Let \(f_n, f : E \to \mathbb{R}\).
Pointwise: \(f_n \to f\) on \(E\) if for every \(x \in E\) and every \(\varepsilon > 0\) there is \(N = N(\varepsilon, x)\) with \(|f_n(x) - f(x)| < \varepsilon\) for all \(n \ge N\).
Uniformly: the same, but with \(N = N(\varepsilon)\) not depending on \(x\).
The whole difference is whether \(N\) may depend on the point. Equivalently, and far more usable in practice:
\[ f_n \to f \text{ uniformly on } E \iff M_n := \sup_{x \in E}\left|f_n(x) - f(x)\right| \longrightarrow 0. \]To test uniform convergence, therefore: find the pointwise limit, form the difference, maximise it over \(E\) — usually by differentiating — and see whether that maximum tends to zero. Every example below follows exactly those four steps.
Given. \(f_n(x) = x^{n}\) on \([0,1]\).
Step 1 — the pointwise limit. For \(0 \le x < 1\), \(x^{n} \to 0\); at \(x = 1\), \(f_n(1) = 1\) for every \(n\). So
\[ f(x) = \begin{cases} 0, & 0 \le x < 1, \\ 1, & x = 1. \end{cases} \]Reading a few values at \(x = 0.5\): \(0.25\) at \(n = 2\), \(0.0009765625\) at \(n = 10\), and below \(10^{-30}\) at \(n = 100\) — convergence is quick there.
Step 2 — the supremum of the difference. On \([0,1)\) the difference is \(x^{n}\), whose supremum is
\[ M_n = \sup_{0 \le x < 1} x^{n} = 1 \quad \text{for every } n, \]approached as \(x \to 1^{-}\). It does not tend to \(0\), so the convergence is not uniform.
Step 3 — a second proof, with no computation. Every \(f_n\) is continuous and the limit \(f\) is not. By Theorem A below, a uniform limit of continuous functions is continuous. So the convergence cannot be uniform. \(\blacksquare\)
Interpretation. Convergence is fast in the interior and arbitrarily slow near \(x = 1\): to get \(x^{n} < 0.01\) at \(x = 0.5\) needs \(n = 7\), but at \(x = 0.99\) it needs \(n = 459\). No single \(N\) serves the whole interval, and that is exactly what non-uniform means. On \([0, a]\) with \(a < 1\) the supremum is \(a^{n} \to 0\) and the convergence is uniform — the failure is localised at the endpoint.
Given. \(f_n(x) = \dfrac{x}{1 + n x^{2}}\) on \(\mathbb{R}\).
Step 1 — the pointwise limit. For fixed \(x \ne 0\) the denominator grows without bound, so \(f_n(x) \to 0\); and \(f_n(0) = 0\) for every \(n\). Hence \(f \equiv 0\).
Step 2 — maximise \(|f_n|\) by differentiating. By the quotient rule,
\[ f_n'(x) = \frac{\left(1 + n x^{2}\right) - x(2nx)}{\left(1 + n x^{2}\right)^{2}} = \frac{1 - n x^{2}}{\left(1 + n x^{2}\right)^{2}}. \]Setting the numerator to zero gives \(x = \pm 1/\sqrt{n}\).
Step 3 — the value at the maximum. At \(x = 1/\sqrt n\),
\[ f_n\!\left(\frac{1}{\sqrt n}\right) = \frac{1/\sqrt n}{1 + n \cdot \frac{1}{n}} = \frac{1/\sqrt n}{2} = \frac{1}{2\sqrt n}. \]Step 4 — let \(n\) grow.
| \(n\) | location of the peak \(1/\sqrt n\) | \(M_n = 1/(2\sqrt n)\) |
|---|---|---|
| 1 | 1.000000 | 0.500000 |
| 4 | 0.500000 | 0.250000 |
| 25 | 0.200000 | 0.100000 |
| 100 | 0.100000 | 0.050000 |
\(M_n \to 0\), so the convergence is uniform on all of \(\mathbb{R}\).
Step 5 — and yet the derivatives do not converge. From Step 2,
\[ f_n'(0) = \frac{1 - 0}{(1 + 0)^{2}} = 1 \quad \text{for every } n, \]while the limit function is \(f \equiv 0\), whose derivative is \(0\). So
\[ \lim_{n} f_n'(0) = 1 \ne 0 = f'(0). \]Interpretation. Uniform convergence of \(f_n\) does not carry derivatives. The theorem in section 3 therefore assumes uniform convergence of the derivatives instead — and this example is why it must.
Statement. If each \(f_n\) is continuous on \(E\) and \(f_n \to f\) uniformly, then \(f\) is continuous on \(E\).
Proof — the \(\varepsilon/3\) argument. Fix \(p \in E\) and \(\varepsilon > 0\).
Step 1. By uniform convergence choose \(N\) with \(|f_N(x) - f(x)| < \varepsilon/3\) for every \(x \in E\). This is where uniformity is used, and it is the only place.
Step 2. \(f_N\) is continuous at \(p\), so choose \(\delta > 0\) with \(|f_N(x) - f_N(p)| < \varepsilon/3\) whenever \(d(x,p) < \delta\).
Step 3. For such \(x\), insert \(f_N\) twice and use the triangle inequality:
\[ |f(x) - f(p)| \le \underbrace{|f(x) - f_N(x)|}_{< \varepsilon/3} + \underbrace{|f_N(x) - f_N(p)|}_{< \varepsilon/3} + \underbrace{|f_N(p) - f(p)|}_{< \varepsilon/3} < \varepsilon. \]So \(f\) is continuous at \(p\), and \(p\) was arbitrary. \(\blacksquare\)
Where pointwise convergence fails. Step 1 would give an \(N\) depending on \(x\), and the \(N\) in the first and third terms could then differ from the one in the second, so the three bounds would not combine. Example 4.1 shows the conclusion genuinely fails.
The M-test is the workhorse: it reduces a question about functions to a question about a series of constants. For instance \(\sum_n \dfrac{\sin nx}{n^{2}}\) converges uniformly on \(\mathbb{R}\), because \(\left|\dfrac{\sin nx}{n^{2}}\right| \le \dfrac{1}{n^{2}}\) and \(\sum 1/n^{2} = \pi^{2}/6 = 1.644934\) converges — the same series evaluated in Probability Theory, Unit 3.
Statement. If \(f_n \in \mathcal{R}(\alpha)\) on \([a,b]\) and \(f_n \to f\) uniformly, then \(f \in \mathcal{R}(\alpha)\) and
\[ \int_a^b f \, d\alpha = \lim_{n \to \infty} \int_a^b f_n \, d\alpha. \]Why it works. With \(M_n = \sup|f_n - f| \to 0\), the linear bound from Unit 2 gives
\[ \left|\int_a^b f_n\,d\alpha - \int_a^b f\,d\alpha\right| \le M_n\left[\alpha(b) - \alpha(a)\right] \to 0, \]because \(\alpha(b) - \alpha(a)\) is a fixed finite number. The interchange is that one inequality.
Given. \(f_n(x) = n x\left(1 - x^{2}\right)^{n}\) on \([0,1]\).
Step 1 — the pointwise limit. At \(x = 0\) every \(f_n(0) = 0\). For \(0 < x \le 1\) we have \(0 \le 1 - x^{2} < 1\), and a geometric factor beats the linear factor \(n\), so \(f_n(x) \to 0\). Hence \(f \equiv 0\) and \(\int_0^1 f = 0\).
Step 2 — the integral of each \(f_n\), exactly. Substitute \(u = 1 - x^{2}\), \(du = -2x\,dx\), with \(x = 0 \mapsto u = 1\) and \(x = 1 \mapsto u = 0\):
\[ \int_0^1 n x \left(1 - x^{2}\right)^{n} dx = \frac{n}{2}\int_0^1 u^{n}\,du = \frac{n}{2}\cdot\frac{1}{n+1} = \frac{n}{2n+2}. \]Step 3 — evaluate, and locate the peak. Differentiating shows the maximum is at \(x = 1/\sqrt{2n+1}\):
| \(n\) | \(\int_0^1 f_n\) | peak position | peak height |
|---|---|---|---|
| 1 | 0.250000 | 0.577350 | 0.384900 |
| 5 | 0.416667 | 0.301511 | 0.936074 |
| 50 | 0.490196 | 0.099504 | 3.025106 |
| 500 | 0.499002 | 0.031607 | 9.587695 |
Step 4 — compare the two limits.
\[ \lim_{n \to \infty}\int_0^1 f_n = \lim_{n\to\infty}\frac{n}{2n+2} = \frac12, \qquad \int_0^1 \lim_{n \to \infty} f_n = \int_0^1 0 = 0. \]Step 5 — find the failed hypothesis. The peak height grows without bound — \(0.3849, 0.9361, 3.0251, 9.5877\) — so \(M_n = \sup|f_n - 0| \to \infty\), and the convergence is not uniform. Theorem B does not apply, and its conclusion is false here.
Interpretation. The bump grows taller and slides towards \(0\) while keeping roughly constant area — the same escape-to-infinity as the spike sequence of Probability Theory, Unit 1, Example 1.4, and as the estimator counterexample in Unit 3, Example 3.1 of that course. Three different subjects, one mechanism.
Statement. Suppose each \(f_n\) is differentiable on \([a,b]\), the sequence \(f_n(x_0)\) converges for some \(x_0\), and \(f_n'\) converges uniformly on \([a,b]\). Then \(f_n\) converges uniformly to some \(f\), and
\[ f'(x) = \lim_{n \to \infty} f_n'(x). \]Note what is assumed. It is the derivatives that must converge uniformly, not the functions. Example 4.2 shows why: there \(f_n \to 0\) uniformly, yet \(f_n'(0) = 1\) for every \(n\) while \(f'(0) = 0\). Uniform convergence of \(f_n\) alone is worth nothing here, and convergence at a single point \(x_0\) is all that is needed of the functions themselves.
Statement. If \(f\) is continuous on \([a,b]\), then for every \(\varepsilon > 0\) there is a polynomial \(p\) with
\[ \sup_{x \in [a,b]}\left|f(x) - p(x)\right| < \varepsilon. \]Equivalently: the polynomials are dense in \(C[a,b]\) under the uniform metric of Unit 1. Every continuous function on a closed bounded interval, however jagged, is a uniform limit of polynomials.
Bernstein's constructive proof is a probability argument. Define
\[ B_n(f; x) = \sum_{k=0}^{n} f\!\left(\frac{k}{n}\right)\binom{n}{k} x^{k}(1-x)^{n-k}, \]which is exactly \(E\left[f(S_n/n)\right]\) with \(S_n \sim \text{Bin}(n, x)\). By the weak law of large numbers of Probability Theory, Unit 4, \(S_n/n \xrightarrow{P} x\); uniform continuity of \(f\) on the compact \([0,1]\) — consequence (c) of Unit 1, section 5 — converts that into \(B_n(f;x) \to f(x)\) uniformly in \(x\). A theorem of pure analysis, proved by Chebyshev's inequality.
Statement. Let \(K\) be a compact metric space and \(\mathcal{A}\) a family of real continuous functions on \(K\) that
Then \(\mathcal{A}\) is dense in \(C(K)\) under the uniform metric.
Weierstrass as a special case. On \(K = [a,b]\), the polynomials form an algebra; \(f(x) = x\) separates points; \(f(x) = 1\) vanishes nowhere. All three hypotheses hold, and the conclusion is Weierstrass's theorem.
Where it is used in statistics. Three places, and they are not decorative.