Sequence of Functions

Types of Convergence

Pointwise Convergence

\forall \epsilon \gt 0\; \forall x\in [a,b]\; \exists N \in \mathbb{Z^+}\; \forall n > N\;;\; \big|f_n(x)-f(x)\big| \lt \epsilon

Here $N$ depends on $\epsilon, x$ .

Examples:

$x^n$ on $[0,1]$

Uniformly Convergence

\forall \epsilon \gt 0\; \exists N \in \mathbb{Z^+}\; \forall x\in [a,b]\; \forall n > N\;;\; \big|f_n(x)-f(x)\big| \lt \epsilon

Here $N$ depends on $\epsilon$ only. Implies pointwise convergence.

Examples:

$\frac{x^2}{n}$ on $[0,1]$

Uniform Convergence Tests

Supremum Test

Suppose $u_n(x)$ is sequence of bounded functions. $u_n(x)$ converges to $u(x)$ uniformly iff:

\lim_{n\to\infty} \sup_x\, \lvert u_n(x) - u(x) \rvert = 0

Properties of Uniform Convergence

Continuity

If $u_n(x)$ is continuous and converging to $u(x)$ , then $u(x)$ is also continuous.

Integrability

Explained in Converging Functions | Riemann Integration.

Differentiability

Uniform convergence-differentiation pair doesn’t go as smooth like integration was.

Suppose $u_n(x)$ is a sequence of differentiable functions, and they uniformly converges to $u(x)$ . Differentiability of $u(x)$ is not guaranteed. An example is:

u_n(x) = \sqrt{x^2 + \frac{1}{n}}

$u_n(x)$ is differentiable but $u(x)$ is only differentiable on $\mathbb{R}-\set{0}$ .

Theorem

If (all conditions must be met):

$u_n(x)$ is differentiable on $[a,b]$
$u_n(x_0)$ converges (pointwise) for some $x_0 \in [a,b]$
$u_n'(x)$ converges to $f(x)$ uniformly on $[a,b]$

Then:

$u_n(x)$ converges to $u(x)$ uniformly on $[a,b]$
$u(x)$ is differentiable on $[a,b]$
$u'(x)=f(x)$ OR in other words $u_n'(x)$ converges to $u'(x)$ uniformly

Written by

Sahithyan K.

October 19, 2026 2 min read

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