Real Analysis Comprehensive Proof Set II

When Pointwise Convergence Becomes Uniform

Pointwise convergence controls the sequence separately at each point: the index after which the error is small may depend on the point. Uniform convergence requires a single index to work across the entire domain. In “Real Analysis Comprehensive Proof Set I,” the sequential Arzelà–Ascoli theorem used equicontinuity to turn selected pointwise information into uniform control. Here a different structure does the work: monotonicity of the sequence, continuity of its limit, and compactness of the domain.

Dini’s theorem gives a useful criterion for uniform convergence. It does not say that every pointwise limit of continuous functions is continuous, nor that every pointwise-convergent sequence on a compact space converges uniformly. Its hypotheses work together: continuity makes certain error sets open, monotonicity makes those sets nested, and compactness reduces the cover to finitely many sets.

Dini’s Theorem

Theorem (Dini’s Theorem, Decreasing Version): Let $K$ be a compact topological space. Suppose $f_n:K\to\mathbb R$ is continuous for every positive integer $n$, $f_n(x)\geq f_{n+1}(x)$ for every $x\in K$ and every $n$, and $f_n(x)\to f(x)$ pointwise on $K$, where $f:K\to\mathbb R$ is continuous. Then $f_n\to f$ uniformly on $K$.

Proof: If $K$ is empty, uniform convergence holds vacuously. Suppose $K$ is nonempty, and let $\varepsilon>0$. Since $f_n(x)$ decreases to $f(x)$ at each $x$, we have $f_n(x)\geq f(x)$ for every $n$ and $x$.

For each $n$, define $$ U_n=\{x\in K:f_n(x)-f(x)<\varepsilon\}. $$ The function $f_n-f$ is continuous, so $U_n$ is open in $K$. The sets are increasing: if $x\in U_n$, then $f_{n+1}(x)\leq f_n(x)$, and hence $f_{n+1}(x)-f(x)<\varepsilon$, so $x\in U_{n+1}$. Pointwise convergence to $f$ ensures that for each $x\in K$, there is some $n$ for which $f_n(x)-f(x)<\varepsilon$. Thus the sets $U_n$ cover $K$.

By compactness, finitely many of these open sets cover $K$, say $K=U_{n_1}\cup\cdots\cup U_{n_r}$. Let $N$ be the largest of the indices $n_1,\ldots,n_r$. Because the sets are increasing, each $U_{n_i}$ is contained in $U_N$. Therefore $K\subseteq U_N$. For every $x\in K$ and every $n\geq N$, we have $$ 0\leq f_n(x)-f(x)\leq f_N(x)-f(x)<\varepsilon. $$ The same $N$ works for every $x\in K$. This is uniform convergence. $\square$

The proof turns a pointwise statement into a cover: each point belongs to some set $U_n$, but the index may initially vary with the point. Compactness gives a finite subcover, and the nested structure of the sets means the largest of those finitely many indices works everywhere. Without monotonicity, a finite subcover would not generally produce one set that contains all the others.

Corollary (Dini’s Theorem, Increasing Version): Under the same compactness and continuity assumptions, if $f_n(x)\leq f_{n+1}(x)$ for every $x$ and $n$, and $f_n(x)\to f(x)$ pointwise for a continuous $f$, then $f_n\to f$ uniformly. To see this, apply the decreasing version to the continuous functions $-f_n$, which decrease pointwise to the continuous function $-f$. Uniform convergence of $-f_n$ to $-f$ is equivalent to uniform convergence of $f_n$ to $f$.

Worked Examples

Worked Example: A Decreasing Sequence with a Continuous Limit

On $K=[-1,1]$, define $$ f_n(x)=\sqrt{x^2+\frac1n}. $$ Each $f_n$ is continuous. Since $1/(n+1)<1/n$, we have $f_{n+1}(x)\leq f_n(x)$ for every $x$. Also, $f_n(x)\to |x|$ pointwise, and $x\mapsto |x|$ is continuous. Dini’s theorem therefore gives uniform convergence to $|x|$.

In this example the error can also be bounded directly. Rationalizing gives $$ 0\leq \sqrt{x^2+\frac1n}-|x| =\frac{1/n}{\sqrt{x^2+1/n}+|x|} \leq \frac{1/n}{1/\sqrt n} =\frac1{\sqrt n}. $$ The denominator is at least $1/\sqrt n$, including when $x=0$. Thus for every $x\in[-1,1]$, the error is at most $1/\sqrt n$, which tends to zero independently of $x$. The direct estimate verifies the same uniform convergence and illustrates what the theorem can guarantee even when finding an error bound is less convenient.

Worked Example: A Discontinuous Limit Blocks Dini’s Theorem

Consider $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and $f_{n+1}(x)\leq f_n(x)$ because $0\leq x\leq1$. The pointwise limit is $$ f(x)= \begin{cases} 0,&0\leq x<1,\\ 1,&x=1. \end{cases} $$ This limit is not continuous at $1$, so the continuity hypothesis in Dini’s theorem fails.

The convergence is not uniform. For $n\geq2$, let $x_n=1-1/n$. Then $x_n<1$, so $f(x_n)=0$, while $$ |f_n(x_n)-f(x_n)|=\left(1-\frac1n\right)^n\longrightarrow e^{-1}>0. $$ Consequently the supremum of the error over $[0,1]$ cannot tend to zero. The example also shows why monotonicity and compactness alone are insufficient: the limit must be continuous as well.

Worked Example: A Continuous Limit on a Noncompact Domain

On $[0,\infty)$, set $h_n(x)=e^{-x/n}$. Each $h_n$ is continuous, and for each fixed $x\geq0$, $h_n(x)$ increases to $1$ as $n\to\infty$. The limit $h(x)=1$ is continuous. Nevertheless, the convergence is not uniform: evaluating the $n$-th error at $x=n$ gives $$ |h_n(n)-1|=|e^{-1}-1|=1-e^{-1}. $$ This positive error is independent of $n$. The interval $[0,\infty)$ is not compact, so the compactness step in Dini’s proof is unavailable. Pointwise convergence to a continuous limit, even with monotonicity, need not be uniform on a noncompact domain.

Worked Example: Compactness and a Continuous Limit Do Not Replace Monotonicity

For $n\geq2$, define a function on $[0,1]$ by $$ g_n(x)=\max\{1-n|x-2/n|,0\}. $$ Each $g_n$ is continuous: it is the maximum of the continuous function $1-n|x-2/n|$ and the constant function $0$. Its values are zero outside $[1/n,3/n]$, and $g_n(2/n)=1$.

For each fixed $x>0$, choose $n$ large enough that $3/n<x$. Then $x$ lies outside $[1/n,3/n]$, so $g_n(x)=0$ for all sufficiently large $n$. Also $g_n(0)=0$ for every $n$. Hence $g_n\to0$ pointwise, and the limit is continuous. But $$ \sup_{x\in[0,1]}|g_n(x)|\geq g_n(2/n)=1 $$ for every $n\geq2$, so convergence is not uniform. The functions do not form a monotone sequence: their narrow peaks move across the interval. This example isolates the role of monotonicity in Dini’s theorem.

What to Check Before Applying the Theorem

Dini’s theorem is useful when pointwise limits are straightforward but a direct uniform error estimate is difficult. In an application, verify the hypotheses separately: the domain must be compact, every function in the sequence must be continuous, the sequence must be monotone in the same direction at every point, and the pointwise limit must be continuous. Monotonicity means $f_n(x)\geq f_{n+1}(x)$ for all $x$, or the reverse inequality for all $x$; it is not enough for the direction to vary from point to point.

The conclusion is stronger than pointwise convergence because one index controls the error across the entire domain. It complements the sequential Arzelà–Ascoli theorem: Arzelà–Ascoli obtains a uniformly convergent subsequence under boundedness and equicontinuity, while Dini’s theorem gives uniform convergence of the whole sequence from monotonicity and continuity of its limit. Neither result allows its distinctive hypotheses to be silently omitted.

Check Your Understanding

Use the proof and examples to identify which hypothesis supplies each step of Dini’s argument.

Why is each set $U_n=\{x:f_n(x)-f(x)<\varepsilon\}$ open in $K$?
Where does monotonicity enter the argument after the sets $U_n$ have been defined?
How does compactness turn pointwise convergence into one index that works everywhere?
Which hypothesis fails for $x^n$ on $[0,1]$, and how does the pointwise limit demonstrate that failure?
In the moving-peak example, why does pointwise convergence hold even though the supremum of the error stays at least $1$?