The sequence of [functions](/page/Function) $f_n(x) = x^n$ on $[0, 1]$ converges pointwise: for each fixed $x \in [0, 1)$, $x^n \to 0$, and $f_n(1) = 1 \to 1$. Every $f_n$ is continuous, yet the pointwise limit \begin{align*} f(x) = \begin{cases} 0 & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x = 1 \end{cases} \end{align*} is discontinuous at $x = 1$. The limit of continuous functions is not continuous — pointwise convergence has destroyed a fundamental property. What went wrong? Near $x = 1$, the functions $f_n$ converge *slowly*: for $x = 1 - 1/n$, $f_n(x) = (1 - 1/n)^n \to 1/e \neq 0$, so $f_n(x)$ is still far from $f(x) = 0$ even for large $n$. The convergence is fast at $x = 0$ and arbitrarily slow near $x = 1$. This non-uniformity — different points requiring different values of $N$ to achieve the same accuracy — is the source of the pathology. **Uniform convergence** eliminates this problem by demanding that a *single* $N$ works for *all* points simultaneously. This stronger mode of convergence preserves [continuity](/page/Continuity%20(Real%20Analysis)), commutes with [integration](/page/Riemann%20Integral), and provides the convergence mode that makes [power series](/page/Power%20Series) well-behaved inside their interval of convergence. It is the natural notion of convergence for the [supremum metric](/page/Continuity%20(Metric%20Spaces)) on function spaces: $f_n \to f$ uniformly iff $\|f_n - f\|_\infty \to 0$. ## Definition [definition:Pointwise Convergence] A sequence of functions $f_n: E \to \mathbb{R}$ **converges pointwise** to $f: E \to \mathbb{R}$ if for every $x \in E$: \begin{align*} \lim_{n \to \infty} f_n(x) = f(x). \end{align*} That is, for every $x \in E$ and every $\varepsilon > 0$, there exists $N = N(\varepsilon, x)$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $n \geq N$. [/definition] [definition:Uniform Convergence] A sequence of functions $f_n: E \to \mathbb{R}$ **converges uniformly** to $f: E \to \mathbb{R}$ on $E$ if for every $\varepsilon > 0$, there exists $N = N(\varepsilon)$ (depending only on $\varepsilon$, *not* on $x$) such that \begin{align*} |f_n(x) - f(x)| < \varepsilon \quad \text{for all } x \in E \text{ and all } n \geq N. \end{align*} Equivalently, $\sup_{x \in E} |f_n(x) - f(x)| \to 0$ as $n \to \infty$. [/definition] The crucial difference is a quantifier swap. Pointwise: $\forall x \; \forall \varepsilon \; \exists N \; \ldots$ Uniform: $\forall \varepsilon \; \exists N \; \forall x \; \ldots$ Moving $\forall x$ outside $\exists N$ forbids $N$ from depending on $x$ — the same $N$ must work everywhere. The equivalent formulation via the [supremum](/page/Supremum%20and%20Infimum) norm is often the cleanest: $f_n \to f$ uniformly iff $\|f_n - f\|_\infty = \sup_{x \in E} |f_n(x) - f(x)| \to 0$. This is convergence in the metric of the function space $(B(E), \|\cdot\|_\infty)$ — uniform convergence is the "natural" convergence for the sup-norm, just as ordinary convergence is natural for the absolute value. [example:Pointwise But Not Uniform] $f_n(x) = x^n$ on $[0, 1]$ converges pointwise to the discontinuous [limit](/page/Limit) above, but $\|f_n - f\|_\infty = \sup_{x \in [0,1)} x^n = 1$ for every $n$ (the supremum approaches $1$ as $x \to 1^-$, though is never attained). Since $\|f_n - f\|_\infty \not\to 0$, the convergence is not uniform. By contrast, $f_n(x) = x^n$ on $[0, 1/2]$ converges uniformly to $0$: $\|f_n\|_\infty = (1/2)^n \to 0$. [/example] [example:Uniform Convergence On Compact Subsets] $f_n(x) = x/n$ on $\mathbb{R}$ converges pointwise to $0$, but $\|f_n\|_\infty = \sup_{x \in \mathbb{R}} |x/n| = \infty$ for every $n$. The convergence is not uniform on $\mathbb{R}$. However, on any bounded interval $[-M, M]$, $\|f_n\|_\infty = M/n \to 0$, so the convergence is uniform on compact subsets. This mode — uniform on every compact subset but not globally — is called **locally uniform convergence** and is the standard notion in complex analysis. [/example] ## The [Uniform Limit Theorem](/theorems/258) The most important property of uniform convergence: it preserves [continuity](/page/Continuity%20(Real%20Analysis)). If $f_n: E \to \mathbb{R}$ are continuous and $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$. The argument is the **$\varepsilon/3$ trick**: to show $|f(x) - f(a)| < \varepsilon$, decompose via the triangle inequality as $|f(x) - f_N(x)| + |f_N(x) - f_N(a)| + |f_N(a) - f(a)|$, then control the first and third terms by uniform convergence (choosing one $N$ that works for *all* points) and the middle term by continuity of $f_N$. The crucial point is that uniformity lets us choose $N$ before knowing $x$ — if the convergence were merely pointwise, the $N$ for the first term would depend on $x$, and we could not choose $\delta$ for the middle term without knowing which $x$ we are dealing with. This is the precise mechanism by which the quantifier swap (moving $\forall x$ outside $\exists N$) prevents the pathology of $f_n(x) = x^n$ on $[0, 1]$. The converse fails: a pointwise limit of continuous functions can be continuous without the convergence being uniform. The functions $f_n(x) = x^n(1 - x^n)$ on $[0, 1]$ converge pointwise to $0$ (which is continuous), but the convergence is not uniform ($\max f_n = 1/4$ for all $n$). ## Interchange of Limit and Integral Uniform convergence also commutes with [Riemann integration](/page/Riemann%20Integral): If $f_n: [a, b] \to \mathbb{R}$ are Riemann [integrable](/page/Integral) and $f_n \to f$ uniformly on $[a, b]$, then $f$ is Riemann integrable and \begin{align*} \lim_{n \to \infty} \int_a^b f_n(x) \, dx = \int_a^b f(x) \, dx. \end{align*} The key estimate is $|\int_a^b f_n - \int_a^b f| \leq (b - a) \|f_n - f\|_\infty$, which tends to $0$ by uniform convergence. Without uniform convergence, the interchange can fail: the functions $f_n = n \cdot \mathbf{1}_{[0, 1/n]}$ on $[0, 1]$ satisfy $f_n \to 0$ pointwise, $\int_0^1 f_n = 1$ for all $n$, but $\int_0^1 0 = 0 \neq 1$. The "mass" concentrates and escapes in the limit. ## Interchange of Limit and Derivative [Differentiation](/page/Derivative) is harder: uniform convergence of $f_n$ alone does *not* guarantee that $f_n' \to f'$. The correct hypothesis requires uniform convergence of the *derivatives*. [quotetheorem:260] The key point is asymmetric: we need uniform convergence of $f_n'$ (not just of $f_n$), plus convergence of $f_n$ at a single point. This is because differentiation amplifies oscillation — a sequence that converges uniformly can have wildly divergent derivatives ($f_n(x) = \sin(nx)/\sqrt{n}$ converges uniformly to $0$, but $f_n'(x) = \sqrt{n} \cos(nx)$ diverges). ## The Weierstrass $M$-Test For *series* of functions $\sum_{n=0}^{\infty} g_n(x)$, uniform convergence of the partial sums $S_N(x) = \sum_{n=0}^{N} g_n(x)$ is guaranteed by a simple domination condition. **Weierstrass $M$-Test.** Let $g_n: E \to \mathbb{R}$ be functions and let $M_n \geq 0$ be constants with $|g_n(x)| \leq M_n$ for all $x \in E$ and all $n$. If $\sum_{n=0}^{\infty} M_n$ converges, then $\sum_{n=0}^{\infty} g_n(x)$ converges absolutely and uniformly on $E$. The idea: the tail $|\sum_{k=n+1}^{m} g_k(x)| \leq \sum_{k=n+1}^{m} M_k$ is bounded by the tail of a convergent numerical [series](/page/Series), which tends to $0$ independently of $x$. The partial sums are therefore uniformly [Cauchy](/page/Cauchy%20Sequence), hence uniformly convergent. The $M$-test is the standard tool for establishing uniform convergence of [power series](/page/Power%20Series) on compact subsets: for $|x - c| \leq r < R$, $|a_n(x - c)^n| \leq |a_n| r^n = M_n$, and $\sum |a_n| r^n$ converges (since $r < R$). This is why power series are continuous, differentiable, and integrable term by term inside their interval of convergence. [example:Weierstrass Function Via The $M$-Test] The Weierstrass function $W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ with $0 < a < 1$ converges uniformly on $\mathbb{R}$: $|a^n \cos(b^n \pi x)| \leq a^n = M_n$, and $\sum a^n = 1/(1-a) < \infty$. By the $M$-test, the series converges uniformly. Since each partial sum is continuous (finite sum of continuous functions), the uniform limit theorem gives continuity of $W$. This is how we know $W$ is continuous — despite being nowhere differentiable. [/example] ## Dini's Theorem A remarkable converse to the uniform limit theorem holds under monotonicity: if the convergence is monotone and the limit is continuous, then the convergence is automatically uniform. **Dini's Theorem.** Let $f_n: [a, b] \to \mathbb{R}$ be continuous with $f_1(x) \geq f_2(x) \geq \cdots$ for all $x$ (monotone decreasing). If $f_n \to f$ pointwise on $[a, b]$ and $f$ is continuous, then $f_n \to f$ uniformly. The key insight is that compactness upgrades pointwise convergence to uniform when monotonicity prevents oscillation. The [sets](/page/Set) $U_n = \{x : |f_n(x) - f(x)| < \varepsilon\}$ are open (by continuity), nested (by monotonicity), and cover $[a, b]$ (by pointwise convergence). Compactness of $[a, b]$ extracts a finite subcover; nesting then implies $U_N = [a, b]$ for some single $N$, giving uniform convergence. Dini's theorem explains why monotone approximation schemes (truncations of monotone series, increasing [sequences](/page/Sequence) of step functions approximating a continuous function from below) automatically produce uniform convergence on compact intervals — no separate verification is needed. ## Uniform Convergence and Function Spaces Uniform convergence is convergence in the supremum metric $d_\infty(f, g) = \|f - g\|_\infty$ on the space of bounded functions. The [Cauchy Sequence](/page/Cauchy%20Sequence) page established that the space $C([a, b])$ of continuous functions on $[a, b]$, equipped with $\|\cdot\|_\infty$, is a complete [metric space](/page/Metric%20Space) — a [Banach space](/page/Banach%20Space). The uniform limit theorem is equivalent to the statement that $C([a, b])$ is *closed* in the larger space $B([a, b])$ of all bounded functions: the uniform limit of continuous functions is continuous, so the limit stays in $C([a, b])$. This perspective unifies several results: The **[Weierstrass Approximation Theorem](/theorems/480)** states that every continuous function on $[a, b]$ is the uniform limit of polynomials. In function-space language: the polynomials are *dense* in $(C([a, b]), \|\cdot\|_\infty)$. Since $C([a, b])$ is complete and the polynomials are dense, $C([a, b])$ is the *completion* of the polynomial space under the sup-norm. The **Arzelà-Ascoli Theorem** characterises which subsets of $C([a, b])$ are compact (= sequentially compact) in the sup-norm: a set $\mathcal{F} \subseteq C([a, b])$ is compact iff it is closed, uniformly bounded ($\sup_{f \in \mathcal{F}} \|f\|_\infty < \infty$), and equicontinuous (for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - f(y)| < \varepsilon$ for all $f \in \mathcal{F}$ whenever $|x - y| < \delta$). Equicontinuity is the family-level analogue of uniform continuity — it requires the *same* $\delta$ to work for all functions in $\mathcal{F}$ simultaneously.