Uniform Convergence of Functions

In this lesson you’ll see that a limit of continuous functions need not be continuous, learn the stronger notion - uniform convergence - that repairs this, and understand exactly which operations may be exchanged with a limit and which may not.

A sequence of functions can converge in more than one way, and the difference is not a technicality - it decides whether continuity, integrals, and derivatives survive the limit.

$f_n \to f$ pointwise on $S$ if $f_n(x) \to f(x)$ for every $x \in S$.

Spelled out with quantifiers:

$$\forall x \in S,\ \forall \varepsilon>0,\ \exists N \text{ such that } n \ge N \implies |f_n(x)-f(x)|<\varepsilon$$

The $N$ sits after the $x$, so it may depend on it. That is the entire weakness, and it is enough to destroy nearly everything.

$f_n \to f$ uniformly on $S$ if for every $\varepsilon>0$ there is an $N$ such that $n \ge N$ implies $|f_n(x)-f(x)|<\varepsilon$ for all $x \in S$.

$$\forall \varepsilon>0,\ \exists N,\ \forall n\ge N,\ \forall x \in S: |f_n(x)-f(x)|<\varepsilon$$

Same symbols, one quantifier moved. Now $N$ is chosen before $x$ is seen, so a single index works across the whole set. Uniform convergence implies pointwise convergence; the converse fails.

Equivalently, in terms of the supremum norm,

$$f_n \rightrightarrows f \iff \|f_n - f\| = \sup_{x\in S}|f_n(x)-f(x)| \to 0$$

That reformulation turns a statement about functions into a statement about a single sequence of numbers, which is usually the easiest way to check it.

Geometrically: draw a band of half-width $\varepsilon$ around the graph of $f$. Uniform convergence says that from some index on, the entire graph of $f_n$ lies inside the band. Pointwise convergence only says each individual point eventually gets in, and points may enter at wildly different times.

On $[0,1]$ let $f_n(x)=x^n$. Then

$$f_n(x) \to f(x) = \begin{cases}0 & 0\le x<1\\ 1 & x=1\end{cases}$$

Every $f_n$ is continuous. The limit is not. So pointwise limits of continuous functions need not be continuous, and this single example is why the whole lesson exists.

The convergence is not uniform, because

$$\|f_n-f\| = \sup_{0\le x<1}x^n = 1 \quad \text{for every } n$$

The supremum is never attained, but it is 1 regardless of $n$: the graph always has a stretch climbing from near 0 to near 1, and that stretch just slides right without flattening.

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On $[0, 0.9]$, by contrast, $\|f_n-f\| = 0.9^n \to 0$, so the convergence is uniform there. Uniformity is a property of the sequence together with the set, and shrinking the set can create it.

Continuity. If each $f_n$ is continuous on $S$ and $f_n \rightrightarrows f$, then $f$ is continuous.

Proof. Fix $c$ and $\varepsilon>0$. Pick $n$ with $\|f_n-f\|<\varepsilon/3$. Pick $\delta$ from continuity of that single $f_n$ at $c$, so $|f_n(x)-f_n(c)|<\varepsilon/3$ for $|x-c|<\delta$. Then

$$|f(x)-f(c)| \le |f(x)-f_n(x)|+|f_n(x)-f_n(c)|+|f_n(c)-f(c)| < \varepsilon$$

∎

The three-epsilon argument. Two of the three pieces are controlled by uniformity and one by continuity of a fixed member of the sequence. Note that the middle term needs $|f_n(x)-f_n(c)|$ small for the same $n$ at both points, which is exactly what pointwise convergence cannot deliver.

An equivalent phrasing of the same theorem is that limits may be interchanged:

$$\lim_{x\to c}\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty}\lim_{x\to c}f_n(x)$$

Interchanging two limits is the recurring theme of this lesson, and uniformity is the usual licence to do it.

Integrals. If each $f_n$ is integrable on $[a,b]$ and $f_n \rightrightarrows f$, then $f$ is integrable and

$$\int_a^b f_n \to \int_a^b f$$

Reason. $\left|\int f_n - \int f\right| \le \int|f_n-f| \le \|f_n-f\|(b-a) \to 0$. One line, and the $(b-a)$ shows why a bounded interval matters.

Pointwise convergence is not enough. Let $f_n$ be a spike of height $n$ and width $1/n$ near 0, and 0 elsewhere. Then $f_n \to 0$ pointwise, but $\int_0^1 f_n = 1$ for every $n$. The limit of the integrals is 1, the integral of the limit is 0.

Derivatives. This is the important asymmetry.

Let $f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$. Then $\|f_n\| = 1/\sqrt{n}\to0$, so $f_n \rightrightarrows 0$ uniformly. But

$$f_n'(x)=\sqrt{n}\cos(nx)$$

which is unbounded and converges to nothing. Uniform convergence of $f_n$ says nothing about $f_n'$.

The reason is structural: integration averages and therefore smooths, while differentiation amplifies fine detail, and uniform closeness of the graphs is a statement about values, not slopes.

The correct theorem runs the other way round:

If $f_n \to f$ pointwise on $[a,b]$, each $f_n'$ is continuous, and $f_n' \rightrightarrows g$ uniformly, then $f$ is differentiable and $f'=g$.

The hypothesis is uniform convergence of the derivatives, and then the conclusion about $f$ follows. It only needs $f_n$ to converge at a single point, in fact, with the rest supplied by the fundamental theorem.

For $\sum u_n(x)$, uniform convergence means the partial sums converge uniformly. The workhorse test:

Weierstrass M-test. If $|u_n(x)| \le M_n$ for all $x \in S$ and $\sum M_n$ converges, then $\sum u_n$ converges uniformly on $S$.

The bound $M_n$ must not depend on $x$, which is where the uniformity comes from. For example $\sum \frac{\sin(nx)}{n^2}$ converges uniformly on $\mathbb{R}$ because $|\sin(nx)/n^2| \le 1/n^2$ and $\sum 1/n^2$ converges.

This is what makes power series well behaved: inside the radius of convergence they converge uniformly on every closed subinterval, so they may be differentiated and integrated term by term. That is the justification for every term-by-term manipulation of Taylor series you performed in calculus.

Just as with sequences of numbers, you can detect uniform convergence without knowing the limit:

$f_n$ converges uniformly on $S$ if and only if for every $\varepsilon>0$ there is $N$ with $\|f_n - f_m\| < \varepsilon$ for all $n,m \ge N$.

This is completeness once more, now for the space of bounded functions under the supremum norm. Uniform convergence is precisely convergence in that norm, which is why the theory looks so much like the sequence theory from earlier in the section.

Example 1: Show $x^n$ does not converge uniformly on $[0,1]$.

Solution. The pointwise limit $f$ is 0 on $[0,1)$ and 1 at 1.

$$\|f_n - f\| \ge \sup_{0\le x<1}|x^n - 0| = 1$$

since $x^n \to 1$ as $x \to 1^-$. The supremum norm is 1 for every $n$, so it does not go to 0. Not uniform. ∎

Alternatively, cite the theorem: each $f_n$ is continuous, the limit is not, so the convergence cannot be uniform. That argument is shorter and generalizes.

Example 2: Same functions, smaller set.

Show $x^n \rightrightarrows 0$ on $[0,a]$ for any fixed $a<1$.

Solution. On $[0,a]$ the limit is 0 everywhere, and

$$\|f_n - 0\| = \sup_{0\le x\le a}x^n = a^n \to 0$$

since $0\le a<1$. Uniform. ∎

Uniformity depends on the set. The failure at $x=1$ is entirely local, and excluding any neighbourhood of it fixes the problem.

Example 3: Uniform convergence with a non-obvious supremum.

Let $f_n(x)=\frac{nx}{1+n^2x^2}$ on $[0,\infty)$. Is the convergence uniform?

Solution. For fixed $x>0$, the numerator grows like $n$ and the denominator like $n^2$, so $f_n(x)\to0$; and $f_n(0)=0$. The pointwise limit is 0.

Find the supremum by calculus: setting the derivative to zero gives $x=1/n$, where

$$f_n(1/n)=\frac{1}{1+1}=\frac12$$

So $\|f_n\|=\frac12$ for every $n$. Not uniform, even though every $f_n$ and the limit are continuous.

Continuity of the limit does not imply uniformity. The theorem is one-directional: uniform plus continuous gives continuous, not the reverse.

Example 4: The spike, and why integrals need uniformity.

Let $f_n$ be the triangular spike of height $n$ on $[0, 2/n]$, peaking at $1/n$, and 0 afterwards. Compare $\lim\int f_n$ with $\int\lim f_n$.

Solution. For any fixed $x>0$, once $n>2/x$ the spike has passed and $f_n(x)=0$; also $f_n(0)=0$. So $f_n \to 0$ pointwise.

Each integral is the triangle’s area:

$$\int_0^1 f_n = \tfrac12 \cdot \tfrac{2}{n}\cdot n = 1$$

So $\lim_n \int f_n = 1$ while $\int \lim_n f_n = 0$.

The two operations do not commute here, and the reason is that $\|f_n - 0\| = n \to \infty$, the opposite of uniform.

Example 5: Uniform convergence without convergence of derivatives.

Let $f_n(x)=\frac{\sin(nx)}{\sqrt n}$ on $\mathbb{R}$.

Solution. $\|f_n - 0\| = \frac{1}{\sqrt n}\to0$, so $f_n \rightrightarrows 0$.

But $f_n'(x)=\sqrt n\cos(nx)$, and at $x=0$ this is $\sqrt n \to \infty$. The derivatives diverge.

Uniform convergence of the functions gives nothing about the derivatives. The graphs get flat in amplitude while oscillating ever faster, so slopes blow up.

Example 6: Apply the M-test.

Show $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ converges uniformly on $\mathbb{R}$.

Solution. For every real $x$,

$$\left|\frac{\cos(nx)}{n^2}\right| \le \frac{1}{n^2} = M_n$$

and $\sum 1/n^2$ converges as a $p$-series with $p=2>1$. By the M-test the series converges uniformly on all of $\mathbb{R}$. ∎

Consequently the sum is a continuous function of $x$, and it may be integrated term by term over any bounded interval. Note that term-by-term differentiation is not licensed here: the differentiated series has terms $-\sin(nx)/n$, and $\sum 1/n$ diverges.

Example 7: Term-by-term integration.

Use $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ to obtain a series for $-\ln(1-x)$ on $|x|<1$.

Solution. On $[-a,a]$ with $a<1$, the M-test with $M_n=a^n$ gives uniform convergence. So integration from 0 to $x$ may be done term by term:

$$-\ln(1-x)=\int_0^x\frac{dt}{1-t}=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\sum_{k=1}^{\infty}\frac{x^k}{k}$$

∎

Every term-by-term manipulation of a power series is an appeal to uniform convergence on a closed subinterval, which is why the operation is safe strictly inside the radius and needs separate care at the endpoints.

Fourier series. Whether a Fourier series converges to the function, and in what sense, is the founding question of this whole circle of ideas. The Gibbs phenomenon near a jump is non-uniform convergence made visible: the overshoot shrinks in width but not in height.

Numerical analysis. A method that converges pointwise but not uniformly can be badly wrong on part of the domain at every step. Error bounds are quoted in the supremum norm precisely because that is the uniform notion.

Machine learning. Uniform convergence of empirical averages to true expectations over a class of functions is the central object of statistical learning theory; it is what turns “low training error” into “low true error”.

Signal processing. Reconstructing a signal from finitely many terms is a question about uniform approximation, and the ringing artefacts near sharp edges in images and audio are the same non-uniformity as Gibbs.

PDE solutions. Solutions are frequently built as series or as limits of approximations, and uniform convergence is what allows the limit to be differentiated and to satisfy the equation. Without it the limit may exist and solve nothing.

Numerical integration and simulation. Interchanging a limit with an integral is the routine step in Monte Carlo estimates and in convergence proofs for numerical schemes. The spike example is the standing warning about doing it unjustified.