A limit that forgets to be continuous

Take the functions $x$, $x^2$, $x^3$, and so on, on the interval from 0 to 1. Every one of them is a polynomial — continuous, differentiable as many times as anyone could want, drawn with a single unbroken stroke. Fix any $x$ strictly below 1 and the sequence of values $x, x^2, x^3, \dots$ marches down to zero. At $x = 1$ every one of them is 1.

So the sequence settles, everywhere, on a function that is zero on $[0,1)$ and 1 at the endpoint. That function has a jump in it, and not one member of the sequence did.

Something has gone wrong with a piece of reasoning that felt safe, and locating it exactly is what this essay is about.

The reasoning that fails #

The tempting argument runs: each $f_n$ is continuous; $f_n \to f$; therefore $f$ is continuous. Written out with quantifiers it needs the value of $f$ near a point $a$ to be near $f(a)$, and it tries to get there by going through some $f_n$ — $f(x)$ is near $f_n(x)$, which is near $f_n(a)$, which is near $f(a)$.

Three “nears”, and the argument works only if one $n$ can be chosen that makes the first and third small at the same time. Pointwise convergence promises, for each $x$ separately, that a large enough $n$ works. It does not promise that one $n$ works for every $x$ at once, and near $x = 1$ the required $n$ runs away: to get $x^n$ below a tenth needs $n$ past 22 at $x = 0.9$, past 219 at $x = 0.99$, past 2,301 at $x = 0.999$.

There is no $n$ that does the whole interval. The failure is not in any column of the picture; it is in the demand that one choice serve all the columns.

The number that sees it #

The quantity that separates the two kinds of convergence is the largest vertical gap between a member and the limit, taken over the whole domain:

$$\|f_n - f\|_\infty = \sup_{x} |f_n(x) - f(x)|.$$

For $x^n$ against its limit, that supremum is 1 at every $n$ — approached just to the left of the endpoint, where the member is near 1 and the limit is 0. It never decreases, because there is always a place close enough to 1 for the curve not to have come down yet.

Contrast a sequence that behaves.

A sequence for which the largest gap goes to zero converges uniformly. A sequence for which every column settles converges pointwise. Uniform convergence implies pointwise; the reverse is what $x^n$ refutes.

The supremum here is stated rather than measured, and the reason is a trap worth naming. For $x^n$ the largest gap is reached at no point of the interval at all: at $x = 1$ the member and the limit agree, and just left of 1 the gap is close to 1 without ever getting there. A measurement taken on a grid of a few thousand points therefore reports something a little below 1, and the number it reports falls as $n$ grows — so a family that took its supremum off a grid would have declared a non-uniform sequence uniform. The figures assert the stated value bounds the grid, and, where the value is not attained, that refining the grid closes on it.

The tube, which is the definition #

Uniform convergence has a picture and it is worth drawing rather than describing. Fix a tolerance $\varepsilon$ and draw a band of half-width $\varepsilon$ about the limit function. The sequence converges uniformly if, from some point on, every member lies entirely inside the band — and if that happens for every $\varepsilon$, however small.

Reading the two pictures side by side is the fastest way to hold the distinction. Pointwise convergence says every vertical slice of the picture eventually enters the band. Uniform convergence says the whole curve eventually enters it. The second is a stronger demand, and the first is not enough to conclude anything about the limit’s shape.

What uniform convergence buys #

Once the tube condition holds, the three-near argument goes through, because one $n$ can be chosen for the whole interval. So a uniform limit of continuous functions is continuous — which is the theorem the failed argument was reaching for, with the missing hypothesis supplied.

Three more results have the same shape and the same hypothesis.

Integration passes through. If $f_n \to f$ uniformly on a bounded interval, the integrals converge to the integral. The proof is one line: the difference of the integrals is at most the length of the interval times the largest gap, and the second factor goes to zero.

Differentiation does not. This is the one that surprises. Uniform convergence of $f_n$ says nothing about $f_n'$ — the sequence $\sin(nx)/n$ converges uniformly to zero, with largest gap exactly $1/n$, while its derivatives are $\cos(nx)$, which converge to nothing at all.

Sums of uniformly convergent series can be integrated and rearranged term by term, which is what makes Fourier analysis usable and is where the hypothesis is most often quietly assumed. The corners go first shows the partial sums of a square wave’s Fourier series approaching their target — pointwise everywhere except at the jump, and not uniformly anywhere near it, which is Gibbs’ phenomenon: an overshoot of about 9% of the jump that narrows without ever getting shorter.

Gibbs is the same failure as $x^n$, wearing different clothes. The largest gap does not go to zero; it moves.

A sequence that loses its integral #

Continuity is not the only thing a pointwise limit can drop. Here is a sequence of perfectly ordinary continuous functions, each with area exactly 1 under it, converging pointwise to the zero function.

Fix $x > 0$. Once $n$ exceeds $2/x$ the triangle’s base has shrunk to the left of $x$ entirely, and $f_n(x) = 0$ from then on. At $x = 0$ every $f_n$ is zero already. So the pointwise limit is identically zero, whose integral is 0, while every member integrates to 1.

The largest gap here does not merely fail to shrink; it grows, being $n$. So the sequence is as far from uniform as a sequence can be, and the integral’s disappearance is no surprise once that is measured. What it shows is that the question “does the limit of the integrals equal the integral of the limit?” has an answer that depends on how the convergence happens and not only on whether it happens.

That question is the whole reason Lebesgue’s theory exists, and its answer there is a weaker and much more useful hypothesis than uniformity — domination by a fixed integrable function, which the travelling spike violates because no fixed function dominates a spike of height $n$. Almost none of it left, and still uncountably many builds the measure that theory runs on.

When pointwise is enough after all #

The two notions are not always different, and the conditions under which they coincide are worth knowing because they are exactly the conditions $x^n$ fails.

Dini’s theorem. Suppose the domain is a closed bounded interval, the convergence is monotone — each $f_{n+1}$ is everywhere below $f_n$, say — and the limit function is continuous. Then the convergence is automatically uniform.

Every hypothesis is doing work, and $x^n$ shows which one it is missing. The domain $[0,1]$ is closed and bounded; the convergence is monotone, since $x^{n+1} \le x^n$ throughout; and the limit is not continuous. So Dini’s theorem does not apply, and the conclusion fails. Change the interval to $[0, 0.9]$ and the limit becomes the continuous zero function, Dini applies, and the convergence is uniform — which it is, at rate $0.9^n$.

The proof of Dini’s theorem is a compactness argument, and it is the same shape as every compactness argument: a local choice is made at each point, the local choices are shown to cover the interval, and finitely many of them are extracted and their worst taken. That “finitely many, so take the worst” is where the single $n$ good for the whole interval comes from — it is manufactured, rather than assumed.

The Weierstrass M-test is the other everyday route to uniformity, and it is easier to check than the definition. If a series $\sum g_k$ has $|g_k(x)| \le M_k$ for every $x$, with $\sum M_k$ a convergent series of numbers, then the series of functions converges uniformly. The bound has to hold for every $x$ with one constant per term — the same “one choice for all points” that the definition demands, pushed back to a place where it can be verified without knowing the sum.

Nearly every uniform convergence anybody uses in practice is established by the M-test, and the reason is that it converts a question about a function into a question about a series of numbers. A curve with a corner at every point is built exactly this way: Weierstrass’s nowhere-differentiable function is a series whose $k$-th term is bounded by $a^k$ with $a < 1$, so the M-test gives uniform convergence, which gives continuity — and the differentiability is then destroyed by the other factor in each term, which the M-test never sees.

That is worth stating plainly, because it is the sharpest illustration of the essay’s point. Uniform convergence is enough to inherit continuity and is not enough to inherit anything about slopes, and the standard example of a continuous nowhere-differentiable function is precisely a construction that exploits the gap.

Where uniformity is more than can be asked #

Insisting on uniform convergence everywhere is often too strong to be useful, and analysis has a standard weakening: uniform convergence on every closed bounded subinterval.

$x^n$ has it. On $[0, 0.9]$ the largest gap is $0.9^n$, which goes to zero perfectly well; the trouble is confined to any neighbourhood of the endpoint. So the limit is continuous on $[0,1)$ — which it is, being zero there — and the jump lives exactly where the uniformity fails.

This is the usual state of affairs for power series, which converge uniformly on every closed disc strictly inside their radius of convergence and generally not on the boundary. One point’s worth of information builds Taylor series without needing the distinction; it becomes unavoidable the moment anything is integrated term by term out to the edge.

Where it came from #

Cauchy published a proof in 1821 that a convergent series of continuous functions has a continuous sum. It is wrong, and Fourier series were already supplying counterexamples — a square wave is a convergent series of continuous functions with a discontinuous sum, and it had been in print since 1807.

Abel pointed out the difficulty in 1826, in a footnote, calling it an exception to the theorem. Nobody could say precisely what distinguished the exceptions for another twenty years. Weierstrass and Seidel, independently and around 1847, isolated the condition, and Weierstrass’s teaching made it standard.

The delay is instructive rather than embarrassing. The distinction is invisible without a language for the order of the quantifiers — for every $x$ there is an $n$, against there is an $n$ for every $x$ — and that language did not exist when Cauchy wrote. The theorem was not so much wrong as unstatable, and getting it right required inventing the notation the modern statement is written in.

What the pictures cannot show #

Every figure here draws finitely many members of an infinite sequence, and picks the members to make the trend legible. Choosing $n = 1, 2, 4, 12, 40$ makes the crowding near the endpoint obvious; choosing $n = 1, 2, 3, 4, 5$ would not have. Nothing in the drawing indicates which choice was made or that the choice matters.

The tube figures report the first member that fits and assert that every later one fits too. The second half of that claim is checked on the stated sequences by their closed forms, not by drawing them — an infinite family cannot be drawn, and the honest position is that the picture illustrates one instance of a statement whose content is in the “for all beyond”.

And the supremum for $x^n$ is not attained. The figures say so and cannot draw it: a value approached at no point is exactly the kind of thing a picture has no way of marking, which is why the family computes it from a stated formula and uses the grid only to check that the formula bounds it.

The ladder from here #

Below: a curve with a corner at every point, which is built as a uniform limit and inherits continuity from it, and the corners go first, where the same failure appears at a jump. Sideways: rearranged into any answer, where a different kind of convergence turns out not to be strong enough either, and almost none of it left, and still uncountably many, which supplies the measure the better convergence theorems are stated over. Above: equicontinuity and the Arzelà–Ascoli theorem, dominated convergence, and the completeness of the space of continuous functions in the supremum distance.

What is worth carrying away #

Two quantifiers, and the order they come in. For each $x$ there is an $N$ is a statement about points; there is an $N$ for each $x$ is a statement about the whole function at once, and the second is the one that lets a property survive a limit.

The supremum distance is what makes the second statement a number. Once functions are points in a space, with distance measured by the largest gap, uniform convergence is ordinary convergence in that space — and the theorem that a uniform limit of continuous functions is continuous becomes the statement that the continuous functions form a closed subset. The difficulty that took Cauchy, Abel and Weierstrass twenty-five years to name is, in the right language, the difference between two metrics on the same set.