Real Analysis — 05 · Sequences of Functions, Uniform Convergence and Power Series

Much of analysis builds functions as limits: $e^x$ as a power series, $\sin x$ as a Taylor series, the Weierstrass function of post 03 as a sum of cosines, solutions of differential equations as limits of approximations. The question is always the same: which properties survive the limit? Continuity, integrals and derivatives can all be lost under the obvious "point by point" notion of convergence. This post introduces the stronger notion that repairs most of this, uniform convergence, proves exactly what it preserves (continuity and integrals) and what it does not (derivatives, without an extra hypothesis), and then applies it to power series: radius of convergence, term-by-term calculus, Abel's theorem, and the difference between a Taylor series and the function it came from. It closes the core course with the Weierstrass approximation theorem and a map of where analysis goes next.

Series map: 00 The Real Numbers & Completeness → 01 Sequences & Series → 02 Topology of ℝ & Continuity → 03 Differentiation & Taylor's Theorem → 04 The Riemann Integral → 05 Uniform Convergence & Power Series (this post).

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Learning Outcome

• Define pointwise and uniform convergence of a sequence of functions, and tell them apart with $x^n$ on $[0, 1]$.
• Use the sup-norm test and the Cauchy criterion for uniform convergence.
• Prove that a uniform limit of continuous functions is continuous (the $\varepsilon/3$ argument).
• Prove that integrals commute with uniform limits, and show with an example that pointwise limits break this.
• State the theorem on limits of derivatives and give the counterexample that shows why it needs uniform convergence of the derivatives.
• State and prove the Weierstrass M-test and use it on series of functions.
• Compute the radius of convergence with the Cauchy–Hadamard formula and the ratio test, and check the endpoints.
• Prove that a power series converges uniformly on every closed interval inside its radius, and differentiate and integrate it term by term.
• State Abel's theorem and use it to show $1 - \frac12 + \frac13 - \dots = \ln 2$.
• Explain why a smooth function need not equal its Taylor series, and state the Weierstrass approximation theorem with the Bernstein-polynomial idea.

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Chapter 1 — Pointwise versus uniform convergence

Terminology

• Sequence of functions — functions $f_n : A \to \mathbb{R}$, $n = 1, 2, \dots$, all on the same domain $A \subseteq \mathbb{R}$.
• Pointwise convergence — $f_n \to f$ pointwise on $A$ if for each fixed $x \in A$, the number sequence $f_n(x)$ converges to $f(x)$. In quantifiers: for every $x \in A$ and every $\varepsilon > 0$ there is $N$ (which may depend on both $x$ and $\varepsilon$) with $\lvert f_n(x) - f(x)\rvert < \varepsilon$ for $n \geq N$.
• Uniform convergence — $f_n \to f$ uniformly on $A$ if for every $\varepsilon > 0$ there is $N$ (depending on $\varepsilon$ only) with $\lvert f_n(x) - f(x)\rvert < \varepsilon$ for all $n \geq N$ and all $x \in A$.
• Sup norm — for a bounded $g : A \to \mathbb{R}$, $\lVert g\rVert_\infty = \sup_{x \in A}\lvert g(x)\rvert$.
• Epsilon-tube — the band of width $\varepsilon$ on each side of the graph of $f$. Uniform convergence says: from some $N$ on, the whole graph of every $f_n$ lies inside the tube.

The only difference between the two definitions is the order of "there is $N$" and "for all $x$". Uniform convergence implies pointwise convergence; the converse fails.

Sup-norm test. $f_n \to f$ uniformly on $A$ if and only if $\lVert f_n - f\rVert_\infty \to 0$ (where $\lVert f_n - f\rVert_\infty$ is finite for large $n$). This is the definition rewritten: "$\lvert f_n(x) - f(x)\rvert < \varepsilon$ for all $x$" gives $\sup \leq \varepsilon$, and $\sup < \varepsilon$ gives it for all $x$.

The key example: x to the n on [0, 1]

Let $f_n(x) = x^n$ on $[0, 1]$. For $0 \leq x < 1$, $x^n \to 0$ (post 01); at $x = 1$, $1^n = 1$. So the pointwise limit is

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

Every $f_n$ is continuous, and the limit is not. The convergence is not uniform: the point $x_n = (1/2)^{1/n}$ lies in $[0, 1)$ and $f_n(x_n) - f(x_n) = \frac12$, so $\lVert f_n - f\rVert_\infty \geq \frac12$ for every $n$ (in fact the sup over $[0,1)$ is $1$). The "bad point" slides towards $1$:

$n$	1	2	5	10	50
$x_n = (1/2)^{1/n}$	$0.5$	$0.7071$	$0.8706$	$0.9330$	$0.9862$

For each fixed $x < 1$ you eventually get close to $0$, but no single $N$ works for all $x$ at once.

On a smaller domain the problem disappears. On $[0, \frac12]$, $\lVert x^n - 0\rVert_\infty = (\frac12)^n \to 0$, so $x^n \to 0$ uniformly on $[0, \frac12]$, and indeed on $[0, r]$ for any $r < 1$.

Damping the bad point. Multiply by $(1 - x)$: $k_n(x) = x^n(1 - x)$ on $[0, 1]$. The pointwise limit is $0$ everywhere (including $x = 1$, where $k_n(1) = 0$). Setting $k_n'(x) = x^{n-1}\bigl(n - (n+1)x\bigr) = 0$ gives the maximum at $x = \frac{n}{n+1}$, so

$$\lVert k_n\rVert_\infty = \left(\frac{n}{n+1}\right)^n \frac{1}{n+1} < \frac{1}{n+1} \to 0.$$

So $k_n \to 0$ uniformly on all of $[0, 1]$. For example, $\lVert k_{10}\rVert_\infty = \frac{10^{10}}{11^{11}} \approx 0.0350$. The factor $(1 - x)$ kills the values near $1$ exactly where $x^n$ misbehaved.

A uniform example: smoothing the corner of the absolute value

Let $g_n(x) = \sqrt{x^2 + 1/n^2}$ on $\mathbb{R}$. With $h = 1/n$,

$$0 \leq \sqrt{x^2 + h^2} - \lvert x\rvert = \frac{h^2}{\sqrt{x^2 + h^2} + \lvert x\rvert} \leq \frac{h^2}{h} = h,$$

with equality at $x = 0$. So $\lVert g_n - \lvert x\rvert\rVert_\infty = \frac1n \to 0$: uniform convergence to $\lvert x\rvert$ on all of $\mathbb{R}$. With $\varepsilon = 0.25$, every $g_n$ with $n \geq 5$ lies inside the $\varepsilon$-tube around $\lvert x\rvert$ (its largest error is $\frac15 = 0.2$), while $g_2$ (error $0.5$) does not.

Every $g_n$ is smooth, and the limit $\lvert x\rvert$ has a corner. Uniform convergence preserved continuity but not differentiability. Chapter 2 explains both facts.

The Cauchy criterion for uniform convergence

Theorem. A sequence $(f_n)$ converges uniformly on $A$ if and only if for every $\varepsilon > 0$ there is $N$ with

$$\lvert f_n(x) - f_m(x)\rvert < \varepsilon \quad \text{for all } m, n \geq N \text{ and all } x \in A.$$

Proof. ($\Rightarrow$) If $\lvert f_n - f\rvert < \varepsilon/2$ everywhere for $n \geq N$, then $\lvert f_n - f_m\rvert \leq \lvert f_n - f\rvert + \lvert f - f_m\rvert < \varepsilon$.

($\Leftarrow$) For each fixed $x$, $(f_n(x))$ is a Cauchy sequence of reals, so it converges (completeness, post 01); call the limit $f(x)$. Given $\varepsilon$, take $N$ from the hypothesis. For $n \geq N$ and any $x$, let $m \to \infty$ in $\lvert f_n(x) - f_m(x)\rvert < \varepsilon$ to get $\lvert f_n(x) - f(x)\rvert \leq \varepsilon$. The same $N$ works for every $x$, so the convergence is uniform. $\blacksquare$

The value of the criterion: it proves uniform convergence without knowing the limit in advance. The M-test in Chapter 3 is built on it.

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Chapter 2 — What uniform convergence preserves

Continuity

Theorem. If each $f_n$ is continuous at $c \in A$ and $f_n \to f$ uniformly on $A$, then $f$ is continuous at $c$.

Proof (the $\varepsilon/3$ argument). Let $\varepsilon > 0$. Choose $N$ with $\lvert f_N(x) - f(x)\rvert < \varepsilon/3$ for all $x \in A$. Since $f_N$ is continuous at $c$, choose $\delta$ with $\lvert f_N(x) - f_N(c)\rvert < \varepsilon/3$ for $\lvert x - c\rvert < \delta$. Then for such $x$,

$$\lvert f(x) - f(c)\rvert \leq \lvert f(x) - f_N(x)\rvert + \lvert f_N(x) - f_N(c)\rvert + \lvert f_N(c) - f(c)\rvert < \frac\varepsilon3 + \frac\varepsilon3 + \frac\varepsilon3 = \varepsilon. \qquad \blacksquare$$

The first and third terms need the same $N$ at two different points, $x$ and $c$. That is exactly what uniformity supplies and pointwise convergence does not. The theorem also gives a quick proof that $x^n$ does not converge uniformly on $[0, 1]$: the limit is discontinuous.

A partial converse (Dini's theorem). If $K$ is compact, each $f_n$ is continuous on $K$, $f_n(x) \geq f_{n+1}(x)$ for all $x$ and $n$, and $f_n \to f$ pointwise with $f$ continuous, then the convergence is uniform. (Rudin, Theorem 7.13.) All four hypotheses are needed; $x^n$ on $[0, 1]$ fails only "the limit is continuous".

Integrals

Theorem. If each $f_n$ is Riemann integrable on $[a, b]$ and $f_n \to f$ uniformly on $[a, b]$, then $f$ is integrable and

$$\int_a^b f = \lim_{n \to \infty} \int_a^b f_n.$$

Proof. Let $\varepsilon_n = \lVert f_n - f\rVert_\infty$, which tends to $0$. Then $f_n - \varepsilon_n \leq f \leq f_n + \varepsilon_n$ on $[a, b]$ (so $f$ is bounded). Upper and lower sums respect these inequalities, so the upper and lower integrals of $f$ (post 04) satisfy

$$\int_a^b f_n - \varepsilon_n(b - a) \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq \int_a^b f_n + \varepsilon_n(b - a).$$

The gap between the upper and lower integrals is at most $2\varepsilon_n(b - a) \to 0$, so they are equal and $f$ is integrable. The same inequalities give $\left\lvert \int_a^b f - \int_a^b f_n\right\rvert \leq \varepsilon_n(b - a) \to 0$. $\blacksquare$

Pointwise convergence is not enough. Let $f_n(x) = n x (1 - x^2)^n$ on $[0, 1]$. For $0 < x < 1$, $f_n(x) \to 0$ (a geometric factor beats the factor $n$), and $f_n(0) = f_n(1) = 0$. So $f_n \to 0$ pointwise. But

$$\int_0^1 n x (1 - x^2)^n\,dx = \left[-\frac{n}{2(n+1)}(1 - x^2)^{n+1}\right]_0^1 = \frac{n}{2(n+1)} \to \frac12 \neq 0 = \int_0^1 0\,dx.$$

The convergence is far from uniform: the maximum of $f_n$, at $x = 1/\sqrt{2n + 1}$, grows without bound (about $1.34$ for $n = 10$ and $4.28$ for $n = 100$). A travelling spike keeps area $\approx \frac12$ while its base shrinks.

Derivatives need more

Uniform convergence of $f_n$ says nothing about $f_n'$. Two warnings:

• $g_n(x) = \sqrt{x^2 + 1/n^2} \to \lvert x\rvert$ uniformly (Chapter 1): smooth functions, non-differentiable limit.
• $h_n(x) = \frac{\sin(n^2 x)}{n} \to 0$ uniformly on $\mathbb{R}$ (the sup norm is $\frac1n$), but $h_n'(x) = n\cos(n^2 x)$, and $h_n'(0) = n \to \infty$. The derivatives do not converge to the derivative of the limit, which is $0$.

The fix is to assume uniform convergence of the derivatives.

Theorem (limits of derivatives). Let each $f_n$ be differentiable on $[a, b]$. Suppose $(f_n(x_0))$ converges for some $x_0 \in [a, b]$, and $(f_n')$ converges uniformly on $[a, b]$. Then $(f_n)$ converges uniformly on $[a, b]$ to a function $f$, and $f'(x) = \lim_{n \to \infty} f_n'(x)$ for every $x \in [a, b]$. (Rudin, Theorem 7.17.)

Idea of proof (with the extra assumption that each $f_n'$ is continuous). By FTC 2 (post 04), $f_n(x) = f_n(x_0) + \int_{x_0}^x f_n'$. The integrals converge uniformly by the integral theorem above, so $f_n \to f(x) = f(x_0) + \int_{x_0}^x g$ with $g = \lim f_n'$, which is continuous as a uniform limit. FTC 1 gives $f' = g$. The general proof uses the Mean Value Theorem instead of integrals.

Watch out: "uniformly" in this theorem is about the derivatives. Uniform convergence of $f_n$ alone never licenses swapping $\lim$ and $\frac{d}{dx}$.

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Chapter 3 — Series of functions and the Weierstrass M-test

A series of functions $\sum_{k=1}^{\infty} u_k(x)$ converges pointwise or uniformly on $A$ when its partial sums $S_n(x) = \sum_{k=1}^{n} u_k(x)$ do. The three theorems of Chapter 2 become:

If the series converges uniformly on the interval and	then
each $u_k$ is continuous	the sum is continuous
each $u_k$ is integrable on $[a, b]$	$\int \sum u_k = \sum \int u_k$ (integrate term by term)
each $u_k$ is differentiable, $\sum u_k'$ converges uniformly, $\sum u_k(x_0)$ converges	$(\sum u_k)' = \sum u_k'$ (differentiate term by term)

Theorem (Weierstrass M-test). Suppose $\lvert u_k(x)\rvert \leq M_k$ for all $x \in A$ and all $k$, and $\sum M_k$ converges. Then $\sum u_k$ converges uniformly (and absolutely) on $A$.

Proof. For $n > m$ and every $x \in A$,

$$\lvert S_n(x) - S_m(x)\rvert = \left\lvert \sum_{k=m+1}^{n} u_k(x)\right\rvert \leq \sum_{k=m+1}^{n} M_k.$$

The right side does not depend on $x$, and it is small for large $m$ because $\sum M_k$ converges (Cauchy criterion for number series, post 01). So the partial sums satisfy the uniform Cauchy criterion of Chapter 1. $\blacksquare$

Worked example. $F(x) = \sum_{k=1}^{\infty} \frac{\sin(kx)}{k^2}$. Take $M_k = \frac{1}{k^2}$; $\sum \frac1{k^2}$ converges (its value is $\frac{\pi^2}{6}$). So the series converges uniformly on $\mathbb{R}$ and $F$ is continuous. Term-by-term differentiation would give $\sum \frac{\cos(kx)}{k}$, which diverges at $x = 0$ (harmonic series). So the M-test gives continuity, and says nothing about differentiability.

The same test proves the continuity of the Weierstrass function in post 03: $\lvert a^n\cos(b^n\pi x)\rvert \leq a^n$ with $\sum a^n$ a convergent geometric series.

Common pitfall: the M-test is sufficient, not necessary. $\sum_{k \geq 1} \frac{(-1)^{k+1}}{k}$ (constant functions) converges uniformly, trivially, but no summable $M_k \geq \frac1k$ exists.

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Chapter 4 — Power series

Terminology

• Power series about $x_0$ — a series $\sum_{k=0}^{\infty} a_k (x - x_0)^k$ with real coefficients $a_k$. Below, $x_0 = 0$ for simplicity.
• Radius of convergence — the number $R \in [0, \infty]$ such that the series converges absolutely for $\lvert x\rvert < R$ and diverges for $\lvert x\rvert > R$. At $x = \pm R$ anything can happen.
• Limit superior — $\limsup_{k} b_k = \lim_{n \to \infty} \sup_{k \geq n} b_k$, which always exists in $[-\infty, \infty]$ (monotone convergence, post 01).

Radius of convergence

Theorem (Cauchy–Hadamard). Let $\beta = \limsup_{k \to \infty} \lvert a_k\rvert^{1/k}$. Then the radius of convergence is $R = 1/\beta$ (with $R = \infty$ if $\beta = 0$ and $R = 0$ if $\beta = \infty$).

Proof. Apply the root test (post 01) to $\sum \lvert a_k x^k\rvert$: $\limsup \lvert a_k x^k\rvert^{1/k} = \beta\lvert x\rvert$. If $\beta\lvert x\rvert < 1$ the series converges absolutely; if $\beta\lvert x\rvert > 1$ the terms do not tend to $0$, so it diverges. $\blacksquare$

Ratio formula. If $\lim \left\lvert \frac{a_k}{a_{k+1}}\right\rvert$ exists (in $[0, \infty]$), it equals $R$. When it applies it is easier; when it does not, the root formula still works.

Series	$R$	At $x = R$	At $x = -R$
$\sum x^k$	$1$	diverges	diverges
$\sum x^k / k$ (from $k = 1$)	$1$	diverges (harmonic)	converges (alternating)
$\sum x^k / k^2$ (from $k = 1$)	$1$	converges	converges
$\sum x^k / k!$	$\infty$	—	—
$\sum k!\,x^k$	$0$	—	—

The first three show that all endpoint behaviours occur with the same radius.

When the ratio test fails. Let $a_k = 2^k$ for even $k$ and $a_k = 1$ for odd $k$. The ratios $\lvert a_k / a_{k+1}\rvert$ alternate between values tending to $0$ and to $\infty$, so they have no limit. But $\lvert a_k\rvert^{1/k}$ is $2$ for even $k$ and $1$ for odd $k$, so $\limsup = 2$ and $R = \frac12$.

Uniform convergence inside the radius

Theorem. If $0 < r < R$, the power series converges uniformly on $[-r, r]$. Consequently it converges uniformly on every compact subset of $(-R, R)$, and its sum is continuous on $(-R, R)$.

Proof. Since $r < R$, the series converges absolutely at $x = r$, so $\sum \lvert a_k\rvert r^k$ converges. On $[-r, r]$, $\lvert a_k x^k\rvert \leq \lvert a_k\rvert r^k = M_k$. Apply the M-test. A compact $K \subset (-R, R)$ is closed and bounded, so its max of $\lvert x\rvert$ is some $r < R$ and $K \subseteq [-r, r]$. Every $x \in (-R, R)$ lies in some $[-r, r]$, and continuity is local. $\blacksquare$

Watch out: the convergence need not be uniform on the whole open interval. For $\sum x^k$ on $(-1, 1)$ the tail after $x^n$ is $\sum_{k > n} x^k = \frac{x^{n+1}}{1 - x}$, which is unbounded as $x \to 1^-$ for every $n$.

Term-by-term differentiation and integration

Theorem. Let $f(x) = \sum_{k=0}^{\infty} a_k x^k$ with radius $R > 0$. Then on $(-R, R)$:

$$f'(x) = \sum_{k=1}^{\infty} k\,a_k x^{k-1}, \qquad \int_0^x f(t)\,dt = \sum_{k=0}^{\infty} \frac{a_k}{k+1} x^{k+1},$$

and both new series also have radius $R$. Hence $f$ is infinitely differentiable on $(-R, R)$, and $a_k = \frac{f^{(k)}(0)}{k!}$: a power series is the Taylor series of its own sum.

Proof sketch. Since $k^{1/k} \to 1$, $\limsup (k\lvert a_k\rvert)^{1/k} = \limsup \lvert a_k\rvert^{1/k}$, so the differentiated series has the same radius (similarly for the integrated one). On any $[-r, r]$ with $r < R$ the differentiated series converges uniformly, so the derivative theorem (Chapter 2) applies. Integration follows from the integral theorem. Repeating gives all derivatives, and evaluating the $k$-th derivative at $0$ leaves only $k!\,a_k$. $\blacksquare$

Worked example 1. From $\frac{1}{1 - x} = \sum_{k \geq 0} x^k$ on $(-1, 1)$, differentiate: $\frac{1}{(1 - x)^2} = \sum_{k \geq 1} k x^{k-1}$. Multiply by $x$ and put $x = \frac12$:

$$\sum_{k=1}^{\infty} \frac{k}{2^k} = \frac{1/2}{(1 - 1/2)^2} = 2.$$

Worked example 2. From $\frac{1}{1 + t} = \sum_{k \geq 0} (-1)^k t^k$ on $(-1, 1)$, integrate from $0$ to $x$:

$$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}x^k, \qquad \lvert x\rvert < 1.$$

Worked example 3. $E(x) = \sum_{k \geq 0} \frac{x^k}{k!}$ has $R = \infty$. Differentiating term by term gives $E' = E$, with $E(0) = 1$. So $E(x) = e^x$ (the function $E(x)e^{-x}$ has zero derivative, hence is constant, by post 03).

Abel's theorem at the edge

The formula for $\ln(1 + x)$ was proved for $\lvert x\rvert < 1$. At $x = 1$ the series $1 - \frac12 + \frac13 - \dots$ converges (alternating series test, post 01), but does it converge to $\ln 2$?

Theorem (Abel). Suppose $\sum_{k=0}^{\infty} a_k$ converges. Then $f(x) = \sum a_k x^k$ converges for $\lvert x\rvert < 1$ and

$$\lim_{x \to 1^-} f(x) = \sum_{k=0}^{\infty} a_k.$$

Proof. Let $s = \sum a_k$. Replacing $a_0$ by $a_0 - s$ changes $f$ by the constant $-s$ and makes the sum $0$, so assume $s = 0$. Let $s_n = a_0 + \dots + a_n$; then $s_n \to 0$, so $(s_n)$ is bounded and $\sum s_k x^k$ converges for $\lvert x\rvert < 1$. Summation by parts ($a_k = s_k - s_{k-1}$) gives, for $0 < x < 1$,

$$f(x) = \sum_{k=0}^{\infty} a_k x^k = (1 - x)\sum_{k=0}^{\infty} s_k x^k.$$

Given $\varepsilon > 0$, choose $N$ with $\lvert s_k\rvert < \varepsilon$ for $k > N$. Then

$$\lvert f(x)\rvert \leq (1 - x)\sum_{k=0}^{N}\lvert s_k\rvert + \varepsilon(1 - x)\sum_{k > N} x^k \leq (1 - x)\sum_{k=0}^{N}\lvert s_k\rvert + \varepsilon.$$

The first term tends to $0$ as $x \to 1^-$, so $\lvert f(x)\rvert < 2\varepsilon$ for $x$ close enough to $1$. Hence $f(x) \to 0 = s$. $\blacksquare$

Apply it to $a_k = \frac{(-1)^{k+1}}{k}$: the left side is $\lim_{x \to 1^-}\ln(1 + x) = \ln 2$ by continuity of $\ln$. So

$$1 - \frac12 + \frac13 - \frac14 + \dots = \ln 2 \approx 0.693147.$$

The same argument with $\arctan x = \sum_{k \geq 0} \frac{(-1)^k x^{2k+1}}{2k+1}$ gives $1 - \frac13 + \frac15 - \dots = \frac{\pi}{4}$. (These series converge slowly: the first 10 terms of the $\ln 2$ series sum to $0.6456$.)

Common pitfall: Abel's theorem needs the series at $x = 1$ to converge. For $a_k = (-1)^k$, $f(x) = \frac{1}{1 + x} \to \frac12$ as $x \to 1^-$, but $\sum (-1)^k$ diverges; the limit $\frac12$ is not a sum in the ordinary sense.

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Chapter 5 — Taylor series, approximation, and what comes next

Taylor series versus the function

For a $C^\infty$ function $f$, the Taylor series at $0$ is $\sum \frac{f^{(k)}(0)}{k!}x^k$. Two things can go wrong:

1. The series can have radius $0$ (there are $C^\infty$ functions like this).
2. The series can converge, to a different function.

Case 2 is the function of post 03: $f(x) = e^{-1/x^2}$ with $f(0) = 0$. All derivatives at $0$ vanish, so its Taylor series is $0 + 0x + 0x^2 + \dots$, with radius $\infty$ and sum $0$. It equals $f$ only at $x = 0$.

The link to Taylor's theorem is the remainder: the Taylor series of $f$ converges to $f(x)$ exactly when $R_n(x) \to 0$. For $e^x$ at any $x$, the Lagrange remainder satisfies $\lvert R_n(x)\rvert \leq e^{\lvert x\rvert}\frac{\lvert x\rvert^{n+1}}{(n+1)!} \to 0$, so $e^x$ equals its Taylor series everywhere. A function is real analytic at $x_0$ when it equals a convergent power series on some interval around $x_0$. By the term-by-term theorem, such a power series must be the Taylor series. So: analytic ⇒ $C^\infty$, but not conversely.

The Weierstrass approximation theorem

Not every continuous function is a power series, but every continuous function is uniformly close to a polynomial.

Theorem (Weierstrass, 1885). Let $f$ be continuous on $[a, b]$. For every $\varepsilon > 0$ there is a polynomial $p$ with $\lvert f(x) - p(x)\rvert < \varepsilon$ for all $x \in [a, b]$.

Bernstein's constructive proof (idea). On $[0, 1]$, define

$$B_n f(x) = \sum_{k=0}^{n} f\!\left(\frac{k}{n}\right)\binom{n}{k} x^k (1 - x)^{n-k}.$$

The weights $\binom{n}{k}x^k(1-x)^{n-k}$ are the binomial probabilities of $k$ successes in $n$ trials with success probability $x$. They add up to $1$ and concentrate near $k/n \approx x$ as $n$ grows, so $B_n f(x)$ is a weighted average of values of $f$ near $x$. Uniform continuity of $f$ (post 02) turns this into $\lVert B_n f - f\rVert_\infty \to 0$.

Worked example. For $f(x) = x^2$ a direct computation gives

$$B_n f(x) = x^2 + \frac{x(1 - x)}{n},$$

so the error is $\frac{x(1-x)}{n}$, largest at $x = \frac12$, and $\lVert B_n f - f\rVert_\infty = \frac{1}{4n}$. An error of at most $0.01$ needs $n \geq 25$. The convergence is uniform but slow; Bernstein polynomials are valued for their shape-preserving properties (they are the basis of Bézier curves), not their speed.

Compare with the Weierstrass function of post 03: a continuous, nowhere-differentiable function can still be approximated uniformly, to any accuracy, by polynomials, which are as smooth as functions get. Uniform limits keep continuity and lose everything finer.

A look ahead

• Metric spaces. Uniform convergence is convergence in the metric $d(f, g) = \lVert f - g\rVert_\infty$ on $C[a, b]$. Chapter 1's Cauchy criterion, together with the continuity theorem of Chapter 2, says this space is complete, and Weierstrass says polynomials are dense in it. The same words (complete, dense, compact) from posts 00–02 now describe spaces of functions.
• Lebesgue integration. The spike example $n x(1 - x^2)^n$ shows that pointwise limits and integrals do not commute in general. Lebesgue's theory says exactly when they do (monotone and dominated convergence), without asking for uniformity.
• Fourier series. Replacing powers $x^k$ by $\cos(kx)$ and $\sin(kx)$ gives Fourier series. Their convergence is subtler than for power series: the Fourier series of a discontinuous square wave converges pointwise at every point but never uniformly, which ties back to the continuity theorem of Chapter 2.

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Practice Exercises

Exercise 1. Show that $f_n(x) = \frac{x}{n}$ converges uniformly on $[0, 1]$ but not on $\mathbb{R}$.

Solution

The pointwise limit is $0$. On $[0, 1]$, $\lVert f_n\rVert_\infty = \frac1n \to 0$: uniform. On $\mathbb{R}$, $\sup_x \lvert x/n\rvert = \infty$ for every $n$ (take $x = n$ to get $f_n(n) = 1$): not uniform.

Exercise 2. Let $f_n(x) = x^n$. Prove that $f_n \to 0$ uniformly on $[0, r]$ for each $r < 1$, and explain in one line why it cannot be uniform on $[0, 1)$ even though the limit $0$ is continuous there.

Solution

On $[0, r]$, $\lVert x^n\rVert_\infty = r^n \to 0$. On $[0, 1)$, the points $x_n = (1/2)^{1/n}$ give $x_n^n = \frac12$, so $\sup_{[0,1)} x^n \geq \frac12$ for every $n$. (The continuity theorem only says a discontinuous limit rules out uniformity; a continuous limit does not guarantee it.)

Exercise 3. Use the M-test to show that $\sum_{k=1}^{\infty} \frac{x^k}{k^2}$ converges uniformly on $[-1, 1]$. Is its sum continuous at $x = 1$?

Solution

For $\lvert x\rvert \leq 1$, $\left\lvert \frac{x^k}{k^2}\right\rvert \leq \frac1{k^2} = M_k$, and $\sum \frac1{k^2}$ converges. So the series converges uniformly on $[-1, 1]$ and the sum is continuous there, including at the endpoint $x = 1$ (one-sided).

Exercise 4. Find the radius of convergence of $\sum_{k \geq 1} \frac{k^2}{3^k}x^k$ and of $\sum_{k \geq 0} \frac{(2k)!}{(k!)^2}x^k$.

Solution

First: $\lvert a_k\rvert^{1/k} = \frac{k^{2/k}}{3} \to \frac13$, so $R = 3$. Second, by the ratio formula: $\frac{a_{k+1}}{a_k} = \frac{(2k+2)(2k+1)}{(k+1)^2} \to 4$, so $R = \frac14$.

Exercise 5. Compute $\sum_{k \geq 1} \frac{k}{3^k}$ using term-by-term differentiation.

Solution

From $\sum_{k \geq 1} k x^{k} = \frac{x}{(1 - x)^2}$ for $\lvert x\rvert < 1$, put $x = \frac13$: $\frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac34$.

Exercise 6. For $f_n(x) = n x (1 - x^2)^n$ on $[0, 1]$, compute $\int_0^1 f_n$ for $n = 10$ and explain why this does not contradict the integral theorem.

Solution

$\int_0^1 f_n = \frac{n}{2(n+1)}$, so for $n = 10$ it is $\frac{10}{22} = \frac{5}{11} \approx 0.4545$. The pointwise limit is $0$ with integral $0$. No contradiction: the convergence is not uniform (the maximum of $f_{10}$ is about $1.34$ and the maxima grow without bound), so the theorem does not apply.

Exercise 7. Show that the Bernstein polynomial $B_n f$ of $f(x) = x$ is $x$ itself for every $n$.

Solution

$B_n f(x) = \sum_{k=0}^{n} \frac{k}{n}\binom{n}{k}x^k(1 - x)^{n-k}$. This is $\frac1n$ times the mean of a binomial distribution with $n$ trials and success probability $x$, which is $nx$. So $B_n f(x) = x$. (Bernstein polynomials reproduce linear functions exactly; the first error appears for $x^2$.)

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Cheat-sheet

Idea	Statement
Pointwise	for each $x$, $f_n(x) \to f(x)$; $N$ may depend on $x$
Uniform	$\lVert f_n - f\rVert_\infty \to 0$; one $N$ for all $x$
Key example	$x^n$ on $[0,1]$: pointwise, not uniform; uniform on $[0, r]$, $r < 1$
Uniform Cauchy	$\lvert f_n(x) - f_m(x)\rvert < \varepsilon$ for all $x$, $m, n \geq N$
Continuity	uniform limit of continuous is continuous ($\varepsilon/3$)
Integrals	uniform on $[a,b]$ $\Rightarrow \int f_n \to \int f$
Derivatives	need $f_n'$ uniform plus one convergent $f_n(x_0)$
M-test	$\lvert u_k\rvert \leq M_k$, $\sum M_k < \infty \Rightarrow$ uniform
Radius	$1/R = \limsup \lvert a_k\rvert^{1/k}$; or $R = \lim \lvert a_k/a_{k+1}\rvert$
Inside the radius	uniform on $[-r, r]$ for $r < R$; not always on $(-R, R)$
Term by term	$f'$ and $\int f$ by term, same $R$; $a_k = f^{(k)}(0)/k!$
Abel	$\sum a_k$ converges $\Rightarrow \lim_{x \to 1^-}\sum a_k x^k = \sum a_k$
Smooth, not analytic	$e^{-1/x^2}$: Taylor series is $0$
Weierstrass approximation	continuous on $[a,b]$ ⇒ uniform limit of polynomials
Bernstein for $x^2$	error $x(1-x)/n$, sup $1/(4n)$

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Course core complete — what's next

This post completes the core course: the real numbers (00), limits of sequences and series (01), topology and continuity (02), derivatives (03), integrals (04), and limits of functions (05). The series is built to grow. The posts below are planned, in roughly this order.

Planned future extensions

Post	Topic	What it covers
06	Metric spaces and completeness	open and closed sets, compactness and sequential compactness, completeness of $C[a,b]$, the contraction mapping theorem and Picard's existence theorem for ODEs
07	Multivariable analysis	the derivative as a linear map, the chain rule in $\mathbb{R}^n$, the inverse and implicit function theorems
08	Measure theory and the Lebesgue integral	outer measure, measurable functions, monotone and dominated convergence, why $L^1$ is complete
09	Fourier analysis	Fourier series, Dirichlet and Fejér kernels, pointwise and $L^2$ convergence, Parseval's identity
10	Complex analysis	holomorphic functions, Cauchy's theorem and integral formula, why complex-differentiable implies analytic
11	Functional analysis	normed and Banach spaces, Hilbert spaces, bounded operators, the Arzelà–Ascoli theorem

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References

• Stephen Abbott, Understanding Analysis, 2nd ed., Springer (Undergraduate Texts in Mathematics), 2015 — Chapter 6, "Sequences and Series of Functions" (uniform convergence, power series, Taylor series, the Weierstrass approximation theorem).
• Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976 — Chapter 7 (Theorems 7.8 Cauchy criterion, 7.10 M-test, 7.12 continuity, 7.13 Dini, 7.16 integration, 7.17 differentiation, 7.26 Weierstrass) and Chapter 8 (power series, Abel's theorem 8.2).
• Robert G. Bartle and Donald R. Sherbert, Introduction to Real Analysis, 4th ed., Wiley, 2011 — Chapter 8 (sequences of functions) and Section 9.4 (series of functions and power series).
• Michael Spivak, Calculus, 4th ed., Publish or Perish, 2008 — Chapters 23–24 (infinite series; uniform convergence and power series) and Chapter 20 (Taylor polynomials).
• Jiří Lebl, Basic Analysis I: Introduction to Real Analysis, free textbook — jirka.org/ra, Chapter 6.
• MIT OpenCourseWare, 18.100A Real Analysis (Fall 2020), lecture notes on uniform convergence, power series and the Weierstrass approximation theorem.

Note: this post is in progress. Planned additions: Tauber's partial converse of Abel's theorem, a proof of the Weierstrass approximation theorem via Bernstein polynomials with the Chebyshev inequality, the Arzelà–Ascoli theorem, and a worked Fourier-series example showing the Gibbs phenomenon.