Pointwise and uniform convergence

Lesson 1 showed that pointwise limits of functions can lose continuity, integrals and derivatives. This lesson gives the precise definitions of pointwise and uniform convergence, shows how to test for each, and proves the uniform Cauchy criterion, which decides uniform convergence without knowing the limit.

Two definitions

Let \(A\) be a set of real numbers and let \(f_n: A \to \mathbb{R}\) and \(f: A \to \mathbb{R}\) be functions.

Definition. The sequence \((f_n)\) converges pointwise to \(f\) on \(A\) if for every \(x \in A\) and every \(\varepsilon > 0\) there is an \(N\) such that \(|f_n(x) - f(x)| < \varepsilon\) for all \(n \ge N\).

Definition. The sequence \((f_n)\) converges uniformly to \(f\) on \(A\) if for every \(\varepsilon > 0\) there is an \(N\) such that \(|f_n(x) - f(x)| < \varepsilon\) for all \(n \ge N\) and all \(x \in A\).

The only difference is the position of 'for all \(x\)'. In pointwise convergence \(N\) is chosen after \(x\), so it may depend on \(x\); in uniform convergence \(N\) is chosen before \(x\) and must serve every point at once. Uniform convergence implies pointwise convergence, and the limit is the same, since pointwise limits of real sequences are unique.

Geometrically, uniform convergence says that for every \(\varepsilon > 0\), the graphs of \(f_n\) eventually lie inside the band of vertical half-width \(\varepsilon\) around the graph of \(f\), the set of points \((x, y)\) with \(f(x) - \varepsilon < y < f(x) + \varepsilon\). Pointwise convergence only says that above each single \(x\) the values eventually enter the band.

The sup norm test

For a bounded function \(g\) on \(A\) write \(\|g\|_\infty = \sup_{x \in A} |g(x)|\), the sup norm. The definition of uniform convergence can then be restated in one line.

Theorem 2.1. \(f_n \to f\) uniformly on \(A\) if and only if \(\|f_n - f\|_\infty \to 0\), where for each large \(n\) the function \(f_n - f\) is bounded.

Proof. Suppose \(f_n \to f\) uniformly and let \(\varepsilon > 0\). Choose \(N\) with \(|f_n(x) - f(x)| < \varepsilon\) for all \(n \ge N\) and \(x \in A\). Then for \(n \ge N\), \(\varepsilon\) is an upper bound for \(\{|f_n(x) - f(x)| : x \in A\}\), so \(f_n - f\) is bounded and \(\|f_n - f\|_\infty \le \varepsilon\). Hence \(\|f_n - f\|_\infty \to 0\). Conversely, if \(\|f_n - f\|_\infty \to 0\), given \(\varepsilon > 0\) choose \(N\) with \(\|f_n - f\|_\infty < \varepsilon\) for \(n \ge N\). Then \(|f_n(x) - f(x)| \le \|f_n - f\|_\infty < \varepsilon\) for every \(x\). \(\square\)

In practice the test runs in three steps. First find the pointwise limit \(f\). Then compute or estimate \(M_n = \sup_{x \in A} |f_n(x) - f(x)|\), often by calculus. Finally decide whether \(M_n \to 0\). To prove uniform convergence it is enough to find any bound \(|f_n(x) - f(x)| \le b_n\) valid for all \(x\) with \(b_n \to 0\). To disprove it, it is enough to find points \(x_n \in A\) with \(|f_n(x_n) - f(x_n)|\) not tending to 0.

Worked examples

The last two examples show a common pattern. A pointwise limit fails to be uniform because the region where \(f_n\) differs significantly from \(f\) moves, typically toward a boundary point, as \(n\) grows. Restricting to a set that stays away from that point often restores uniformity.

Measuring the sup distance numerically

A computer cannot take a supremum over infinitely many points, but evaluating on a fine grid gives a lower bound for the sup and, when the functions are smooth, a good estimate. The code compares the three examples above.

import math

def sup_dist(f, g, a, b, pts=20001):
    """Largest |f(x) - g(x)| over an even grid on [a, b]: a lower bound for the sup."""
    return max(abs(f(a + (b - a) * i / (pts - 1)) - g(a + (b - a) * i / (pts - 1))) for i in range(pts))

zero = lambda x: 0.0
print(" n    x^n on [0,1)   x^n on [0,0.9]   x/(1+nx^2) on [-5,5]   nx e^(-nx) on [0,5]")
for n in [1, 5, 25, 125, 625]:
    d1 = sup_dist(lambda x: x ** n, zero, 0, 1 - 1e-9)
    d2 = sup_dist(lambda x: x ** n, zero, 0, 0.9)
    d3 = sup_dist(lambda x: x / (1 + n * x * x), zero, -5, 5)
    d4 = sup_dist(lambda x: n * x * math.exp(-n * x), zero, 0, 5)
    print(f"{n:3d}   {d1:10.4f}    {d2:12.3e}    {d3:16.4f}       {d4:14.4f}")
print("predicted: 1, 0.9^n, 1/(2 sqrt n), 1/e =", round(1 / math.e, 4))

 n    x^n on [0,1)   x^n on [0,0.9]   x/(1+nx^2) on [-5,5]   nx e^(-nx) on [0,5]
  1       1.0000       9.000e-01              0.5000               0.3679
  5       1.0000       5.905e-01              0.2236               0.3679
 25       1.0000       7.179e-02              0.1000               0.3679
125       1.0000       1.907e-06              0.0447               0.3679
625       1.0000       2.521e-29              0.0200               0.3671
predicted: 1, 0.9^n, 1/(2 sqrt n), 1/e = 0.3679

The middle columns tend to 0, matching \(0.9^n\) and \(\frac{1}{2\sqrt{n}}\). The first column stays near 1 and the last at \(\frac{1}{e} \approx 0.3679\), so those sequences do not converge uniformly. When the bump becomes narrower than the grid spacing the estimate can drop, which is why a grid can suggest but never prove uniform convergence; the proofs above are what settle it.

The uniform Cauchy criterion

Often we want to prove that a sequence of functions converges uniformly without knowing its limit in advance, for example when the functions are partial sums of a series. The tool is a uniform version of the Cauchy criterion.

Definition. A sequence \((f_n)\) of functions on \(A\) is uniformly Cauchy if for every \(\varepsilon > 0\) there is an \(N\) such that \(|f_m(x) - f_n(x)| < \varepsilon\) for all \(m, n \ge N\) and all \(x \in A\).

Theorem 2.2 (uniform Cauchy criterion). A sequence \((f_n)\) of functions on \(A\) converges uniformly on \(A\) if and only if it is uniformly Cauchy.

Proof. Suppose \(f_n \to f\) uniformly. Given \(\varepsilon > 0\), choose \(N\) with \(|f_n(x) - f(x)| < \frac{\varepsilon}{2}\) for all \(n \ge N\) and all \(x\). For \(m, n \ge N\) the triangle inequality gives \(|f_m(x) - f_n(x)| \le |f_m(x) - f(x)| + |f(x) - f_n(x)| < \varepsilon\).

Conversely, suppose \((f_n)\) is uniformly Cauchy. For each fixed \(x\), the real sequence \((f_n(x))\) is Cauchy, so by completeness of \(\mathbb{R}\) it converges; call its limit \(f(x)\). This defines the candidate limit \(f\). Now let \(\varepsilon > 0\) and choose \(N\) with \(|f_m(x) - f_n(x)| < \frac{\varepsilon}{2}\) for all \(m, n \ge N\) and all \(x\). Fix \(n \ge N\) and \(x \in A\), and let \(m \to \infty\) in this inequality. Since \(f_m(x) \to f(x)\) and the absolute value is continuous, we get \(|f(x) - f_n(x)| \le \frac{\varepsilon}{2} < \varepsilon\). This holds for every \(n \ge N\) and every \(x\), with \(N\) independent of \(x\), so \(f_n \to f\) uniformly. \(\square\)

The key step is letting \(m \to \infty\) in a non-strict inequality, which preserves it. The same trick appears whenever we prove completeness of a space of functions, and it will reappear in lesson 8 when we show that the space \(C[a, b]\) of continuous functions with the sup distance is complete.

Uniform convergence and algebra

Uniform limits behave well under addition and scalar multiplication: if \(f_n \to f\) and \(g_n \to g\) uniformly on \(A\), then \(f_n + g_n \to f + g\) and \(cf_n \to cf\) uniformly, because \(\|(f_n + g_n) - (f + g)\|_\infty \le \|f_n - f\|_\infty + \|g_n - g\|_\infty\). Products need more care. If \(f_n \to f\) and \(g_n \to g\) uniformly and all the functions are bounded by a common constant \(M\), then \[|f_ng_n - fg| \le |f_n||g_n - g| + |g||f_n - f| \le M\|g_n - g\|_\infty + M\|f_n - f\|_\infty,\] so \(f_ng_n \to fg\) uniformly. Without boundedness this fails: on \(\mathbb{R}\), \(f_n(x) = x + \frac{1}{n}\) converges uniformly to \(x\), but \(f_n(x)^2 = x^2 + \frac{2x}{n} + \frac{1}{n^2}\) does not converge uniformly to \(x^2\), because \(\frac{2x}{n}\) is unbounded in \(x\) for each \(n\).

A uniform limit of bounded functions is bounded. If \(\|f_N - f\|_\infty < 1\) and \(|f_N| \le M\), then \(|f| \le M + 1\). In contrast, the pointwise limit of bounded functions can be unbounded: \(f_n(x) = \min(n, \frac{1}{x})\) on \((0, 1]\) is bounded for each \(n\), but converges pointwise to \(\frac{1}{x}\).

Uniform convergence on subsets

If \(f_n \to f\) uniformly on \(A\), then also on every subset of \(A\), and if \(f_n \to f\) uniformly on \(A\) and on \(B\), then on \(A \cup B\): take the larger of the two values of \(N\). By induction this extends to finite unions, but not to infinite ones. The powers \(x^n\) converge uniformly on \([0, 1 - \frac{1}{k}]\) for each \(k\), but not on the union \([0, 1)\). This motivates a weaker notion that is extremely common in practice.

Definition. The sequence \(f_n\) converges locally uniformly, or uniformly on compact sets, on an open interval \(I\) if it converges uniformly on every closed bounded subinterval \([a, b] \subseteq I\).

The powers \(x^n\) converge locally uniformly to 0 on \((-1, 1)\), and \(nxe^{-nx}\) converges locally uniformly to 0 on \((0, \infty)\). Power series, as lesson 5 shows, always converge locally uniformly inside their interval of convergence, though not usually uniformly on the whole interval. Because continuity and differentiability are local properties, locally uniform convergence is enough for most of the theorems in lesson 3.