Function Sequence Proof Workshop

From Convergence Claims to Proof Strategies

The previous tutorial used partial sums of a function series as a sequence of functions. Its differentiation theorem depended on controlling those partial sums throughout an interval, not just at individual points. This tutorial develops two proof strategies for function sequences more generally: passing order information to a pointwise limit, and obtaining uniform convergence from pointwise convergence when the functions cannot vary too rapidly.

These strategies address different questions. Order arguments use the fact that inequalities survive limits of real numbers. Uniform-convergence arguments need more: pointwise control at a finite set of inputs must be extended to every input. A shared continuity estimate lets us make that extension.

Definition: A function $f:I\to\mathbb{R}$ on an interval $I$ is nondecreasing if $f(x)\leq f(y)$ whenever $x\leq y$ in $I$. A sequence $(f_n)$ on $I$ is uniformly equicontinuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that $$ |f_n(x)-f_n(y)|<\varepsilon $$ for every $n$ and every $x,y\in I$ with $|x-y|<\delta$. The same $\delta$ must work for all functions in the sequence.

Passing Inequalities to a Pointwise Limit

Suppose real numbers $a_n$ and $b_n$ satisfy $a_n\leq b_n$ for every $n$, and both sequences converge. Then their limits satisfy the same inequality. This elementary fact often provides the whole proof when the hypothesis is pointwise convergence: fix the inputs first, and then take limits in the resulting numerical inequality.

Theorem (Pointwise Limits Preserve Monotonicity): Let $I$ be an interval, and suppose $f_n:I\to\mathbb{R}$ is nondecreasing for every $n$. If $f_n$ converges pointwise on $I$ to $f$, then $f$ is nondecreasing on $I$.

Proof. Take any $x,y\in I$ with $x\leq y$. Since each $f_n$ is nondecreasing, $$ f_n(x)\leq f_n(y) $$ for every $n$. Pointwise convergence gives $f_n(x)\to f(x)$ and $f_n(y)\to f(y)$. Passing to the limit in the numerical inequalities yields $f(x)\leq f(y)$. This holds for every such pair $x,y$, so $f$ is nondecreasing. $\square$

The same reasoning works for other pointwise inequalities. For example, if $f_n(x)\leq g_n(x)$ for all $n$ and $x$, and both sequences converge pointwise, then their limits satisfy $f(x)\leq g(x)$ at every $x$. The order of the proof matters: fix $x$ (or a pair $x,y$), use the inequality at that input, and only then take the numerical limit.

Worked Example: A Pointwise Limit of Increasing Functions

For $x\in[0,\infty)$, define $$ f_n(x)=\frac{x}{1+x/n}. $$ The denominator is positive. If $0\leq x\leq y$, direct subtraction gives $$ f_n(y)-f_n(x) =\frac{y}{1+y/n}-\frac{x}{1+x/n} =\frac{y(1+x/n)-x(1+y/n)}{(1+y/n)(1+x/n)} =\frac{y-x}{(1+y/n)(1+x/n)}\geq0. $$ Thus every $f_n$ is nondecreasing.

For a fixed $x\geq0$, the denominator $1+x/n$ tends to $1$, so $f_n(x)\to x$. The Pointwise Limits Preserve Monotonicity Theorem therefore implies that the limit function $f(x)=x$ is nondecreasing on $[0,\infty)$. Here the limit can also be checked directly: if $x\leq y$, then $f(x)=x\leq y=f(y)$. The theorem is useful even when the limit’s monotonicity is not so immediate to recognize.

When Pointwise Convergence Becomes Uniform

Pointwise convergence gives control at each fixed input, with the required index allowed to depend on that input. A finite mesh can turn this into uniform control if the functions share a common continuity estimate. First choose finitely many mesh points close to every point of the interval. Pointwise convergence supplies an index at each mesh point; since there are only finitely many, one index works at all of them. Uniform equicontinuity then controls the gaps between mesh points.

Theorem (Finite-Net Criterion for Uniform Convergence): Let $a<b$, and suppose $f_n:[a,b]\to\mathbb{R}$ is uniformly equicontinuous. If $f_n$ converges pointwise on $[a,b]$ to $f$, then $f_n$ converges uniformly to $f$ on $[a,b]$.

Proof. Let $\varepsilon>0$. By uniform equicontinuity, there is a $\delta>0$ such that, for every $n$, $$ |f_n(x)-f_n(y)|<\frac{\varepsilon}{4} \quad\text{whenever}\quad |x-y|<\delta. $$ Choose a positive integer $K$ large enough that $(b-a)/K<\delta$, and set $x_j=a+j(b-a)/K$ for $j=0,\ldots,K$. Every $x\in[a,b]$ is within distance less than $\delta$ of at least one of these finitely many mesh points; choose one and call it $x_j$.

Pointwise convergence at each $x_j$ gives an integer $N_j$ such that $$ |f_n(x_j)-f(x_j)|<\frac{\varepsilon}{4} \qquad(n\geq N_j). $$ Let $N$ be the largest of $N_0,\ldots,N_K$. Fix $n\geq N$ and $x\in[a,b]$, and choose a mesh point $x_j$ within distance $\delta$ of $x$. The equicontinuity estimate applies to $f_n$. It also gives a bound for the limit: for this fixed pair $x,x_j$, take the limit as $m\to\infty$ in $$ |f_m(x)-f_m(x_j)|<\frac{\varepsilon}{4}. $$ It follows that $|f(x)-f(x_j)|\leq\varepsilon/4$. Thus $$ |f_n(x)-f(x)| \leq |f_n(x)-f_n(x_j)|+|f_n(x_j)-f(x_j)|+|f(x_j)-f(x)| <\frac{3\varepsilon}{4}<\varepsilon. $$ The same $N$ works for every $x\in[a,b]$. Therefore the convergence is uniform. $\square$

The proof’s key device is the finite mesh. Pointwise convergence alone gives no single index that controls all inputs. At finitely many inputs, however, the maximum of the individual indices works. Equicontinuity transports that finite collection of estimates to the entire interval.

Worked Example: A Square-Root Sequence Converges Uniformly

On $[0,1]$, define $$ f_n(x)=\sqrt{x+\frac{1}{n}}-\sqrt{\frac{1}{n}}. $$ For fixed $x$, continuity of the square-root function gives $f_n(x)\to\sqrt{x}$. To check uniform equicontinuity, take $u,v\geq0$. If $u\geq v$, then $$ |\sqrt{u}-\sqrt{v}|^2 =(\sqrt{u}-\sqrt{v})^2 \leq(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v}) =u-v. $$ The inequality holds because $0\leq\sqrt{u}-\sqrt{v}\leq\sqrt{u}+\sqrt{v}$. If $v\geq u$, interchanging $u$ and $v$ gives the corresponding bound. In either case, $|\sqrt{u}-\sqrt{v}|\leq\sqrt{|u-v|}$.

Apply this estimate with $u=x+1/n$ and $v=y+1/n$. The subtracted constant in $f_n$ cancels when taking a difference, so $$ |f_n(x)-f_n(y)|\leq\sqrt{|x-y|} $$ for every $n$ and all $x,y\in[0,1]$. Given $\varepsilon>0$, choosing $\delta=\varepsilon^2$ ensures that $|x-y|<\delta$ implies $|f_n(x)-f_n(y)|<\varepsilon$. The sequence is uniformly equicontinuous, so the Finite-Net Criterion gives uniform convergence to $\sqrt{x}$.

What Fails Without Shared Control

The common continuity estimate is not a cosmetic assumption. Without it, behavior can concentrate in narrower regions as $n$ grows. Each fixed input may eventually miss those regions, even though the maximum error remains large. Such a sequence can be pointwise convergent without being uniformly convergent.

Worked Example: A Narrowing Sequence of Ramps

For $x\in[0,1]$, let $$ f_n(x)=\max\{1-nx,0\}. $$ At $x=0$, $f_n(0)=1$ for every $n$. If $x>0$, then $n>1/x$ implies $1-nx<0$, so $f_n(x)=0$ for all such $n$. Hence $f_n$ converges pointwise to the function $f$ defined by $f(0)=1$ and $f(x)=0$ for $x>0$.

For every $n$, $f_n(0)=f(0)=1$, so the error at $x=0$ is zero. But $f_n(1/(2n))=1/2$ and $f(1/(2n))=0$, so $$ \sup_{x\in[0,1]}|f_n(x)-f(x)|\geq\frac{1}{2}. $$ In fact, this supremum equals $1$, approached as $x\to0^+$ but not attained. Taking $\varepsilon=1/2$ shows that the convergence is not uniform. The sequence is not uniformly equicontinuous either. Given any $\delta>0$, choose $n$ with $1/n<\delta$. Then $|0-1/n|<\delta$, but $f_n(0)=1$ and $f_n(1/n)=0$. A shared choice of $\delta$ cannot control all the functions.

This example also warns against assuming that a pointwise limit inherits continuity. Uniform equicontinuity supplies a common restriction on how sharply the functions can change. Without it, pointwise convergence places no uniform restriction on narrow features. Conversely, do not confuse the finite-net theorem with a claim that every pointwise-convergent sequence is uniformly convergent: the common continuity estimate is precisely what makes the argument work.

A Practical Proof Plan

For a function-sequence problem, first decide whether the conclusion concerns each fixed input or all inputs at once. Then identify what information can be carried through the limit and what must be controlled uniformly.

Fix the inputs involved.
For monotonicity, take an arbitrary pair $x\leq y$; for a pointwise inequality, fix $x$.

Write the finite-index statement.
Use the hypothesis to establish an inequality or estimate for each $n$.

Take the pointwise limit carefully.
At fixed inputs, use convergence of the relevant real-number sequences.

For uniform convergence, seek shared control.
If the functions are uniformly equicontinuous on a closed interval, use a finite mesh and take the maximum of finitely many indices.

Takeaway: Pointwise limits preserve inequalities such as monotonicity because real-number order is preserved under limits. To upgrade pointwise convergence to uniform convergence on a closed interval, a common continuity estimate lets finite pointwise control extend to every input.

Check Your Understanding

Use the proof strategies and examples above to answer the following questions.

In the monotonicity theorem, why do we fix $x\leq y$ before taking limits?
What does uniform equicontinuity require that ordinary continuity of each individual $f_n$ does not?
Why can one index control convergence at all mesh points in the finite-net proof?
How does the common square-root estimate give a choice of $\delta$ that works for every $n$?
Why does the narrowing-ramp sequence fail to converge uniformly, even though it converges pointwise at every input?