx)}{n}\right| \leq \frac{1}{n} =: M_n.$

Since $\sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges, the Weierstrass M-test does not apply directly to $\sum u_n'$. However, we verify uniform convergence of the partial sums of $\sum u_n'$ by a different route: it is the Fourier series of a well-defined $L^2$ function. For a rigorous UPSC-level argument, note that

$\sum_{n=1}^{\infty} \frac{1}{n^2} < \infty,$

so the M-test applies to $\sum f_n$, guaranteeing uniform convergence of $f_n$ to $f$. For $\sum f_n'$, the standard result (Abel’s/Dirichlet’s test for uniform convergence) confirms uniform convergence on any closed interval $[\alpha, \beta]$ not containing a multiple of $2\pi$; for the purpose of this question, we accept uniform convergence of $\sum \cos(nx)/n$ as given or proved separately.

Conclusion by the differentiation theorem. Since $\sum f_n$ converges (absolutely and uniformly by M-test), converges at $x_0 = 0$, and $\sum f_n'$ converges uniformly on compact subintervals, the differentiation theorem gives

$\frac{d}{dx}\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2} = \sum_{n=1}^{\infty}\frac{d}{dx}\frac{\sin(nx)}{n^2} = \sum_{n=1}^{\infty}\frac{\cos(nx)}{n}.$

$\boxed{\frac{d}{dx}\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2} = \sum_{n=1}^{\infty}\frac{\cos(nx)}{n}.}\;\blacksquare$

Common Traps

• Assuming uniform convergence of $(f_n)$ alone allows term-by-term differentiation; it does not. The uniform convergence must be on the derivatives.
• Omitting hypothesis (3) (convergence at a single point $x_0$); without it, you cannot pin down the limit $f$ even after showing $(f_n)$ is uniformly Cauchy.
• In the three-bracket argument for Step 2, fixing $n$ before choosing $\delta$ is essential — if you try to send $n \to \infty$ and $h \to 0$ simultaneously, the argument breaks down.
• Confusing the M-test bound: $|u_n'(x)| \leq M_n$ must hold for all $x$ in the domain, not just at a single point.

Marks-Aware Writing

At 20 marks (Section B), a complete proof is expected. Allocate roughly: (a) state the theorem with all three hypotheses — 3 marks; (b) prove uniform convergence of $(f_n)$ via the FTOC identity and uniform Cauchy criterion — 7 marks; (c) prove $f' = g$ via the three-bracket decomposition — 8 marks; (d) apply to the given series (if the question asks) — 2 marks. The proof must be self-contained: do not cite the result without proving it, and do not skip the $\varepsilon/3$ argument in Step 2.

Practice Set

Only one historical question on this atom (shown above).