Real Analysis Comprehensive Proof Set I

From Pointwise Selection to Uniform Control

A useful proof often has two distinct stages: first select a subsequence with convergence at enough points, then show that this convergence controls the entire domain. The diagonal selection result established in “The Diagonalization Strategy” handles the first stage when the values at each selected point form a bounded real sequence. Equicontinuity supplies the bridge from those selected points to all points of an interval.

This tutorial develops that proof pattern as a sequential form of the Arzelà–Ascoli theorem. The key is not merely to select convergence at a dense set. Dense-set convergence alone need not imply uniform convergence. We will prove that a common continuity scale lets a finite collection of dense points control the whole interval.

Select on a countable dense set.
Boundedness gives convergent subsequences of function values at the first point, then the second, and so on; diagonal selection handles all the points at once.

Reduce to finitely many points.
Compactness of the interval lets finitely many neighborhoods of dense-set points cover it.

Transfer control to the interval.
Uniform equicontinuity bounds the change from any point to a nearby selected point.

Conclude uniform convergence.
The selected functions are uniformly Cauchy, so completeness of the real numbers gives a uniform limit.

A Dense-Set Criterion

For a sequence of functions, convergence at a dense set is a collection of pointwise statements. To make it useful for uniform convergence, we need a uniform way to compare values at nearby points. The following lemma isolates exactly that step.

Definition: A sequence of functions $f_n:[a,b]\to\mathbb R$ is uniformly equicontinuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that for every positive integer $n$ and all $x,y\in[a,b]$, $$ |x-y|<\delta \quad\Longrightarrow\quad |f_n(x)-f_n(y)|<\varepsilon. $$ The same $\delta$ must work for every function in the sequence and every pair of points in the interval.

Lemma (Dense-Set Uniform Cauchy Criterion): Let $D$ be a dense subset of $[a,b]$, and suppose $(f_n)$ is uniformly equicontinuous on $[a,b]$. If $(f_n(q))$ converges in $\mathbb R$ for every $q\in D$, then $(f_n)$ is uniformly Cauchy on $[a,b]$, and therefore converges uniformly to a real-valued function.

Proof: Let $\varepsilon>0$. By uniform equicontinuity, there is a $\delta>0$ such that $$ |x-y|<\delta\quad\Longrightarrow\quad |f_n(x)-f_n(y)|<\frac{\varepsilon}{3} $$ for every $n$. Since $D$ is dense, the open intervals of radius $\delta$ centered at its points cover $[a,b]$. The interval is compact by the Heine–Borel Theorem, so finitely many such intervals cover it. Write their centers as $q_1,\ldots,q_r\in D$. For each $x\in[a,b]$, at least one of these centers, say $q_i$, satisfies $|x-q_i|<\delta$.

At each of the finitely many points $q_i$, the numerical sequence $f_n(q_i)$ converges and hence is Cauchy. There is therefore an $N$ such that whenever $n,m\geq N$, all $i=1,\ldots,r$ satisfy $$ |f_n(q_i)-f_m(q_i)|<\frac{\varepsilon}{3}. $$ For any $x\in[a,b]$, choose a center $q_i$ with $|x-q_i|<\delta$. The triangle inequality gives $$ |f_n(x)-f_m(x)| \leq |f_n(x)-f_n(q_i)|+|f_n(q_i)-f_m(q_i)|+|f_m(q_i)-f_m(x)| <\varepsilon. $$ The choice of $N$ does not depend on $x$, so $(f_n)$ is uniformly Cauchy.

For each fixed $x$, the real sequence $f_n(x)$ is Cauchy and therefore converges; define $f(x)=\lim_{n\to\infty}f_n(x)$. To see that convergence is uniform, let $\eta>0$, and use the uniform Cauchy property with $\eta/2$. There is an $N$ such that $|f_n(x)-f_m(x)|<\eta/2$ for every $x$ whenever $n,m\geq N$. Fix $n\geq N$ and $x$, and let $m$ tend to infinity. Then $$ |f_n(x)-f(x)|\leq\frac{\eta}{2}<\eta. $$ This bound holds for every $x$, proving uniform convergence. $\square$

The proof uses compactness only to obtain finitely many centers. Convergence at each individual point of a dense set initially gives a different index threshold for each point. The finite cover reduces the number of thresholds that must be coordinated to a finite number, so one index works for all of them.

The Sequential Arzelà–Ascoli Theorem

We now combine the dense-set criterion with diagonal selection. Pointwise boundedness means that for every fixed $x$, the set of values $\{f_n(x):n\geq1\}$ is bounded. It does not require one bound to work at every $x$; the theorem below assumes the stronger, convenient condition of uniform boundedness.

Theorem (Arzelà–Ascoli Theorem, Sequential Form on a Closed Interval): Let $a<b$, and let $(f_n)$ be a uniformly bounded, uniformly equicontinuous sequence of functions $f_n:[a,b]\to\mathbb R$. Then $(f_n)$ has a subsequence that converges uniformly on $[a,b]$. Its limit is continuous.

Proof: Choose a countable dense subset $D=\{q_1,q_2,\ldots\}$ of $[a,b]$. For example, one may take the points $a+(b-a)r$, where $r$ ranges over the rational numbers in $[0,1]$. Uniform boundedness implies that for each fixed $k$, the real sequence $(f_n(q_k))_{n\geq1}$ is bounded.

Apply the Diagonal Subsequence Theorem established earlier in the course to the bounded real sequences $x_n^{(k)}=f_n(q_k)$. It gives a strictly increasing sequence of indices $n_j$ such that, for every $k$, the sequence $(f_{n_j}(q_k))_{j\geq1}$ converges. The subsequence $(f_{n_j})$ remains uniformly equicontinuous because the same $\delta$ that works for every $f_n$ also works for every selected function.

Apply the Dense-Set Uniform Cauchy Criterion to $(f_{n_j})$ and $D$. It follows that this subsequence converges uniformly to a real-valued function $f$. Uniform equicontinuity implies that each $f_{n_j}$ is continuous: given a point $x$ and an error tolerance, the common $\delta$ controls $|f_{n_j}(y)-f_{n_j}(x)|$ whenever $|y-x|<\delta$. Finally, the Uniform Limits Preserve Continuity theorem established earlier shows that $f$ is continuous. $\square$

Worked Examples

Worked Example: A Bounded Oscillatory Perturbation

On $[0,1]$, define $$ f_n(x)=x^2+\frac{\sin(nx)}{n}. $$ Since $0\leq x^2\leq1$ and $|\sin(nx)|/n\leq1$, we have $|f_n(x)|\leq2$ for every $x$ and $n$. Thus the sequence is uniformly bounded.

For $x,y\in[0,1]$, the Mean Value Theorem applied to $f_n$ gives $$ |f_n(x)-f_n(y)|\leq \sup_{t\in[0,1]}|f_n'(t)|\,|x-y|. $$ Here $$ f_n'(t)=2t+\cos(nt), \qquad |f_n'(t)|\leq 2+1=3. $$ Consequently, $|f_n(x)-f_n(y)|\leq3|x-y|$ for every $n$, so the sequence is uniformly equicontinuous. The theorem guarantees a uniformly convergent subsequence. In fact, the whole sequence converges uniformly to $x^2$, because $$ |f_n(x)-x^2|=\frac{|\sin(nx)|}{n}\leq\frac{1}{n} $$ for every $x\in[0,1]$, and $1/n\to0$.

Worked Example: Power Functions Do Not Have a Uniformly Convergent Subsequence

Consider $f_n(x)=x^n$ on $[0,1]$. This sequence is uniformly bounded, since $0\leq x^n\leq1$. But it is not uniformly equicontinuous. For $n\geq2$, set $x_n=1-1/n$ and $y_n=1$. Then $|x_n-y_n|=1/n\to0$, while $$ |f_n(x_n)-f_n(y_n)| =1-\left(1-\frac1n\right)^n \longrightarrow 1-e^{-1}>0. $$ Thus arbitrarily close inputs can have function values separated by an amount that does not tend to zero.

There is also a direct proof that no subsequence converges uniformly. For every fixed $x\in[0,1)$, $x^n\to0$; at $x=1$, $x^n=1$ for every $n$. Every subsequence therefore has the same pointwise limit, the function that is $0$ on $[0,1)$ and $1$ at $1$. That function is discontinuous at $1$. If any subsequence converged uniformly, the Uniform Limits Preserve Continuity theorem would make its limit continuous, a contradiction. This example shows why boundedness alone cannot replace equicontinuity.

Worked Example: An Integral Family with a Uniformly Convergent Subsequence

Let $(g_n)$ be any sequence of continuous functions on $[0,1]$ satisfying $|g_n(t)|\leq1$ for every $n$ and $t$, and define $$ F_n(x)=\int_0^x g_n(t)\,dt. $$ The integral bound gives $|F_n(x)|\leq x\leq1$, so $(F_n)$ is uniformly bounded. For $x,y\in[0,1]$, additivity of the integral and the absolute-value bound give $$ |F_n(x)-F_n(y)| =\left|\int_y^x g_n(t)\,dt\right| \leq |x-y|. $$ When $x<y$, the integral from $y$ to $x$ is interpreted with reversed endpoints, and the same bound follows by changing the sign. Thus the sequence is uniformly equicontinuous, with $\delta=\varepsilon$ for any tolerance $\varepsilon>0$.

The Arzelà–Ascoli theorem now guarantees a uniformly convergent subsequence of $(F_n)$. The conclusion does not require the functions $g_n$ themselves to converge. It comes from the common bound on their integrals, which prevents the functions $F_n$ from changing too rapidly. For the specific choice $g_n(t)=\sin(nt)$, one can also check directly that $$ F_n(x)=\frac{1-\cos(nx)}{n}, \qquad |F_n(x)|\leq\frac{2}{n}, $$ so in this case the whole sequence converges uniformly to zero.

How to Use the Proof Pattern

The argument is useful when direct formulas for a convergent subsequence are unavailable. Rather than trying to guess the subsequence, separate the task into two obligations: secure convergence on a countable dense set, then use equicontinuity and a finite cover to promote that convergence to uniform control. Uniform boundedness supports the first obligation by making each sequence of sampled values bounded.

Keep the hypotheses distinct. Boundedness controls the range of each function but says nothing by itself about how rapidly values can change across the interval. Equicontinuity controls those changes but does not, by itself, bound the values at one reference point. Both are used in the theorem: uniform boundedness enables diagonal selection, while uniform equicontinuity makes the selected subsequence uniformly Cauchy.

The theorem asserts existence of a uniformly convergent subsequence, not convergence of the original sequence. The sequence $x^n$ illustrates a failure caused by lack of equicontinuity. In the integral example, the common bound on $g_n$ supplies a Lipschitz estimate for $F_n$; such estimates are an efficient way to verify uniform equicontinuity. In any application, write out the bound and check that its constants do not depend on $n$.

Check Your Understanding

For each question, identify the role played by the hypothesis or estimate in the proof.

Why does convergence of $f_n(q)$ at every point of a dense set not, by itself, prove uniform convergence?
Where does compactness of $[a,b]$ enter the Dense-Set Uniform Cauchy Criterion?
In the Arzelà–Ascoli proof, which hypothesis allows diagonal selection at each dense-set point?
Why does the estimate $|F_n(x)-F_n(y)|\leq|x-y|$ prove uniform equicontinuity independently of $n$?
What contradiction rules out a uniformly convergent subsequence of $x^n$ on $[0,1]$?