Interchange of Limits: Continuity, Integration, Differentiation

Much of analysis is the question of whether two limiting operations commute. Continuity of a function is a limit ( $lim_{x_{k} \to x} f (x_{k}) = f (x)$ ); so are integration and differentiation; and convergence of a sequence of functions is another. Whenever a limit function inherits a property that each $f_{n}$ has, the inheritance is a claim that two limits may be swapped. That claim is often false.¹

The failure is visible already for numbers. With $x_{n, k} = \frac{n}{n + k}$ ,

0 = n \to \infty lim k \to \infty lim \frac{n}{n + k} \neq = k \to \infty lim n \to \infty lim \frac{n}{n + k} = 1,

because the inner limits are taken in different orders. For sequences of functions the two orders are value of the limit function versus limit of the values, and uniform convergence is the hypothesis that reconciles them.

Continuity of the limit is the claim that this square of limits commutes; without uniform convergence the marked equality can fail.

Continuity of the limit

Pointwise convergence does not preserve continuity. The powers $x^{2 n}$ , a discontinuous pointwise limit, are one witness; a simpler one uses straight lines.

Each tent

f_{n}

is continuous, but the peak at

0

never descends while the ramp narrows, so the pointwise limit jumps at the origin.

Uniform convergence rules this out. It keeps every $f_{n}$ inside a fixed band around $f$ , and continuity of the $f_{n}$ then carries over to $f$ .

This is the three-epsilons argument: two of the thirds come from uniform convergence (at $x_{m}$ and at $x$ ), and the middle third from the continuity of a single well-chosen $f_{k}$ . Uniformity is what lets one index $k$ serve at both $x_{m}$ and $x$ simultaneously.

Integral of the limit

The next swap is between the limit and the integral. Pointwise convergence fails it, and can fail it badly: the limit of the integrals need not equal the integral of the limit, and the limit function need not even be integrable.

The spike keeps unit-half area while its base shrinks toward

0

; at each fixed

x

the height eventually leaves, so the pointwise limit is

0

but every integral stays

\frac{1}{2}

Uniform convergence prevents this. The spike could stay uniformly close to $0$ only by keeping a bounded height, and then its area would have to shrink with its base.

The uniform bound $∥ f - f_{n} ∥_{[a, b]}$ multiplies against the fixed length $b - a$ to control the whole integral at once. This is why the theorem needs a bounded interval: on $[a, b]$ the factor $b - a$ is finite. Over an unbounded interval or improper integral uniform convergence alone is not enough.²

Derivative of the limit

Uniform convergence is enough to swap the limit with an integral, but not with a derivative. Differentiation is the delicate operation, because a small uniform change in a function can hide a large change in its slope.

The curves

sin (n x) / n

are pressed toward

0

uniformly, yet their slopes

cos (n x)

stay order one and never settle.

To address this, require the derivatives themselves to converge uniformly. Then the fundamental theorem of calculus, applied to the derivatives, transfers their uniform convergence back up to the functions.

The anchor value ${f_{n} (c)}$ is not optional. Uniform convergence of the derivatives determines $f$ only up to a constant, and $f_{n} (c) \to ℓ$ supplies that constant. Without it, $f_{n} (x) = x / n$ has $f_{n}^{'} = 1/ n \to 0$ uniformly while $f_{n} \to 0$ merely pointwise.

Summary of hypotheses

Interchange	Sufficient hypothesis	Fails under pointwise
$lim$ with continuity	$f_{n} \to f$ uniformly	tent $\to$ jump at $0$
$lim$ with $\int$	$f_{n} \to f$ uniformly, on $[a, b]$	moving spike, area $\frac{1}{2} \neq = 0$
$lim$ with $\frac{d}{d x}$	$f_{n}^{'} \to g$ uniformly, ${f_{n} (c)}$ converges	$sin (n x) / n \to 0$ , $f_{n}^{'} = cos (n x)$

The pattern is that uniform convergence of the object being differentiated or integrated is what counts. For an integral, that object is $f_{n}$ itself; for a derivative, it is $f_{n}^{'}$ . All three swaps apply at once to power series, which converge uniformly on closed subintervals together with all their derivatives, so they may be integrated and differentiated term by term.

Lebl, §6.2. The iterated-limit example $\frac{n}{n + k}$ shows that even for numbers the order of two limits matters; sequences of functions are the setting where the interchange arises constantly. ↩
Lebl, §6.2, footnote to Theorem 6.2.4. Weaker hypotheses than uniform convergence suffice — pointwise convergence with a uniform bound $∥ f_{n} ∥_{[a, b]} \leq M$ and $f$ integrable — but the proof needs the Lebesgue integral, beyond this course. Over infinite intervals uniform convergence can still fail to pass the limit through the integral. ↩