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Uniform convergence

Uniform convergence controls approximation error across an entire domain. Compare x^n with x/n, distinguish pointwise limits, and learn when continuity and integration pass to the limit.

Uniform convergence means that a sequence of functions approaches a limit function with one error tolerance working across the entire stated domain. For a quant building a pricing approximation, it distinguishes accuracy at each fixed spot or parameter value from an error guarantee that holds wherever the model is used.

Let fn:D→Rf_n:D\to\mathbb{R} approximate f:D→Rf:D\to\mathbb{R}, where nn indexes increasing resolution or approximation order. Uniform convergence on DD means

∀ε>0  ∃N  ∀n≥N  ∀x∈D:∣fn(x)−f(x)∣<ε.\forall\varepsilon>0\;\exists N\;\forall n\ge N\;\forall x\in D: \quad \lvert f_n(x)-f(x)\rvert<\varepsilon.
The same NN works for every xx. By comparison, pointwise convergence lets NN depend on xx: fix an input first, then wait long enough for its error to shrink. The order of “for every input” and “there exists NN” changes the strength of the promise. See Lebl's definitions.

Turn a whole curve into one error number

Define the supremum error

En=sup⁡x∈D∣fn(x)−f(x)∣.E_n=\sup_{x\in D}\lvert f_n(x)-f(x)\rvert.
The supremum is the least upper bound; an input attaining it need not exist. Uniform convergence is equivalent to En→0E_n\to0. This is often called convergence in the uniform norm or supremum norm, written ∥fn−f∥∞→0\lVert f_n-f\rVert_\infty\to0 when the errors are bounded. The criterion controls absolute error in the function's output units, not relative error or derivatives.

A finite grid estimates errors at selected points. Without an additional bound between grid points, its largest observed error does not certify the supremum over a continuous domain.

A boundary that never settles uniformly

On D=[0,1]D=[0,1], take fn(x)=xnf_n(x)=x^n. At every fixed x<1x<1, the powers approach zero; at x=1x=1, they remain one. The pointwise limit is

f(x)={0,0≤x<1,1,x=1.f(x)=\begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases}

Every fnf_n is continuous, but its limit has a jump. For each finite nn, inputs just below one still have xnx^n arbitrarily close to one while their limit value is zero. Consequently,

sup⁡x∈[0,1]∣xn−f(x)∣=1for every n.\sup_{x\in[0,1]}\lvert x^n-f(x)\rvert=1 \quad\text{for every }n.
The error is zero at x=1x=1. Its supremum of one is approached from below and never attained. Calling it the “maximum error at the endpoint” would miss the mechanism.
At n=100n=100, 0.99100≈0.3660320.99^{100}\approx0.366032. More generally, choose xn=(1/2)1/n<1x_n=(1/2)^{1/n}<1: then fn(xn)−f(xn)=1/2f_n(x_n)-f(x_n)=1/2 for every nn. The troublesome input moves toward the boundary as resolution increases. This does not contradict convergence at each fixed input. Angenent's notes develop this example and the uniform-distance test.

Now compare gn(x)=x/ng_n(x)=x/n on the same domain. Its limit is zero and

sup⁡x∈[0,1]∣gn(x)∣=1n⟶0.\sup_{x\in[0,1]}\lvert g_n(x)\rvert=\frac1n\longrightarrow0.

At n=100n=100, every error is at most 0.010.01. For a strict tolerance ε\varepsilon, any integer N>1/εN>1/\varepsilon works simultaneously at all inputs.

On the dimensionless interval [0,1], curves x^n for n=5,20,100 approach a discontinuous pointwise limit but retain supremum error 1. Curves x/n converge uniformly to zero, with error at most 0.01 when n=100.
On the dimensionless interval [0,1], curves x^n for n=5,20,100 approach a discontinuous pointwise limit but retain supremum error 1. Curves x/n converge uniformly to zero, with error at most 0.01 when n=100.
Original deterministic examples, with different vertical scales. The left panel's dashed zero line, open circle at (1,0), and filled point at (1,1) describe the pointwise limit. The right panel's whole curve approaches zero.
The domain is part of the claim. Restrict xnx^n to [0,a][0,a] for a fixed a<1a<1, with a>0a>0: its supremum error becomes an→0a^n\to0. Removing only the endpoint, to use [0,1)[0,1), leaves the supremum at one and does not restore uniform convergence.

Which operations are safe?

Uniform convergence preserves continuity: if every fnf_n is continuous on the stated domain and convergence is uniform there, then ff is continuous. A sequence of continuous approximations to a discontinuous target therefore cannot converge uniformly across the jump.
It also permits integration through a limit on a fixed finite interval. If each fnf_n is Riemann integrable on [a,b][a,b] and converges uniformly to ff, then ff is integrable and
∣∫abfn(x) dx−∫abf(x) dx∣≤(b−a)En⟶0.\left\lvert\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right\rvert \le (b-a)E_n\longrightarrow0.
This follows by bounding the absolute integrand error by EnE_n. It is useful when an approximated curve is accumulated over maturities or states. The finite-interval condition matters; uniform convergence alone does not justify an improper integral over an infinite interval. These preservation results are established in Lebl, §6.2.
Differentiation needs more. Even un(x)=sin⁡(nx)/nu_n(x)=\sin(nx)/n converges uniformly to zero on R\mathbb{R}, since ∣un(x)∣≤1/n\lvert u_n(x)\rvert\le1/n. Yet un′(0)=1u_n'(0)=1 for every nn, while the limit's derivative is zero. A bound on pricing error alone cannot establish convergence of a hedge sensitivity. One sufficient theorem on a bounded interval assumes continuously differentiable approximations, uniformly convergent derivatives, and convergence of the function values at one point.
For the integral comparison used above, the Calculus I definite-integral lesson develops linearity and comparison. Its public preview is available; the complete lesson requires membership.

This article concerns deterministic function sequences. Convergence in probability instead controls the probability of a random error exceeding a threshold. Uniform convergence of empirical risk adds a supremum over a model class to a statistical convergence statement; its sampling assumptions are a separate issue.

Practice: change the domain

For hn(x)=x/nh_n(x)=x/n, what is the smallest integer NN such that ∣hn(x)∣<0.02\lvert h_n(x)\rvert<0.02 for every n≥Nn\ge N and every x∈[0,3]x\in[0,3]? Does the same sequence converge uniformly to zero on [0,∞)[0,\infty)?

Answer. On [0,3][0,3], the supremum is 3/n3/n, so strict inequality requires n>150n>150 and the smallest NN is 151. On [0,∞)[0,\infty), no finite nn bounds all errors: choosing x=nx=n already gives error one. It converges pointwise to zero there, but not uniformly. All inputs and outputs in this exercise are dimensionless.

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