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One index for the whole domain, measured by the supremum of the gap, with the uniform Cauchy criterion that produces the limit.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state uniform convergence with its quantifiers in the right order, decide it by computing the supremum distance as a number depending on the index alone, and deny it by producing a point that moves with the index. You will also be able to use the uniform Cauchy criterion, which establishes uniform convergence without the limit function being known in advance.
Pointwise convergence, and the moving bump that defeats it. Also uniform continuity from unit 3, which moved exactly the same quantifier and for exactly the same reason. If that lesson made sense, this one is the same idea applied to an index instead of a delta.
The supremum distance between two functions is the largest gap between their values over the domain. Convergence is uniform when that distance tends to zero, and locally uniform when it does so on every closed bounded piece of the domain without doing so on the whole.
Pointwise: for every $x$ and every $\varepsilon > 0$ there is $N$ with $|f_n(x) - f(x)| < \varepsilon$ for $n \ge N$. The index may depend on $x$.
Uniform: for every $\varepsilon > 0$ there is $N$ such that $|f_n(x) - f(x)| < \varepsilon$ for every $n \ge N$ and every $x$. The index may not depend on $x$.
As quantifiers: $\forall x\,\forall \varepsilon\,\exists N$ against $\forall \varepsilon\,\exists N\,\forall x$. Moving the existential left strengthens the statement, as it did for uniform continuity.
The supremum distance. Define $$\|f_n - f\|_\infty = \sup_{x \in S} |f_n(x) - f(x)|.$$ Then $f_n \to f$ uniformly on $S$ exactly when $\|f_n - f\|_\infty \to 0$ as a sequence of numbers. That turns a statement with three quantifiers into a single limit, and it is how uniform convergence is checked in practice.
The uniform Cauchy criterion. If $\sup_x|f_m(x) - f_n(x)| \to 0$ as $m, n$ grow, then $(f_n)$ converges uniformly to some $f$. Like the Cauchy criterion for numbers, it produces the limit rather than testing against one, and it is what lets a function be constructed as a uniform limit.
Another way: picture
Draw a band of half-width $\varepsilon$ about the graph of the limit — a tube following the curve. Uniform convergence says that from some index on, the whole graph of every later term lies inside the tube. Pointwise says only that each vertical line is eventually inside it, and the moving bump is a graph that pokes out of the tube in a place that keeps shifting.
Another way: steps
To decide uniform convergence on a set $S$:
If step 3 leaves $x$ in the answer, step 3 is not finished.
Uniform convergence is a property of a sequence on a set, and shrinking the set can create it.
$x^n$ on $[0,1]$: at $x = 2^{-1/n}$ the value is $1/2$, so $\|f_n - f\|_\infty \ge 1/2$ for every $n$. Not uniform.
$x^n$ on $[0, r]$ for any $r < 1$: the gap is largest at $r$, so $\|f_n - 0\|_\infty = r^n \to 0$. Uniform.
So the same sequence is uniformly convergent on every closed subinterval of $[0,1)$ and not on $[0,1)$ itself. That situation — uniform on every compact piece, not on the whole — is so common that it has a name, locally uniform convergence, and it is exactly what a power series does inside its radius.
The practical consequence: a theorem needing uniform convergence can usually still be applied, by applying it on a compact piece at a time and noting that continuity, say, is a local property.
| Sequence, on the set | Gap largest at | Supremum distance | Uniform |
|---|---|---|---|
| $x/n$ on $[0,c]$ | $x = c$ | $c/n$ | yes |
| $x/n$ on $[0,\infty)$ | nowhere | infinite | no |
| $x^n$ on $[0,1)$ | near $1$ | $1$ | no |
| $x^n$ on $[0,r]$, $r<1$ | $x = r$ | $r^n$ | yes |
| $\dfrac{nx}{1+n^2x^2}$ on $[0,1]$ | $x = 1/n$ | $1/2$ | no |
| $\dfrac{\sin(nx)}{\sqrt n}$ on $\mathbb{R}$ | anywhere | $1/\sqrt n$ | yes |
The last row is the one to keep: it converges uniformly to $0$, and its derivatives $\sqrt n \cos(nx)$ converge to nothing at all. Uniform convergence of the functions says nothing whatever about the derivatives, which is the subject of the next lesson.
An index per point. That is pointwise convergence written out. Uniformity is one index for the whole domain, and the test is whether the supremum of the gap tends to zero.
Forgetting to name the set. A sequence is uniformly convergent on a set. Saying a sequence converges uniformly without naming one is an incomplete statement, and often the shrinking of the set is what makes it true.
Leaving the point in the supremum. If the answer to step 3 still mentions $x$, the maximisation has not been done, and nothing can be concluded.
Expecting it to control the derivatives. It does not, and $\sin(nx)/\sqrt n$ is the standing counterexample. Uniform convergence of the functions is a hypothesis about the functions only.
Let $f_n(x) = \dfrac{x}{1 + nx^2}$ on $\mathbb{R}$. Pointwise limit: $0$ at every $x$.
Find the limit first.
Maximise the gap: $f_n'(x) = 0$ at $x = 1/\sqrt n$, where the value is $\dfrac{1}{2\sqrt n}$.
Maximise over the whole domain.
So $\|f_n\|_\infty = \dfrac{1}{2\sqrt n} \to 0$: uniform on all of $\mathbb{R}$, unbounded domain and all.
A number in $n$ alone, tending to zero.
Let $f_n(x) = x^n$ on $[0,1)$, with pointwise limit $0$.
The limit is as simple as it could be.
For each $n$ take $x_n = 2^{-1/n}$, which lies in the interval; then $f_n(x_n) = 1/2$.
One point per index, chosen after the index.
So $\|f_n - 0\|_\infty \ge 1/2$ for every $n$, and the convergence is not uniform.
The supremum does not move.
Pointwise: at $x = 0$ every term is $0$; for fixed $x > 0$ the denominator grows like $n^2x^2$, so the limit is $0$.
Find the pointwise limit, by cases.
But at $x = 1/n$ the second term of the denominator vanishes, and the value is exactly $1$.
Test a point that moves with the index.
So the supremum distance is $1$ at every index and the convergence is not uniform — the same moving bump as before, wearing a different formula. Once the shape is recognised, the point to test is the one where the special behaviour sits.
Put the steps of deciding whether $f_n(x) = \dfrac{x}{n}$ converges uniformly on the interval from $0$ to $2$ into order.
Number the steps in order (write the number in the box):
For $f_n(x) = \dfrac{x}{n}$ on the interval from $0$ to $8$, with limit $0$, what is the supremum distance at index $7$?
Answer:
For $f_n(x) = \dfrac{x}{n}$ on the interval from $0$ to $6$, with limit $0$, give the supremum distance at the indices $1$, $2$ and $4$.
| supremum distance | |
|---|---|
| at index one | |
| at index two | |
| at index four |
Each sequence below converges pointwise on the interval stated. Match it to whether the convergence is uniform there, and why.
| uniform: the supremum distance tends to zero | not uniform: the supremum distance is infinite at every index | not uniform: the supremum distance stays at one | |
|---|---|---|---|
| $x/n$ on the interval from $0$ to $4$ | |||
| $x/n$ on the whole half-line | |||
| $x^n$ on the interval from $0$ to $1$ | |||
| $x^n$ on the interval from $0$ to $1/2$ |
Here is an argument that $f_n(x) = x^n$ converges uniformly on the interval from $0$ to $1$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
A sequence of functions satisfies: for every $\varepsilon > 0$ there is $N$ with $\sup_x |f_m(x) - f_n(x)| < \varepsilon$ for all $m, n \ge N$. What follows?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of deciding whether $f_n(x) = \dfrac{x}{n}$ converges uniformly on the interval from $0$ to $7$ into order.
Number the steps in order (write the number in the box):
You can compute a supremum distance, decide uniform convergence from it, and show a sequence not uniformly convergent by choosing a point per index. Say in your own words why the domain is part of the statement. Next: what this stronger hypothesis actually buys.
10. Your turn: is $f_n(x) = \dfrac{x^2}{x^2 + (1 - nx)^2}$ uniformly convergent on $[0,1]$?, step 3