Pointwise vs uniform convergence, M-test, Dini

In one sentence

Pointwise convergence is a promise made separately to each point; uniform convergence is a single promise made to all of them at once — and the exam lives entirely in the gap between.

Why the exam asks it

Because it is the cleanest place to test whether you reason with quantifiers or with pictures. Almost every candidate can state both definitions. Far fewer can look at $f_n(x) = x^n$ and say which one holds, and Part C is built to separate those two groups. The topic recurs because a single well-chosen sequence exposes the difference in four options.

The idea

Fix a sequence of functions $f_n$ on a set $E$, and a function $f$.

Pointwise convergence says: pick any $x$ you like, and the numbers $f_n(x)$ converge to $f(x)$. Formally, for each $x$ and each $\varepsilon > 0$ there is an $N$ with $|f_n(x) - f(x)| < \varepsilon$ for all $n \ge N$.

Read the order of the quantifiers carefully, because everything turns on it. The $N$ is allowed to depend on $x$. Choose an awkward point and you may need to go much further along the sequence before the values settle.

Uniform convergence removes that freedom. It demands one $N$ that works for every $x$ at once: for each $\varepsilon > 0$ there is an $N$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $n \ge N$ and all $x \in E$.

Equivalently, and far more usefully in an exam:

That single line is the tool. Uniform convergence is a statement about the worst point, and the sup is how you find it. When a question asks whether convergence is uniform, computing that supremum is almost always faster than any argument about $N$.

The example the whole topic is built on

Take $f_n(x) = x^n$ on $[0, 1]$.

At any fixed $x < 1$, the powers shrink to zero. At $x = 1$, every term is $1$. So the pointwise limit exists and is

Each $f_n$ is a polynomial — as smooth as anything gets. The limit has a jump. That alone should tell you something has been lost.

Now the sup. On $[0,1)$, $\sup_x |x^n - 0| = 1$ for every $n$, because $x^n$ gets arbitrarily close to $1$ as $x$ approaches $1$. The supremum does not shrink. It is not merely large — it is constant, and equal to $1$, no matter how far along the sequence you go.

So the convergence is pointwise and not uniform, and the failure is not marginal.

Where intuition breaks

This is the part worth rereading, because each of these is a distractor someone wrote on purpose.

"The curves are getting flatter, so they must be converging uniformly." They are getting flatter on any fixed sub-interval $[0, a]$ with $a < 1$ — and on such an interval the convergence genuinely is uniform, since $\sup = a^n \to 0$. The trap is that the behaviour on $[0,1)$ is not the union of the behaviours on the pieces. Uniformity is not a local property you can glue together.

"The limit is discontinuous, so the convergence can't be uniform." Correct conclusion, invalid reasoning — and the exam knows the difference. The theorem runs one way: a uniform limit of continuous functions is continuous. So a discontinuous limit rules uniform convergence out. But a continuous limit proves nothing. This is a converse assumed, and there are sequences converging pointwise to a perfectly continuous limit with the convergence still not uniform.

"It fails at $x = 1$, so remove that point and it will be fine." Tempting, and wrong. On $[0,1)$ the limit is now the constant $0$, which is continuous — yet the supremum is still $1$. Deleting the bad point removed the discontinuity without removing the failure. What breaks uniformity is that the approach to $1$ is never uniform, not the value at $1$.

"$\sum f_n$ converges at every point, so the Weierstrass M-test applies." The M-test needs $|f_n(x)| \le M_n$ for all $x$, with $\sum M_n$ convergent — a bound uniform in $x$. Pointwise convergence of the series supplies no such bound. This is a hypothesis dropped, the single most common way to lose a Part C mark on this topic.

"Dini's theorem makes monotone pointwise convergence uniform." Only with every hypothesis present: the domain compact, the limit continuous, each $f_n$ continuous, and the convergence monotone in $n$. Drop compactness and it fails. Drop continuity of the limit and it fails. $x^n$ on $[0,1]$ is monotone and the domain is compact — and the convergence is not uniform, because the limit is not continuous. The exam sets this up precisely because three of the four conditions look satisfied at a glance.

The exam's angle

The question almost never says "is this uniform?". It gives you a sequence and four statements, and asks which are true. Expect at least one option to be true pointwise and false uniformly.

The reliable procedure, in order:

1. Find the pointwise limit. Fix $x$, let $n \to \infty$. Watch the endpoints — that is where the limit usually changes character.
2. If the limit is discontinuous while the $f_n$ are continuous, uniform convergence is dead immediately. Say so and move on.
3. Otherwise compute $\sup_x |f_n(x) - f(x)|$. Differentiate if you must; often the supremum sits at an endpoint or at a maximum that drifts as $n$ grows.
4. Check the domain the option specifies. The same sequence is frequently uniform on $[0, a]$ and not on $[0, 1)$, and a well-set paper will offer you both.

Step 4 is where marks are actually lost. Two options differing only in their interval are not a typographical accident.

The night before

• Uniform convergence means $\sup_x |f_n - f| \to 0$. Compute the sup; do not argue about $N$.
• A uniform limit of continuous functions is continuous. The converse is false.
• Uniformity depends on the domain. Shrink the interval and it can appear; the same sequence is often uniform on $[0,a]$ and not on $[0,1)$.
• The M-test needs a bound independent of $x$. Pointwise information never supplies one.
• Dini needs all four: compact domain, continuous $f_n$, continuous limit, monotone in $n$.
• The one example to carry in: $x^n$ on $[0,1]$. Pointwise yes, uniform no, sup constantly $1$.