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Counterexample to a strengthening of the uniform limit theorem, in which pointwise convergence, rather than uniform convergence, is assumed. The continuous green functions converge to the non-continuous red function. This can happen only if convergence is not uniform.

In mathematics, the uniform limit theorem states that the uniform limit of any sequence of continuous functions is continuous.

Statement

[edit]

More precisely, let X be a topological space, let Y be a metric space, and let ƒn : X → Y be a sequence of functions converging uniformly to a function ƒ : X → Y. According to the uniform limit theorem, if each of the functions ƒn is continuous, then the limit ƒ must be continuous as well.

This theorem does not hold if uniform convergence is replaced by pointwise convergence. For example, let ƒn : [0, 1] → R be the sequence of functions ƒn(x) = xn. Then each function ƒn is continuous, but the sequence converges pointwise to the discontinuous function ƒ that is zero on [0, 1) but has ƒ(1) = 1. Another example is shown in the adjacent image.

In terms of function spaces, the uniform limit theorem says that the space C(X, Y) of all continuous functions from a topological space X to a metric space Y is a closed subset of YX under the uniform metric. In the case where Y is complete, it follows that C(X, Y) is itself a complete metric space. In particular, if Y is a Banach space, then C(X, Y) is itself a Banach space under the uniform norm.

The uniform limit theorem also holds if continuity is replaced by uniform continuity. That is, if X and Y are metric spaces and ƒn : X → Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous.

Proof

[edit]

In order to prove the continuity of f, we have to show that for every ε > 0, there exists a neighbourhood U of any point x of X such that:

Consider an arbitrary ε > 0. Since the sequence of functions (fn) converges uniformly to f by hypothesis, there exists a natural number N such that:

Moreover, since fN is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:

In the final step, we apply the triangle inequality in the following way:


Hence, we have shown that the first inequality in the proof holds, so by definition f is continuous everywhere on X.

Uniform limit theorem in complex analysis

[edit]

There are also variants of the uniform limit theorem that are used in complex analysis, albeit with modified assumptions.

Theorem.[1] Let be an open and connected subset of the complex numbers. Suppose that is a sequence of holomorphic functions that converges uniformly to a function on every compact subset of . Then is holomorphic in , and moreover, the sequence of derivatives converges uniformly to on every compact subset of .

Theorem.[2] Let be an open and connected subset of the complex numbers. Suppose that is a sequence of univalent[3] functions that converges uniformly to a function . Then is holomorphic, and moreover, is either univalent or constant in .

Notes

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References

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Uniform limit theorem

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In mathematics, particularly in real analysis, the uniform limit theorem asserts that if a sequence of continuous functions on a domain converges uniformly to a limit function, then the limit function is also continuous on that domain.[1] More precisely, for a set EE in the real numbers and a sequence of functions {fn}\{f_n\} where each fn:E→Rf_n: E \to \mathbb{R} is continuous, if fnf_n converges uniformly to f:E→Rf: E \to \mathbb{R}, then ff is continuous at every point in EE.[2] This result holds more generally for metric spaces, where the domain is a metric space XX and the functions map to R\mathbb{R}, ensuring the preservation of continuity under uniform convergence.[1] The theorem is a cornerstone of the theory of function sequences, distinguishing uniform convergence from mere pointwise convergence, which does not guarantee the continuity of the limit—for instance, a sequence of continuous functions may converge pointwise to a discontinuous function on a dense set.[2] The proof typically employs an ϵ/3\epsilon/3-argument: for any ϵ>0\epsilon > 0, uniform convergence provides an index NN such that ∣fn(x)−f(x)∣<ϵ/3|f_n(x) - f(x)| < \epsilon/3 for all x∈Ex \in E and n>Nn > N; continuity of fNf_N then yields a δ>0\delta > 0 ensuring ∣fN(x)−fN(x0)∣<ϵ/3|f_N(x) - f_N(x_0)| < \epsilon/3 for ∣x−x0∣<δ|x - x_0| < \delta, and the triangle inequality combines these to show ∣f(x)−f(x0)∣<ϵ|f(x) - f(x_0)| < \epsilon.[1] This uniform control over the entire domain is essential, as it prevents pathological behaviors that arise in non-uniform limits. Beyond continuity, the uniform limit theorem underpins further results in analysis, such as the uniform convergence of Riemann integrals of continuous functions preserving integrability and the ability to interchange limits with differentiation or integration under suitable conditions on compact sets.[2] It appears prominently in foundational texts on mathematical analysis and extends to more abstract settings, including sequences of analytic functions or operators in functional analysis, where uniform limits retain key structural properties.[1]

Background Concepts

Pointwise Convergence

In the context of real analysis, pointwise convergence describes a fundamental mode of convergence for sequences of functions. Consider a sequence of functions {fn}\{f_n\} defined on a domain D⊆RD \subseteq \mathbb{R} with values in R\mathbb{R}. The sequence converges pointwise to a limit function f:D→Rf: D \to \mathbb{R} if, for every x∈Dx \in D and every ε>0\varepsilon > 0, there exists an integer N=N(x,ε)∈NN = N(x, \varepsilon) \in \mathbb{N} such that ∣fn(x)−f(x)∣<ε|f_n(x) - f(x)| < \varepsilon for all n>Nn > N.[3] This condition ensures that the sequence {fn(x)}\{f_n(x)\} of real numbers converges to f(x)f(x) at each individual point x∈Dx \in D, but the choice of NN may depend on xx, allowing the rate of convergence to vary across the domain.[4] A classic example illustrates this concept. Define fn(x)=xnf_n(x) = x^n for n∈Nn \in \mathbb{N} on the domain D=[0,1]D = [0, 1]. For each fixed x∈[0,1)x \in [0, 1), lim⁡n→∞xn=0\lim_{n \to \infty} x^n = 0 since ∣x∣<1|x| < 1, while at x=1x = 1, lim⁡n→∞1n=1\lim_{n \to \infty} 1^n = 1. Thus, {fn}\{f_n\} converges pointwise to the function f(x)=0f(x) = 0 if 0≤x<10 \leq x < 1 and f(1)=1f(1) = 1.[5] This limit function is discontinuous at x=1x=1, highlighting how pointwise convergence can produce limits that differ qualitatively from the individual terms, each of which is continuous on [0,1][0,1]. Pointwise convergence preserves basic arithmetic operations on limits, such as linearity and scalar multiplication, in the sense that if {fn}→f\{f_n\} \to f and {gn}→g\{g_n\} \to g pointwise, then {afn+bgn}→af+bg\{a f_n + b g_n\} \to a f + b g pointwise for constants a,b∈Ra, b \in \mathbb{R}.[4] However, it does not generally preserve integrability or differentiability; for instance, the pointwise limit of integrable functions may fail to be integrable, and the integral of the limit need not equal the limit of the integrals.[6] Similarly, the pointwise limit of differentiable functions may not be differentiable. This makes pointwise convergence weaker than uniform convergence, which imposes a uniform rate across the domain.[7] The notion of pointwise convergence emerged in the 19th century, particularly in the study of Fourier series, where Peter Gustav Lejeune Dirichlet proved the first theorem on pointwise convergence in 1829, establishing conditions under which the partial sums of a Fourier series converge to the original function at points of continuity.[8]

Uniform Convergence

Uniform convergence is a mode of convergence for a sequence of functions {fn}\{f_n\} to a function ff on a domain DD that ensures the rate of convergence is controlled uniformly across the entire domain, independent of the point in DD. Specifically, {fn}\{f_n\} converges uniformly to ff on DD if for every ϵ>0\epsilon > 0, there exists an integer NN (depending only on ϵ\epsilon, not on any particular x∈Dx \in D) such that for all n>Nn > N and all x∈Dx \in D, ∣fn(x)−f(x)∣<ϵ|f_n(x) - f(x)| < \epsilon.[9] This is equivalent to the condition that sup⁡x∈D∣fn(x)−f(x)∣→0\sup_{x \in D} |f_n(x) - f(x)| \to 0 as n→∞n \to \infty. Unlike pointwise convergence, which only requires the inequality to hold for each fixed xx with NN possibly depending on xx, uniform convergence provides a stronger, global control that preserves key analytical properties of the functions.[10] A practical criterion for establishing uniform convergence, particularly for series of functions ∑gn(x)\sum g_n(x), is the Weierstrass M-test: if there exist positive constants MnM_n such that ∣gn(x)∣≤Mn|g_n(x)| \leq M_n for all x∈Dx \in D and all nn, and ∑Mn<∞\sum M_n < \infty, then ∑gn(x)\sum g_n(x) converges uniformly (and absolutely) on DD.[11] Several important properties follow from uniform convergence. A sequence {fn}\{f_n\} is uniformly Cauchy on DD if for every ϵ>0\epsilon > 0, there exists NN such that for all m,n>Nm, n > N and all x∈Dx \in D, ∣fm(x)−fn(x)∣<ϵ|f_m(x) - f_n(x)| < \epsilon; in the context of real-valued functions on a domain, uniformly Cauchy sequences converge uniformly to some limit function.[12] Moreover, if each fnf_n is bounded on DD, then the uniform limit ff is also bounded on DD.[12] Similarly, if each fnf_n is uniformly continuous on DD, then the uniform limit ff is uniformly continuous on DD.[13] On compact subsets of the domain, uniform convergence implies pointwise convergence, but the converse does not hold in general.[9]

Continuity of Functions

A function f:D→Rf: D \to \mathbb{R}, where D⊆RD \subseteq \mathbb{R}, is continuous at a point x0∈Dx_0 \in D if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that whenever ∣x−x0∣<δ|x - x_0| < \delta and x∈Dx \in D, it follows that ∣f(x)−f(x0)∣<ϵ|f(x) - f(x_0)| < \epsilon.[14] The function ff is continuous on DD if it is continuous at every point in DD. Uniform continuity strengthens this notion: ff is uniformly continuous on DD if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 independent of x0x_0 such that for all x,y∈Dx, y \in D with ∣x−y∣<δ|x - y| < \delta, ∣f(x)−f(y)∣<ϵ|f(x) - f(y)| < \epsilon.[15] Continuous functions exhibit several key properties. On a compact set K⊆RK \subseteq \mathbb{R}, which by the Heine-Borel theorem is closed and bounded, every continuous function is uniformly continuous and bounded.[16][17] Additionally, continuous functions satisfy the intermediate value theorem: if ff is continuous on the closed interval [a,b][a, b] and dd lies strictly between f(a)f(a) and f(b)f(b), then there exists c∈(a,b)c \in (a, b) such that f(c)=df(c) = d.[14] The composition of continuous functions is continuous: if f:D→Rf: D \to \mathbb{R} is continuous at bb and g:E→Dg: E \to D satisfies lim⁡x→ag(x)=b\lim_{x \to a} g(x) = b with a∈Ea \in E, then f∘gf \circ g is continuous at aa.[18] Examples illustrate these concepts. Polynomial functions, such as f(x)=x2+3x−1f(x) = x^2 + 3x - 1, are continuous on all of R\mathbb{R} because they are finite sums of continuous power functions.[14] Rational functions, like f(x)=x2−1x−2f(x) = \frac{x^2 - 1}{x - 2}, are continuous on their domains, excluding points where the denominator vanishes.[19] In contrast, the Heaviside step function f(x)=0f(x) = 0 for x<0x < 0 and f(x)=1f(x) = 1 for x≥0x \geq 0 is discontinuous at x=0x = 0, as small perturbations around 0 yield values jumping between 0 and 1.[19] The epsilon-delta formulation of continuity was formalized by Karl Weierstrass in the 19th century, building on earlier ideas from Augustin-Louis Cauchy and Bernard Bolzano.[2]

The Uniform Limit Theorem

Statement

The uniform limit theorem states that if {fn}\{f_n\} is a sequence of continuous real-valued functions defined on a nonempty subset D⊆RD \subseteq \mathbb{R} and fnf_n converges uniformly to a function f:D→Rf: D \to \mathbb{R}, then ff is continuous on DD. This result extends to more general settings, such as metric spaces. Specifically, if (X,d)(X, d) is a metric space, D⊆XD \subseteq X is nonempty, and {fn:D→R}\{f_n: D \to \mathbb{R}\} is a sequence of continuous functions that converges uniformly to f:D→Rf: D \to \mathbb{R}, then ff is continuous on DD.[20] An equivalent formulation uses the ϵ\epsilon-NN definition of uniform convergence: for every ϵ>0\epsilon > 0, there exists N∈NN \in \mathbb{N} such that for all n>Nn > N and all x∈Dx \in D, ∣fn(x)−f(x)∣<ϵ|f_n(x) - f(x)| < \epsilon. Under this condition, combined with the continuity of each fnf_n (which means for every x0∈Dx_0 \in D and ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that ∣x−x0∣<δ|x - x_0| < \delta implies ∣fn(x)−fn(x0)∣<ϵ|f_n(x) - f_n(x_0)| < \epsilon for each nn), the limit function ff satisfies the ϵ\epsilon-δ\delta definition of continuity at every point in DD.[12] The theorem applies analogously to complex-valued functions on subsets of C\mathbb{C} equipped with the standard topology, where uniform convergence of continuous functions yields a continuous limit. Uniformity of convergence is a necessary condition for preserving continuity, as pointwise convergence of continuous functions may result in a discontinuous limit.[12]

Proof

To prove the uniform limit theorem, consider a sequence of functions {fn}\{f_n\} defined on a subset EE of a metric space XX, where each fn:E→Yf_n: E \to Y is continuous (with YY another metric space), and suppose fnf_n converges uniformly to a function f:E→Yf: E \to Y. The goal is to show that ff is continuous at every point p∈Ep \in E.[1] Fix p∈Ep \in E and ϵ>0\epsilon > 0. Since fn→ff_n \to f uniformly on EE, there exists an integer NN such that for all n>Nn > N and all x∈Ex \in E,
dY(fn(x),f(x))<ϵ3, d_Y(f_n(x), f(x)) < \frac{\epsilon}{3},
where dYd_Y is the metric on YY. This NN is independent of any particular point in EE, which is a key feature of uniform convergence.[1] Now fix n=N+1>Nn = N+1 > N, so fN+1f_{N+1} is continuous at pp. Thus, there exists δ>0\delta > 0 such that if x∈Ex \in E and dX(x,p)<δd_X(x, p) < \delta, then
dY(fN+1(x),fN+1(p))<ϵ3, d_Y(f_{N+1}(x), f_{N+1}(p)) < \frac{\epsilon}{3},
where dXd_X is the metric on XX. This δ\delta depends on the continuity of fN+1f_{N+1} at pp but not on ff or the convergence.[1] For any x∈Ex \in E with dX(x,p)<δd_X(x, p) < \delta, apply the triangle inequality in YY:
dY(f(x),f(p))≤dY(f(x),fN+1(x))+dY(fN+1(x),fN+1(p))+dY(fN+1(p),f(p)). d_Y(f(x), f(p)) \leq d_Y(f(x), f_{N+1}(x)) + d_Y(f_{N+1}(x), f_{N+1}(p)) + d_Y(f_{N+1}(p), f(p)).
Substituting the established bounds yields
dY(f(x),f(p))<ϵ3+ϵ3+ϵ3=ϵ. d_Y(f(x), f(p)) < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.
Since ϵ>0\epsilon > 0 and p∈Ep \in E are arbitrary, ff is continuous on EE. The uniformity ensures that the choice of NN (and thus the fixed fN+1f_{N+1}) applies globally, allowing the local continuity of fN+1f_{N+1} to propagate to ff without dependence on the evaluation point.[1] This argument extends naturally to general metric spaces, as the proof relies only on the metrics dXd_X and dYd_Y and the triangle inequality, without invoking completeness of the spaces.[1]

Applications and Extensions

In Real Analysis

In real analysis, uniform convergence of a sequence of functions enables the interchange of limits and integrals under suitable conditions. Specifically, if $ {f_n} $ is a sequence of Riemann integrable functions on a compact interval [a,b][a, b] that converges uniformly to $ f $, then $ f $ is Riemann integrable, and
∫abf(x) dx=lim⁡n→∞∫abfn(x) dx. \int_a^b f(x) \, dx = \lim_{n \to \infty} \int_a^b f_n(x) \, dx.
This result follows from the fact that uniform convergence preserves the boundedness and continuity needed for Riemann integrability on compact sets, allowing the limit to pass inside the integral without altering the value.[21] Uniform convergence also justifies term-by-term differentiation of sequences of differentiable functions. If $ {f_n} $ converges pointwise to $ f $ on an interval, each $ f_n $ is differentiable, and the derivatives $ {f_n'} $ converge uniformly to some function $ g $, then $ f $ is differentiable and $ f' = g $. This theorem ensures that the derivative of the limit equals the limit of the derivatives, provided the uniform condition on the derivatives holds. A prominent application arises with power series: within the open disk of convergence (for real variables, the interval of radius $ R $), the series for the derivative converges uniformly to the derivative of the sum function, permitting term-by-term differentiation.[22][23] A related theorem strengthening the conditions for uniform convergence is Dini's theorem, which applies to monotone sequences on compact sets. If $ {f_n} $ is a monotone (increasing or decreasing) sequence of continuous real-valued functions on a compact set $ K \subseteq \mathbb{R} $ that converges pointwise to a continuous function $ f $, then the convergence is uniform on $ K $. This result bridges pointwise and uniform convergence in scenarios where monotonicity provides additional control, often used in proofs involving approximations or series expansions.[24][25] The uniform limit theorem underpins these developments by guaranteeing continuity of the limit function, which is essential for applications like the Weierstrass approximation theorem. This theorem states that any continuous function on a compact interval can be uniformly approximated by polynomials, relying on uniform convergence to ensure the approximating polynomials' limits preserve continuity and enable operations such as integration and differentiation. Such tools are foundational for analyzing function spaces and solving differential equations in real analysis.[26]

In Complex Analysis

In complex analysis, the uniform limit theorem asserts that if a sequence of holomorphic functions {fn}\{f_n\} on a domain Ω⊆C\Omega \subseteq \mathbb{C} converges uniformly on every compact subset of Ω\Omega to a function ff, then ff is holomorphic on Ω\Omega.[27] This result leverages the topological structure of the complex plane, where uniform convergence on compacts ensures the limit inherits the analytic properties of the approximants, extending the real-variable version to preserve holomorphy rather than mere continuity.[28] The proof proceeds via Cauchy's integral formula. For any z∈Ωz \in \Omega, select a simple closed contour γ\gamma in Ω\Omega enclosing zz such that γ\gamma and its interior lie in Ω\Omega. Each fnf_n satisfies
fn(z)=12πi∫γfn(ζ)ζ−z dζ. f_n(z) = \frac{1}{2\pi i} \int_\gamma \frac{f_n(\zeta)}{\zeta - z}\, d\zeta.
The set γ\gamma is compact, so uniform convergence of {fn}\{f_n\} on γ\gamma justifies interchanging the limit and integral:
f(z)=lim⁡n→∞fn(z)=12πi∫γf(ζ)ζ−z dζ, f(z) = \lim_{n \to \infty} f_n(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z}\, d\zeta,
expressing ff in a form that confirms its holomorphy on Ω\Omega.[27] This interchange relies on the boundedness of the integrand due to uniform convergence and the continuity of the path.[29] A key application arises in power series: the partial sums are holomorphic polynomials that converge uniformly on compact subsets within the radius of convergence, so their limit—the infinite series sum—is holomorphic inside the disk.[28] The theorem also relates to Morera's theorem, providing a mechanism to verify holomorphy; since uniform limits preserve the vanishing of contour integrals over closed paths (by the same limit-interchange argument), the limit satisfies Morera's condition and is thus holomorphic.[29]

Examples and Limitations

Illustrative Examples

One classic example demonstrating the uniform limit theorem involves the sequence of functions fn(x)=xnf_n(x) = x^n on the interval [0,1−δ][0, 1 - \delta] where 0<δ<10 < \delta < 1. Here, each fnf_n is continuous on this compact interval. Pointwise, fn(x)→0f_n(x) \to 0 for all x∈[0,1−δ]x \in [0, 1 - \delta], since ∣x∣≤1−δ<1|x| \leq 1 - \delta < 1. To verify uniform convergence, compute the supremum norm: ∥fn−0∥∞=sup⁡x∈[0,1−δ]∣x∣n=(1−δ)n\|f_n - 0\|_\infty = \sup_{x \in [0, 1 - \delta]} |x|^n = (1 - \delta)^n. For any ϵ>0\epsilon > 0, choose N>ln⁡ϵln⁡(1−δ)N > \frac{\ln \epsilon}{\ln (1 - \delta)}; then for n>Nn > N, (1−δ)n<ϵ(1 - \delta)^n < \epsilon, so the convergence is uniform. The limit function f(x)=0f(x) = 0 is continuous, illustrating that uniform convergence preserves continuity.[30] Another illustrative case arises with partial sums of Fourier series for a periodic function with isolated discontinuities. Consider a 2π2\pi-periodic function ff that is piecewise smooth, with jumps at finitely many points. On a compact interval [a,b][a, b] contained in (−π,π)(-\pi, \pi) and avoiding these discontinuities, the partial sums sn(x)=∑k=−nnckeikxs_n(x) = \sum_{k=-n}^n c_k e^{ikx} (where ckc_k are the Fourier coefficients) converge uniformly to f(x)f(x). Uniformity follows from the localization principle and the fact that ff is continuous and smooth on [a,b][a, b], allowing application of integration by parts to the Dirichlet kernel, yielding ∥sn−f∥∞→0\|s_n - f\|_\infty \to 0 as n→∞n \to \infty. Since each sns_n is a trigonometric polynomial (hence continuous), the uniform limit ff restricted to [a,b][a, b] is continuous there.[31] A third example is the approximation of a continuous function f:[0,1]→Rf: [0,1] \to \mathbb{R} by Bernstein polynomials Bn(f;x)=∑k=0nf(kn)(nk)xk(1−x)n−kB_n(f; x) = \sum_{k=0}^n f\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}. Each Bn(f)B_n(f) is a polynomial, thus continuous on [0,1][0,1]. By Bernstein's theorem, Bn(f)→fB_n(f) \to f uniformly on [0,1][0,1], with ∥Bn(f)−f∥∞→0\|B_n(f) - f\|_\infty \to 0 as n→∞n \to \infty, proven via probabilistic interpretation or direct estimation using uniform continuity of ff. The limit ff is continuous, exemplifying how uniform convergence of continuous approximants yields a continuous limit.[32]

Counterexamples

A classic counterexample demonstrating that pointwise convergence of continuous functions does not preserve continuity involves the sequence $ f_n(x) = x^n $ defined on the closed interval [0,1][0, 1]. For each fixed $ x \in [0, 1) $, $ \lim_{n \to \infty} f_n(x) = 0 $, while $ f_n(1) = 1 $ for all $ n $, so the pointwise limit is the function $ f(x) = 0 $ if $ x \in [0, 1) $ and $ f(1) = 1 $. This limit function $ f $ is discontinuous at $ x = 1 $, despite each $ f_n $ being a continuous polynomial on [0,1][0, 1].[33] The convergence fails to be uniform because $ |f_n - f|\infty = \sup{x \in [0,1]} |f_n(x) - f(x)| = \sup_{x \in [0,1)} x^n = 1 $ for every $ n $, which does not tend to 0 as $ n \to \infty $. Another counterexample uses a sequence of "tent" or ramp functions to approximate a step discontinuity. Consider $ f_n(x) = \min(nx, 1) $ on [0,1][0, 1], which rises linearly from $ f_n(0) = 0 $ to $ f_n(1/n) = 1 $ and remains 1 thereafter. Pointwise, $ f_n(0) = 0 \to 0 $, while for any fixed $ x > 0 $, $ f_n(x) = 1 $ for all sufficiently large $ n $, yielding the limit $ f(x) = 0 $ at $ x = 0 $ and $ f(x) = 1 $ for $ x \in (0, 1] $, which is discontinuous at $ x = 0 $. Each $ f_n $ is continuous as a piecewise linear function, but the convergence is not uniform since $ |f_n - f|_\infty = 1/2 $, attained at $ x = 1/(2n) $ where $ |f_n(x) - f(x)| = |1/2 - 1| = 1/2 $, and this supremum does not approach 0. These examples highlight the key insight that without uniformity, the convergence can leave persistent "bumps" or transitions near points of potential discontinuity in the limit function, where the rate of approach to the limit varies significantly across the domain. In both cases, the slow shrinkage of these features near $ x = 1 $ or $ x = 0 $ prevents the supremum error from vanishing, allowing the limit to inherit a jump discontinuity. The uniform limit theorem guarantees preservation of continuity regardless of whether the domain is compact, as the proof relies only on the ε/3 argument at each point using the uniform bound on the tail.[33] However, on non-compact sets, uniform convergence of continuous functions may fail to preserve other properties, such as boundedness or integrability over unbounded intervals, even though continuity holds.

References

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  2. [PDF] An Introduction to Real Analysis - UC Davis Mathematics
  3. 6.1 Pointwise and uniform convergence
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  5. Real Analysis: 8.1. Pointwise Convergence - MathCS.org
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  7. Sequences of Functions | An Introduction to Real Analysis
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  9. Uniform Convergence -- from Wolfram MathWorld
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  15. [PDF] Section 5.3 Properties of Continuous Functions 215
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  17. Continuity and the Intermediate Value Theorem
  18. Calculus I - Continuity - Pauls Online Math Notes
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  20. Theorem 8.2.7: Uniform Convergence and Integration - MathCS.org
  21. Theorem 8.2.11: Uniform Convergence and Differentiation
  22. [PDF] Uniform Convergence
  23. [PDF] Math 118B Solutions
  24. [PDF] Pointwise and Uniform Convergence 1. Defn
  25. [PDF] Weierstrass Approximation Teorem Notes Nov 27, 2011 ... - UAF CS
  26. [PDF] Cauchy's theorem, Cauchy's formula, corollaries 1. Path integrals
  27. [PDF] complex analysis: lecture 23 - OSU Math
  28. [PDF] 2 Cauchy's Theorem and Its - Princeton University
  29. Vitali's theorem without uniform boundedness - Project Euclid
  30. [PDF] 5 Convergence of Fourier series - UCSB Math
  31. [PDF] Bernstein Polynomials and Approximation - Penn Math
  32. [PDF] Introduction to Analysis
  33. [PDF] Counter-examples of Sequence of Functions
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