A sequence of functions f_n can "converge to f"
in two genuinely different senses, and the gap between them is where a great deal of analysis lives.
The weaker notion looks at one input at a time; the stronger one controls the whole graph at once.
Pointwise convergence. For each fixed x, the
numbers f_n(x) converge to f(x):
\forall x \;\; \forall \varepsilon > 0 \;\; \exists N \;(\text{depending on } x) : \; n \ge N \Rightarrow |f_n(x) - f(x)| < \varepsilon.
Uniform convergence. One threshold N works for
every x simultaneously. Equivalently, the worst gap over the whole
domain goes to zero:
\|f_n - f\|_\infty = \sup_x |f_n(x) - f(x)| \;\longrightarrow\; 0.
The difference is exactly the order of the quantifiers: in pointwise convergence
N may depend on x; in uniform convergence it may
not. That single swap is what lets limits be exchanged with continuity and integration.
Centrepiece: f_n(x) = x^n on [0,1]
This is the canonical example where pointwise convergence holds but uniform convergence fails — and
the failure produces a discontinuous limit out of perfectly continuous pieces.
Step 1 — find the pointwise limit, point by point. Split the domain at
x = 1:
0 \le x < 1 \;\Rightarrow\; x^n \to 0, \qquad x = 1 \;\Rightarrow\; 1^n = 1 \to 1.
Step 2 — assemble the limit function. Collecting both cases,
f(x) = \lim_{n\to\infty} x^n = \begin{cases} 0, & 0 \le x < 1, \\ 1, & x = 1. \end{cases}
Each f_n(x) = x^n is continuous, yet the limit
f has a jump at 1 — it is
discontinuous.
Step 3 — measure the worst gap. Is the convergence uniform? Compute the sup-distance
to the limit on [0, 1), where f = 0:
\sup_{0 \le x < 1} |x^n - 0| = \sup_{0 \le x < 1} x^n = 1.
Step 4 — read the verdict. The supremum is 1 for
every n (push x close enough to
1 and x^n is still near
1). So
\|f_n - f\|_\infty = 1 \;\not\to\; 0,
and the convergence is not uniform. The "bad point" simply slides toward
1 as n grows — there is always some
x where f_n hasn't yet settled. Watch this in the
figure below: the curves pin to the floor on the left but rear up to height
1 right before x = 1, no matter how large
n is.
Why it matters. Pointwise convergence let continuity leak away. The
uniform limit theorem says that under uniform convergence this cannot happen:
a uniform limit of continuous functions is continuous. Contrapositively, the moment the limit is
discontinuous — as here — the convergence cannot have been uniform.
Suppose f_n \to f uniformly on a set
S (resp. on [a, b]). Then:
-
Continuity passes to the limit: if each f_n is
continuous, so is f. (The "\tfrac{\varepsilon}{3}
argument": close in n twice and in x once.)
-
Limits and integrals swap: on [a, b],
\lim_{n\to\infty} \int_a^b f_n = \int_a^b \lim_{n\to\infty} f_n = \int_a^b f.
The error is bounded by (b - a)\,\|f_n - f\|_\infty \to 0.
-
Pointwise is not enough: for f_n(x) = x^n on
[0, 1], the pointwise limit is discontinuous, so the convergence is
not uniform — and continuity is lost.
For an infinite series of functions \sum_n g_n(x), checking
uniform convergence by hand is painful. The Weierstrass M-test reduces it to a
single numerical series. If you can dominate each term by a constant,
|g_n(x)| \le M_n \text{ for all } x \in S, \qquad \text{and} \qquad \sum_{n=1}^{\infty} M_n < \infty,
then \sum_n g_n converges uniformly (and absolutely) on
S. The idea: the tail of the function series is squeezed by the tail of the
convergent constant series \sum M_n, which is small independent of
x — exactly the uniform control we need. This is the standard route to
showing a
power series
converges uniformly on any closed disc strictly inside its radius of convergence, so it may be
differentiated and integrated term by term.