Arzelà–Ascoli and equicontinuity

In one sentence

Boundedness gives you a convergent subsequence for numbers but not for functions; equicontinuity is the extra hypothesis that repairs it, and Arzelà–Ascoli is the repair.

Why the exam asks it

Because it is the cleanest place to show that intuition from $\mathbb{R}$ does not transfer to function spaces. Bolzano–Weierstrass feels like a fact about boundedness. In $C[a,b]$ it is false, and the counterexample is a function everyone already knows.

The idea

Arzelà–Ascoli: a sequence in $C[a,b]$ that is uniformly bounded and equicontinuous has a uniformly convergent subsequence.

Uniformly bounded: one constant $M$ with $|f_n(x)| \le M$ for all $n$ and all $x$.

Equicontinuous: for every $\varepsilon > 0$ there is a $\delta > 0$ that works for every $f_n$ at once — $|x - y| < \delta$ implies $|f_n(x) - f_n(y)| < \varepsilon$ for all $n$. It is uniform continuity, made uniform across the sequence as well as across the domain.

That second "uniform" is the whole content. Each $f_n$ being uniformly continuous is not enough; they must share a single $\delta$.

Why both hypotheses are needed

Drop boundedness: take $f_n \equiv n$, the constant functions. Perfectly equicontinuous — any $\delta$ works, since each function is constant — and no subsequence converges.

Drop equicontinuity: take $f_n(x) = x^n$ on $[0,1]$. Uniformly bounded by $1$. But the pointwise limit is discontinuous at $1$, and a uniform limit of continuous functions must be continuous, so no subsequence can converge uniformly.

What fails is equicontinuity near $x = 1$: to keep $|x^n - 1|$ small you need $\delta$ shrinking with $n$, and no single $\delta$ serves them all.

The related failures worth carrying

Two more sequences answer a large share of the options in this area.

$f_n = n \cdot 1_{(0, 1/n)}$ — the collapsing spike.

Every point $x > 0$ is eventually to the right of the spike, so $f_n \to 0$ pointwise. But $\int_0^1 f_n = 1$ for every $n$. Pointwise convergence does not let you pass the limit through an integral.

$f_n(x) = \sin(nx)/n$ — flattening, but never smooth.

$\sup |f_n| = 1/n \to 0$, so this converges uniformly to zero. Yet $f_n'(x) = \cos(nx)$, whose amplitude is always $1$ and which converges nowhere. Uniform convergence controls the values and says nothing about the derivatives.

Where intuition breaks

"Uniformly bounded in $C[0,1]$, so a uniformly convergent subsequence exists." $x^n$. This is Bolzano–Weierstrass imported into a space where it does not hold — a hypothesis dropped, and the missing hypothesis is equicontinuity.

Why it is tempting: in $\mathbb{R}^n$ bounded really does give a convergent subsequence, and $C[a,b]$ looks like just another space. It is infinite-dimensional, and closed bounded sets there are not compact.

"Each $f_n$ is continuous, so the family is equicontinuous." Equicontinuity is a statement about the family, not about its members. $x^n$ is a sequence of perfectly continuous — indeed smooth — functions that is not equicontinuous.

"$f_n \to f$ uniformly, so $f_n' \to f'$." $\sin(nx)/n$. Uniform convergence of the functions gives you continuity of the limit, term-by-term integration, and the exchange of limits — but nothing about derivatives. For derivatives you need $f_n'$ to converge uniformly, and then you get the conclusion.

"$f_n \to 0$ pointwise on $[0,1]$, so $\int_0^1 f_n \to 0$." The collapsing spike. What rescues it is uniform convergence, or domination — the bounded convergence theorem, if you have measure theory available.

"Equicontinuous and pointwise bounded is not enough; you need uniform boundedness." On a compact interval these are the same: pointwise boundedness plus equicontinuity gives uniform boundedness. Options asserting the stronger hypothesis is necessary are wrong, and this is worth knowing because it looks like the cautious answer.

The exam's angle

1. Check both hypotheses separately. Most wrong options supply exactly one.
2. To disprove equicontinuity, find the point where $\delta$ must shrink. For $x^n$ it is $x = 1$; for $\sin(nx)$ it is everywhere.
3. If the claim is about derivatives, uniform convergence of $f_n$ is not the relevant hypothesis. Look for uniform convergence of $f_n'$.
4. If the claim is about integrals, ask whether convergence is uniform or merely pointwise.

The night before

• Arzelà–Ascoli: uniformly bounded and equicontinuous $\Rightarrow$ a uniformly convergent subsequence.
• Equicontinuity: one $\delta$ for the whole family, not one per function.
• $x^n$ on $[0,1]$: bounded, not equicontinuous, no uniformly convergent subsequence.
• $f_n \equiv n$: equicontinuous, not bounded.
• $n \cdot 1_{(0,1/n)}$: pointwise $\to 0$, integrals all $1$.
• $\sin(nx)/n$: uniformly $\to 0$, derivatives $\cos(nx)$ converge nowhere.
• Closed bounded sets in $C[a,b]$ are not compact. That is the whole reason this theorem exists.