Uniform convergence of function sequences

1. Definitions, built from scratch

2. The sup-norm criterion (the working tool)

Formula Sup-norm characterization

Define the uniform deviation $M_n := \sup_{x\in E}\,|f_n(x)-f(x)|.$ Then $f_n \to f \text{ uniformly on } E \iff M_n \to 0 \text{ as } n\to\infty.$

Derivation. ( $\Rightarrow$ ) If uniform, then for $\varepsilon>0$ pick $N$ so $n\ge N \Rightarrow |f_n(x)-f(x)|<\varepsilon/2$ for all $x$ . Taking sup over $x$ : $M_n \le \varepsilon/2 < \varepsilon$ . Hence $M_n\to 0$ . ( $\Leftarrow$ ) If $M_n\to 0$ , given $\varepsilon$ choose $N$ with $n\ge N\Rightarrow M_n<\varepsilon$ . Then for every $x$ , $|f_n(x)-f(x)|\le M_n<\varepsilon$ . That is uniform convergence. $\blacksquare$

Figure — Uniform convergence of function sequences

3. The classic counterexample (must memorize)

Worked example

$f_n(x)=x^n$ on $[0,1]$ Step 1 — find pointwise limit. Why? We must know $f$ before testing uniformity.

For $0\le x<1$ : $x^n\to 0$ .
For $x=1$ : $1^n = 1\to 1$ . So $f(x)=\begin{cases}0 & 0\le x<1\\ 1 & x=1\end{cases}$ .

Step 2 — note $f$ is discontinuous at $x=1$ , even though every $f_n$ is continuous. Why this matters: if convergence were uniform, the limit would have to be continuous (Theorem in §4). The break is our clue it is NOT uniform.

Step 3 — compute $M_n$ . Why? Direct sup test. $M_n=\sup_{x\in[0,1]}|x^n - f(x)|.$ For $x\in[0,1)$ the gap is $x^n$ , whose sup as $x\to 1^-$ is $1$ . So $M_n=1$ for all $n$ . Since $M_n=1\not\to 0$ , convergence is not uniform. $\blacksquare$

Forecast-then-verify: Before computing, you might guess "it converges nicely to $0$ ." The hidden trouble is the point sneaking toward $x=1$ , where $x^n$ stays near $1$ . The sup catches that point.

4. Why uniform convergence is worth the trouble

Formula Continuity passes to the limit

If each $f_n$ is continuous on $E$ and $f_n\to f$ uniformly, then $f$ is continuous.

Derivation (the $\varepsilon/3$ argument). Fix $x_0$ and $\varepsilon>0$ .

Uniformity: choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x$ .
Continuity of $f_N$ at $x_0$ : choose $\delta$ so $|x-x_0|<\delta \Rightarrow |f_N(x)-f_N(x_0)|<\varepsilon/3$ .
Then for $|x-x_0|<\delta$ : $|f(x)-f(x_0)| \le \underbrace{|f(x)-f_N(x)|}_{<\varepsilon/3}+\underbrace{|f_N(x)-f_N(x_0)|}_{<\varepsilon/3}+\underbrace{|f_N(x_0)-f(x_0)|}_{<\varepsilon/3} < \varepsilon. \ \blacksquare$ WHY uniformity is essential: step 1 needs one $N$ that controls $f-f_N$ at both $x$ and $x_0$ simultaneously — exactly what uniform gives.

5. Cauchy criterion (test without knowing the limit)

Common mistakes

Recall & Feynman

Recall Active recall — cover the answers

State both definitions; what's the ONE difference? ::: Quantifier order; $N(\varepsilon,x)$ vs $N(\varepsilon)$ .
The single computational test for uniformity? ::: $M_n=\sup_x|f_n-f|\to 0$ .
Why is $x^n$ on $[0,1]$ not uniform? ::: $M_n=1$ always; limit is discontinuous at $1$ .

Recall Feynman: explain to a 12-year-old

Imagine a class of runners (each $x$ is a runner) all heading to a finish line $f(x)$ . Pointwise: every runner eventually arrives, but some are super slow — you can't say "in 10 minutes everyone's done." Uniform: there's a single stopwatch time after which every runner is already inside a tiny circle of the finish. One time fits all. That single shared deadline is what lets you safely say nice things about the whole crowd at once.

Connections

Pointwise convergence — the weaker cousin.
Weierstrass M-test — sufficient condition for uniform convergence of series.
Continuity preserved under uniform limits — the $\varepsilon/3$ theorem.
Interchange of limit and integral / Dominated convergence theorem — measure-theoretic upgrade.
Cauchy sequences in metric spaces — uniform Cauchy = completeness of $C(E)$ under sup-norm.
Equicontinuity and Arzelà–Ascoli.

Pointwise vs uniform — the key difference

Order of quantifiers: pointwise allows

N=N(\varepsilon,x)

; uniform requires one

N=N(\varepsilon)

for all

x

Sup-norm test for uniform convergence

f_n\to f

uniformly

\iff M_n=\sup_x|f_n(x)-f(x)|\to 0

x^n

uniformly convergent on

[0,1]

No;

M_n=1

for all

n

and the limit jumps at

x=1

x^n

uniformly convergent on

[0,a]

a<1

Yes;

M_n=a^n\to0

Which property does uniform (not pointwise) convergence preserve?

Continuity of the limit function (also integral interchange).

Counterexample that uniform conv. of

f_n

doesn't give conv. of

f_n'

f_n=\sin(nx)/n\to0

uniformly but

f_n'=\cos(nx)

diverges.

Uniform Cauchy criterion statement

Uniform convergence

\iff \forall\varepsilon\,\exists N: m,n\ge N\Rightarrow\sup_x|f_n-f_m|<\varepsilon

Why does

\int f_n\to\int f

under uniform conv.?

|\int f_n-\int f|\le(b-a)M_n\to0

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, function sequence $f_n$ ka convergence do flavour me aata hai. Pointwise me har point $x$ apni alag speed se limit tak pahunchta hai — kisi point ke liye jaldi, kisi ke liye bahut der. Yaha $N$ depend karta hai $\varepsilon$ aur $x$ dono pe. Uniform me ek single $N$ hota hai jo saare points ke liye ek saath kaam karta hai — bas $\varepsilon$ pe depend, $x$ pe nahi. Yahi ek sentence ka difference hai: quantifier ka order ulta ho jaata hai.

Practically test karne ke liye sirf ek cheez yaad rakho: $M_n = \sup_x |f_n(x)-f(x)|$ nikaalo. Agar $M_n \to 0$ to uniform, warna nahi. Classic example $x^n$ on $[0,1]$ : pointwise limit $0$ banta hai except $x=1$ pe jaha $1$ , isliye limit discontinuous ho jaata hai aur $M_n=1$ hamesha — to uniform NAHI. Lekin agar domain $[0,a]$ le lo ( $a<1$ ), to $M_n=a^n\to0$ , uniform ho gaya. Matlab uniformity domain pe bhi depend karti hai, sirf formula pe nahi.

Yeh important kyu? Kyunki uniform convergence "strong glue" hai — agar har $f_n$ continuous hai aur convergence uniform hai, to limit $f$ bhi continuous rahega (famous $\varepsilon/3$ proof). Saath hi integral aur limit ko swap kar sakte ho: $\int f_n \to \int f$ . Pointwise me yeh guarantees nahi milti — corner ya jump sneak kar sakta hai. Exam tip: pehle pointwise limit $f$ nikaalo, fir $M_n$ ka sup lo — bas yahi 80/20 hai.