Pointwise vs. Uniform Convergence of Sequences of Functions

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2019)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~18 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

This atom appeared once in 13 years (2019, 15 marks) and anchors the analysis of function sequences. The distinction between pointwise and uniform convergence is foundational: pointwise convergence preserves almost nothing about the limit function, while uniform convergence preserves continuity, integrability, and (with an extra condition) differentiability. UPSC typically asks you to determine which type of convergence holds for a given sequence and to prove your answer — a task with a predictable structure once you know the supremum test for non-uniform convergence.

Minimum Theory

Pointwise convergence. A sequence of functions $f_n:E\to\mathbb{R}$ converges pointwise to $f:E\to\mathbb{R}$ if, for every $x \in E$,

$f_n(x) \to f(x) \quad \text{as } n \to \infty.$

Equivalently, $\forall\,x \in E,\; \forall\,\varepsilon > 0,\; \exists\,N = N(x,\varepsilon)$ such that $n \geq N \Rightarrow |f_n(x) - f(x)| < \varepsilon$.

Uniform convergence. $f_n \to f$ uniformly on $E$ if

$\sup_{x \in E} |f_n(x) - f(x)| \to 0 \quad \text{as } n \to \infty.$

Equivalently, $\forall\,\varepsilon > 0,\; \exists\,N = N(\varepsilon)$ (independent of $x$) such that $n \geq N \Rightarrow |f_n(x) - f(x)| < \varepsilon$ for all $x \in E$.

Hierarchy. Uniform convergence $\Rightarrow$ pointwise convergence. The converse is false.

Supremum test (non-uniformity). $f_n \not\to f$ uniformly if $\sup_{x \in E}|f_n(x) - f(x)| \not\to 0$; equivalently, if there exist $\varepsilon_0 > 0$ and a sequence $x_n \in E$ such that $|f_n(x_n) - f(x_n)| \geq \varepsilon_0$ for all $n$.

Standard example: $f_n(x) = x^n$ on $[0,1]$.

$f(x) = \lim_{n\to\infty} x^n = \begin{cases} 0 & 0 \leq x < 1, \\ 1 & x = 1. \end{cases}$

Convergence is pointwise but not uniform. Proof of non-uniformity: $\sup_{x \in [0,1]}|x^n - f(x)| = \sup_{x \in [0,1)} x^n$. For any $n$, choose $x_n = (1/2)^{1/n} \in [0,1)$; then $x_n^n = 1/2$, so the supremum is at least $1/2$ for all $n$. Hence the supremum does not tend to $0$.

Key theorem (uniform convergence preserves continuity). If each $f_n$ is continuous on $E$ and $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$.

Proof. Let $x_0 \in E$ and $\varepsilon > 0$. By uniform convergence, choose $N$ so that $|f_N(x) - f(x)| < \varepsilon/3$ for all $x$. By continuity of $f_N$, choose $\delta > 0$ so that $|x - x_0| < \delta \Rightarrow |f_N(x) - f_N(x_0)| < \varepsilon/3$. Then for $|x - x_0| < \delta$:

$|f(x) - f(x_0)| \leq |f(x) - f_N(x)| + |f_N(x) - f_N(x_0)| + |f_N(x_0) - f(x_0)| < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \;\blacksquare$

Contrapositive. If $f_n$ are continuous but $f$ is discontinuous, then convergence is not uniform.

Question Archetypes

determine-type (1 question(s); 2019)

Recognition Cues

• A specific sequence $f_n$ on a closed or half-open interval is given.
• Asked to “investigate,” “determine,” or “examine” convergence.
• Both pointwise and uniform must be addressed.

Solution Template

1. Compute the pointwise limit $f(x) = \lim_{n\to\infty} f_n(x)$ for each $x$ in the domain.
2. Check whether $f$ is continuous. If not, uniform convergence is impossible (use the contrapositive of the theorem above).
3. If $f$ is continuous, compute $M_n = \sup_{x \in E}|f_n(x) - f(x)|$ and determine whether $M_n \to 0$.
4. State the conclusion with justification.

Worked Example

2019 Paper 2, 2019-P2-Q4b (15 marks)

Let $f_n(x) = x^n$ for $x \in [0,1]$ and $n \in \mathbb{N}$. Find the pointwise limit $f$ of $(f_n)$. Is the convergence uniform? Justify your answer.

Pointwise limit.

For $x \in [0,1)$: $|x| < 1$, so $x^n \to 0$ as $n \to \infty$.

For $x = 1$: $1^n = 1 \to 1$.

Hence the pointwise limit is

$f(x) = \begin{cases} 0, & 0 \leq x < 1, \\ 1, & x = 1. \end{cases}$

Convergence is pointwise but not uniform.

Argument 1 — via continuity. Each $f_n(x) = x^n$ is continuous on $[0,1]$. The pointwise limit $f$ is discontinuous at $x = 1$ (since $f(1) = 1$ but $\lim_{x \to 1^-} f(x) = 0$). By the theorem: “uniform convergence of continuous functions implies continuity of the limit,” the convergence cannot be uniform.

Argument 2 — via supremum (direct). For $x \in [0,1)$,

$|f_n(x) - f(x)| = x^n.$

For any $n \in \mathbb{N}$, choose $x_n = \left(\frac{1}{2}\right)^{1/n} \in [0,1)$. Then

$|f_n(x_n) - f(x_n)| = x_n^n = \frac{1}{2}.$

Therefore

$M_n = \sup_{x \in [0,1]} |f_n(x) - f(x)| \geq \frac{1}{2} \quad \text{for all } n \in \mathbb{N}.$

Since $M_n \not\to 0$, the convergence is not uniform on $[0,1]$.

$\boxed{f_n \to f \text{ pointwise on } [0,1],\text{ but not uniformly.}}\;\blacksquare$

Common Traps

• Computing only the pointwise limit and stopping; the question explicitly asks about uniform convergence and its justification.
• Saying “convergence is not uniform because the limit is discontinuous” without citing the theorem that connects uniform convergence to continuity of the limit.
• Using argument 1 alone without argument 2 (or vice versa) in an exam; presenting both is safer at 15 marks, but argument 1 is cleaner and sufficient if stated precisely.
• Confusing $f_n \to 0$ uniformly on $[0, r]$ for any $r < 1$ (true!) with $f_n \to f$ uniformly on $[0,1]$ (false); the interval matters.

Marks-Aware Writing

At 15 marks (Section B), the expected answer has two parts: (a) derive the pointwise limit explicitly — roughly 4 marks; (b) prove or disprove uniform convergence rigorously — roughly 11 marks. For the uniformity part, state the method you use (discontinuity of limit, or direct supremum computation), carry it out in full, and state the conclusion explicitly. A sketchy argument loses marks even if the conclusion is correct.

Practice Set

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