Sequences and series of functions

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Chapter 7: Sequences and series of functions In the present chapter we confine our attention to complex-valued functions (including the real-valued ones, of course), although many of the theorems and proofs which follow extend without difficulty to vector-valued functions, and even to mappings into general metric spaces. We choose to stay within this simple framework in order to focus attention on the most important aspects of the problems that arise when limit processes are interchanged.

Contents
Contents
 1.  Discussion of main problem
 2.  Uniform cnovergence
 3.  Uniform convergence and continuity
 4.  Uniform convergence and integration
 5.  Uniform convergence and differentiation
 6.  Equicontinuous families of functions
 7.  The Stone-Weierstrass theorem

1. Discussion of main problem The main problem which arises is to determine whether important properties of functions are preserved under the limit operations. For instance, if the functions f _ n are continuous, or differentiable, or integrable, is the same true of the limit function? What are the relations between f^\prime _ n and f^\prime, say, or between the integrals of f _ n and that of f?

To say that f is continuous at a limit point x means \displaystyle\lim _ {t\to x}f(t)=f(x). Hence, to ask whether the limit of a sequence of continuous functions is continuous is the same as to ask whether \displaystyle \lim _ {t\to x}\lim _ {n\to\infty}f _ n(t)=\lim _ {n\to\infty}\lim _ {t\to x}f _ n(t), i.e., whether the order in which limit processes are carried out is immaterial.

We shall now show by means of several examples that limit processes cannot in general be interchanged without affecting the result. Afterward, we shall prove that under certain conditions the order in which limit operations are carried out is immaterial.

Our first example, and the simplest one, concerns a "double sequence".

{Example}

{Example 1} Let s _ {m,n}=\dfrac{m}{m+n}. Then \displaystyle\lim _ {n\to\infty}\lim _ {m\to\infty}s _ {m,n}=1. On the other hand, \displaystyle\lim _ {m\to\infty}\lim _ {n\to\infty}s _ {m,n}=0.

{Example 2} Let f(x)=x^2/(1+x^2)^n\, (x\in\mathbb{R}) and consider f(x)=\sum f _ n(x). Then
f(x)=\begin{cases}
0,&x=0 \\
1+x^2,&x\neq0.
\end{cases}
so that a convergent series of continuous functions may have a discontinuous sum.

{Example 3}
\[\lim _ {m\to\infty}\lim _ {n\to\infty}(\cos m!\pi x)^{2n}=\begin{cases}
0, & x \mbox{ irrational}, \\
1, & x \mbox{ ratinoal}.
\end{cases}\]
We have thus obtained an everywhere discontinuous limit function, which is not Riemann-integrable.

{Example 4} Let f _ n(x)=\sin nx/\sqrt{n}\, (x\in\mathbb{R}), and f(x)=\lim f _ n(x)=0. Then f^\prime(x)=0, and f^\prime(x)=\sqrt{n}\cos nx, so that \{f^\prime _ n\} does not converge to f^\prime.

{Example 5} Let f _ n(x)=n^2x(1-x^2)^n\, (0\leqslant x\leqslant1). We have \lim f _ n(x)=0, but \int f _ n(x)\, dx=n^2/(2n+2)\to\infty.

If f _ n(x)=nx(1-x^2)^n\, (0\leqslant x\leqslant1), \lim f _ n(x)=0 still holds, but now we have
\[\lim _ {n\to\infty}\int _ {0}^{1}f _ n(x)\, dx=\lim _ {n\to\infty}\frac{n}{2n+2}=\frac{1}{2},\quad \int _ {0}^{1}\lim _ {n\to\infty}f _ n(x)\, dx=0.\]

Thus the limit of the integral need not be equal to the integral of the limit,
even if both are finite.

After these examples, which show what can go wrong if limit processes are interchanged carelessly, we now define a new mode of convergence, stronger than pointwise convergence, which will enable us to arrive at positive results.
2. Uniform cnovergence The Cauchy criterion for uniform convergence is as follows.

The following criterion is sometimes useful.

For series, there is a very convenient test for uniform convergence, due to Weierstrass.

Note that the converse is not asserted (and is, in fact, not true).
3. Uniform convergence and continuity

This very important result is an immediate corollary of Theorem theorem731. The converse is not true, that is, a sequence of continuous functions may converge to a continuous function, although the convergence is not uniform. But there is a case in which we can assert the converse.

Proof. Put g _ n=f _ n-f. Then g _ n is continuous, and g _ n\to0 pointwise, and g _ n\geqslant g _ {n+1}. We have to prove that g _ n\to0 uniformly on K. Let \varepsilon>0 be given. Let K _ n be the set of all x\in K with g _ n(x)\geqslant\varepsilon. Then K _ n must be compact. We can show that x\notin K _ n if n is sufficiently large, so \bigcap K _ n is empty. Hence K _ N is empty for some N. It follows that 0\leqslant g _ n(x)<\varepsilon for all x\in K and for all n\geqslant N.

Let us note that compactness is really needed here. For instance, if f _ n(x)=1/(nx+1)\, (0<x<1) then f _ n(x)\to0 monotonically in (0,1), but the convergence is not uniform.

[Note that boundedness is redundant if X is compact. Thus \mathcal{C}(X) consists of all complex continuous functions on X if X is compact.]

We associate with each f\in\mathcal{C}(X) its supremum norm |f|=\sup _ {x\in X}|f(x)|. Since f is assumed to be bounded, |f|<\infty. If we define the distance between f\in\mathcal{C}(X) and g\in\mathcal{C}(X) to be |f-g|, it follows that we have made \mathcal{C}(X) into a metric space.

Theorem theorem722 can be rephrased as follows:

A sequence \{f _ n\} converges to f with respect to the metric of \mathcal{C}(X) if and only if f _ n\to f uniformly on X.

Accordingly, closed subsets of \mathcal{C}(X) are sometimes called uniformly closed, the closure of a set \mathcal{A}\subset\mathcal{C}(X) is called its uniform closure, and so on.

We have already seen, in Example 4, that uniform convergence of \{f _ n\} implies nothing about the sequence \{f^\prime _ n\}. Thus stronger hypotheses are required for the assertion that f^\prime _ n\to f^\prime if f _ n\to f.

If the continuity of the functions f^\prime _ n is assumed in addition to the above hypotheses, then a much shorter proof can be based on Theorem theorem741 and the fundamental theorem of calculus.

Proof. Let \varepsilon> 0 be given. Choose N such that n \geqslant N, m \geqslant N, implies |f _ n(x _ 0)-f _ m(x _ 0)|<{\varepsilon}/{2} and |f^\prime _ n(t)-f^\prime _ m(t)|<\varepsilon/[2(b-a)]\, (a\leqslant t\leqslant b). If we apply the mean value theorem to the function f _ n-f _ m, it shows that
\[|f _ n(x)-f _ m(x)-f _ n(t)+f _ m(t)|\leqslant\frac{|x-t|\varepsilon}{2(b-a)}\leqslant\frac{\varepsilon}{2}\]
for any x and t on [a,b], if n \geqslant N, m \geqslant N.
\begin{align*}
\phantom{\Longrightarrow} & |f _ n(x)-f _ m(x)|\leqslant|f _ n(x)-f _ m(x)-f _ n(x _ 0)+f _ m(x _ 0)|+|f _ n(x _ 0)-f _ m(x _ 0)| \\
\Longrightarrow & |f _ n(x)-f _ m(x)|<\varepsilon\quad(a\leqslant x\leqslant b,\, n\geqslant N,\, m\geqslant N),
\end{align*}
so that \{f _ n\} converges uniformly on [a,b]. Let f(x)=\lim f _ n(x)\, (a\leqslant x\leqslant b). Let us now fix a point x on [a,b] and define
\[\phi _ n(t)=\frac{f _ n(t)-f _ n(x)}{t-x},\quad\phi(t)=\frac{f(t)-f(x)}{t-x}\]
for a\leqslant t\leqslant b, t\neq x. Then \lim _ {t\to x}\phi _ n(t)=f^\prime _ n(x), and
\[|\phi _ n(t)-\phi _ m(t)|\leqslant\frac{\varepsilon}{2(b-a)}\quad (n\geqslant N,\, m\geqslant N),\]
so that \{\phi _ n\} converges uniformly, for t\neq x. Since \{f _ n\} converges to f, we conclude that \lim\phi _ n(t)=\phi(t) uniformly for a\leqslant t\leqslant b, t\neq x.

If we now apply Theorem theorem731 to \{\phi _ n\}, we have
\[\lim _ {t\to x}\phi(t)=\lim _ {n\to\infty}f' _ n(x);\]
and this is what we want to prove.

Proof. Define \varphi(x)=|x|\, (-1\leqslant x\leqslant1) and extend the definition of \varphi(x) to all real x by requiring that \varphi(x+2)=\varphi(x). Then, for all s and t, |\varphi(s)-\varphi(t)|\leqslant|s-t|. In particular, \varphi is continuous on \mathbb{R}^1. Define
\[f(x)=\sum _ {n=0}^{\infty}\Big(\frac34\Big)^n\varphi(4^nx).\]
This series converges uniformly on \mathbb{R}^1, so f is continuous on \mathbb{R}^1.

Now fix a real number x and a positive integer m. Put \delta _ m=\pm\frac12\cdot4^{-m} where the sign is so chosen that no integer lies between 4^mx and 4^m(x+\delta _ m). This can be done, since 4^m|\delta _ m|=\frac12. Define \gamma _ n=[\varphi(4^n(x+\delta _ m))-\varphi(4^nx)]/\delta _ m. When n>m, then 4^n\delta _ m is an even integer, so that \gamma _ n=0. When 0\leqslant n\leqslant m, we have |\gamma _ n|\leqslant 4^n. Since |\gamma _ m|=4^m, we conclude that
\[\Big|\frac{f(x+\delta _ m)-f(x)}{\delta _ m}\Big|=\Big|\sum _ {n=0}^{m}\Big(\frac34\Big)^n\gamma _ n\Big|\geqslant3^m-\sum _ {n=0}^{m-1}3^n= \frac12 (3^m+1).\]
As m\to\infty, \delta _ m\to0. It follows that f is not differentiable at x.
6. Equicontinuous families of functions In Chap. 3 we saw that every bounded sequence of complex numbers contains a convergent subsequence, and the question arises whether something similar is true for sequences of functions.

Now if \{f _ n\} is pointwise bounded on E and E _ 1 is a countable subset of E, it is always possible to find a subsequence \{f _ {n _ k}\} such that \{f _ {n _ k}\} converges for every x \in E _ 1; This can he done by the diagonal process which is used in the proof of Theorem Theorem761.

However, even if \{f _ n\} is a uniformly bounded sequence of continuous functions on a compact set E, there need not exist a subsequence which converges pointwise on E.

{Example}Let f _ n(x)=\sin nx. Suppose there exists a sequence \{n _ k\} such that \{\sin n _ kx\} converges, for every x\in[0,2\pi]. In that case we must have \lim(\sin n _ kx-\sin n _ {k+1}x)=0\, (0\leqslant x\leqslant 2\pi); hence \lim(\sin n _ kx-\sin n _ {k+1}x)^2=0\, (0\leqslant x\leqslant 2\pi). By Lebesgue's theorem concerning integration of boundedly convergent sequences, it implies
\[\lim _ {k\to\infty}\int _ {0}^{2\pi}(\sin n _ kx-\sin n _ {k+1}x)^2\, dx=0.\]
But a simple calculation shows that
\[\int _ {0}^{2\pi}(\sin n _ kx-\sin n _ {k+1}x)^2\, dx=2\pi.\]

Another question is whether every convergent sequence contains a uniformly convergent subsequence. Our next example will show that this need not be so, even if the sequence is uniformly bounded on a compact set (Example 5 shows that a sequence of bounded functions may converge without being uniformly bounded, but it is trivial to see that uniform convergence of a sequence of bounded functions implies uniform boundedness.)

{Example}Let f _ n(x)=x^2/[x^2+(1-nx)^2]\, (0\leqslant x\leqslant1). Then |f _ n(x)|\leqslant1, so that \{f _ n\} is uniformly bounded on [0, 1]. Also \lim f _ n(x)=0\, (0\leqslant x\leqslant1), but f _ n(1/n)=1, so that no subsequence can converge uniformly on [0,1].

The concept which is needed in this connection is that of equicontinuity, given in the following definition.

It is clear that every member of an equicontinuous family is uniformly continuous.

Theorems Theorem762 and Theorem763 will show that there is a very close relation between equicontinuity, on the one hand, and uniform convergence of sequences of continuous functions, on the other. But first we describe a selection process which has nothing to do with continuity.

Proof. Theorem Theorem761. Let \{x _ i\}, i = 1, 2, 3, \dots , be the points of E, arranged in a sequence. Since \{f _ n(x _ 1)\} is bounded, there exists a subsequence, which we shall denote by \{f _ {1,k}\}, such that \{f _ {1,k}(x _ 1)\} converges as k\to\infty. Let us now consider sequence S _ 1,S _ 2,S _ 3,\dots, which we represent by the array
\[
\begin{array}{cccccc}
S _ 1: & f _ {1,1} & f _ {1,2} & f _ {1,3} & f _ {1,4} & \cdots \\
S _ 2: & f _ {2,1} & f _ {2,2} & f _ {2,3} & f _ {2,4} & \cdots \\
S _ 3: & f _ {3,1} & f _ {3,2} & f _ {3,3} & f _ {3,4} & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots
\end{array}
\]
and which have the following properties:

(a) S _ n is a subsequence of S _ {n-1}, for n=2,3,4,\dots.

(b) \{f _ {n,k}(x _ n)\} converges, as k\to\infty (the boundedness of \{f _ n(x _ n)\} makes it possible to choose S _ n in this way).

(c) The order in which the functions appear is the same in each sequence; i.e., if one function precedes another in S _ 1, they are in the same relation in every S _ n , until one or the other is deleted. Hence, when going from one row in the above array to the next below, functions may move to the left but never to the right.

We now go down the diagonal of the array; i.e., we consider the sequence

\[S:\quad f _ {1,1}\quad f _ {2,2}\quad f _ {3,3}\quad f _ {4,4}\quad \cdots.\]

By (c), the sequence S (except possibly its first n - 1 terms) is a subsequence of S _ n, for n = 1, 2, 3, \dots. Hence (b) implies that \{f _ {n,n}(x _ i)\} converges, as n \to\infty, for every x _ i \in E.

Theorem Theorem762. Let \varepsilon>0 be given. Since \{f _ n\} converges uniformly, there is an integer N such that |f _ n-f _ N|<\varepsilon\, (n>N). Since continuous functions are uniformly continuous on compact sets, there is a \delta > 0 such that |f _ i(x)-f _ i(y)|<\varepsilon if 1\leqslant i\leqslant N and d(x,y)<\delta. If n>N and d(x,y)<\delta, it follows that |f _ n(x)-f _ n(y)|\leqslant|f _ n(x)-f _ N(x)|+|f _ N(x)-f _ N(y)|+|f _ N(y)-f _ n(y)|<3\varepsilon. The combination of these results proves the theorem.

Theorem Theorem763. (a) Let \varepsilon>0 be given and choose \delta>0 so that |f _ n(x)-f _ n(y)|<\varepsilon for all n, provided that d(x,y)<\delta. Since K is compact, there are finitely many points p _ 1,\dots,p _ r in K such that to every x\in K corresponds at least one p _ i with d(x,p _ i)<\delta. Since \{f _ n\} is pointwise bounded, there exist M _ i<\infty such that |f _ n(p _ i)|<M _ i for all n. If M =\max (M _ 1, \dots , M _ r), then |f _ n(x)|<M + \varepsilon for every x \in K.

(b) We firstly give some definitions and lemma.

Let E be a countable dense subset of K. Theorem Theorem761 shows that \{f _ n\} has a subsequence \{f _ {n _ i}\} such that \{f _ {n _ i}(x)\} converges for every x\in E. Put f _ {n _ i}=g _ i, to simplify the notation. We shall prove that \{g _ i\} converges uniformly on K.

Let \varepsilon>0, and pick \delta>0 as in the beginning of this proof. Let V(x,\delta) be the set of all y\in K with d(x,y)<\delta. Since E is dense in K, and K is compact, there are finitely many points x _ 1, \dots , x _ m in E such that K\subset V(x _ 1,\delta)\cup\dots\cup V(x _ m,\delta). Since \{g _ i(x)\} converges for every x\in E, there is an integer N such that |g _ i(x _ s)-g _ j(x _ s)|<\varepsilon whenever i\geqslant N, j\geqslant N, 1\leqslant s\leqslant m. If x\in K, then x\in V(x _ s,\delta) for some s, so that |g _ i(x)-g _ j(x _ s)|<\varepsilon for every i. If i\geqslant N and j\geqslant N,
\[|g _ i(x)-g _ j(x)|\leqslant|g _ i(x)-g _ i(x _ s)|+|g _ i(x _ s)-g _ j(x _ s)|+|g _ j(x _ s)-g _ j(x)|<3\varepsilon.\]
This completes the proof.
7. The Stone-Weierstrass theorem The theorem originally discovered by Weierstrass is the following form.

Proof. We may assume, without loss of generality, that [a, b]= [0, 1]. We may also assume that f(0)=f(1)=0. For if the theorem is proved for this case, consider g(x)=f(x)-f(0)-x[f(1)-f(0)]\, (0\leqslant x\leqslant1). Here g(0)=g(1)=0, and if g can be obtained as the limit of a uniformly convergent sequence of polynomials, it is clear that the same is true for f, since f-g is a polynomial.

Furthermore, we define f(x) to be zero for x outside [0, 1]. Then f is uniformly continuous on the whole line.

We put Q _ n(x)=c _ n(1-x^2)^n\, (n=1,2,3,\dots), where c _ n is chosen so that
\[\int _ {-1}^{1}Q _ n(x)\, dx=1\quad (n=1,2,3,\dots).\]
We need some information about the order of magnitude of c _ n. Since
\begin{align*}
\int _ {-1}^{1}(1-x^2)^n\, dx & =2\int _ {0}^{1}(1-x^2)^n\, dx\geqslant2\int _ {0}^{1/\sqrt{n}}(1-x^2)^n\, dx \\
& \geqslant2\int _ {0}^{1/\sqrt{n}}(1-nx^2)\, dx=\frac{4}{3\sqrt{n}}>\frac{1}{\sqrt{n}},
\end{align*}
so c _ n<\sqrt{n}.

For \delta>0, this implies Q _ n(x)\leqslant\sqrt{n}(1-\delta^2)^n\, (\delta\leqslant|x|\leqslant1), so that Q _ n\to0 uniformly in \delta\leqslant|x|\leqslant1.

Now set
\[P _ n(x)=\int _ {-1}^{1}f(x+t)Q _ n(t)\, dt=\int _ {0}^{1}f(t)Q _ n(t-x)\, dt\quad(0\leqslant x\leqslant1).\]
The last integral is clearly a polynomial in x. Thus \{P _ n\} is a sequence of polynomials, which are real if f is real.

Given \varepsilon>0, we choose \delta>0 such that |y-x|<\delta implies |f(y)-f(x)|<\varepsilon/2. Let M=\sup |f(x)|. Note that Q _ n(x)\geqslant0, we see that for 0\leqslant x\leqslant1,
\begin{align*}
|P _ n(x)-f(x)| & =\Big|\int _ {-1}^{1}[f(x+t)-f(x)]Q _ n(t)\, dt\Big|\leqslant\int _ {-1}^{1}|f(x+t)-f(x)|Q _ n(t)\, dt \\
& \leqslant2M\int _ {-1}^{\delta}Q _ n(t)\, dt+\frac{\varepsilon}{2}\int _ {-\delta}^{\delta}Q _ n(t)\, dt+2M\int _ {\delta}^{1}Q _ n(t)\, dt\\
& \leqslant 4M\sqrt{n}(1-\delta^2)^n+\frac\varepsilon2<\varepsilon
\end{align*}
for large enough n, which proves the theorem.

It is instructive to sketch the graphs of Q _ n for a few values of n. We can see that Q _ n are peaks around 0 (ignoring what happens outside [0,1]) that get narrower and narrower. A classic approximation idea is to do a convolution integral with peaks. As Q _ n is a peak, the integral only sees the values of f that are very close to x and it does a sort of average of them. When the peak gets narrower, we do this average closer to x and hence we expect to get
close to the value of f(x).

Also, note that we needed uniform continuity of f to deduce uniform convergence of \{P _ n\}.

In the proof of Theorem theorem774 we shall not need the full strength of Theorem theorem771, but only the following special case, which we state as a corollary.

We shall now isolate those properties of the polynomials which make the Weierstrass theorem possible.

For example, the set of all polynomials is an algebra, and the Weierstrass theorem may be stated by saying that the set of continuous functions on [a, b] is the uniform closure of the set of polynomials on [a, b].

The algebra of all polynomials in one variable clearly has these properties on \mathbb{R}^1. An example of an algebra which does not separate points is the set of all even polynomials, say on [-1, 1], since f(- x) = f(x) for every even function f.

The following theorem will illustrate these concepts further.

We now have all the material needed for Stone's generalization of the Weierstrass theorem.

Theorem theorem774 does not hold for complex algebras. However, the conclusion of the theorem does hold, even for complex algebras, if an extra condition is imposed on \mathcal{A}, namely, that \mathcal{A} be self-adjoint. This means that for every f\in\mathcal{A} its complex conjugate \bar f must also belong to \mathcal{A}.

Proof. Theorem theorem773. The assumptions show that \mathcal{A} contains functions g, h, and k such that g(x _ 1)\neq g(x _ 2), h(x _ 1)\neq0, k(x _ 2)\neq0. Put u=gk-g(x _ 1)k, v=gh-g(x _ 2)h. Then u\in\mathcal{A}, v\in\mathcal{A}, u(x _ 1)=v(x _ 2)=0, u(x _ 2)\neq0, and v(x _ 1)\neq0. Therefore f=\dfrac{c _ 1v}{v(x _ 1)}+\dfrac{c _ 2u}{u(x _ 2)} has the desired properties.

Theorem theorem774.

Step 1.\quad If f\in\mathcal{B}, then |f|\in\mathcal{B}.

Step 2.\quad If f\in\mathcal{B} and g\in\mathcal{B}, then \max(f,g)\in\mathcal{B} and \min(f,g)\in\mathcal{B}.

Step 3.\quad Given a real function f, continuous on K, a point x\in K, and \varepsilon > 0, there exists a function g _ x\in\mathcal{B} such that g _ x(x) =f(x) and g _ x(t)>f(t)-\varepsilon\, (t\in K).

Since \mathcal{A}\subset\mathcal{B} and \mathcal{A} satisfies the hypothesis of Theorem theorem773 so does \mathcal{B}. Hence, for every y\in K, we can find a function h _ y\in\mathcal{B} such that h _ y(x)=f(x), h _ y(y)=f(y). By the continuity of h _ y, there exists an open set J _ y, containing y, such that h _ y(t)>f(t)-\varepsilon\, (t\in J _ y). Since K is compact, there is a finite set of points y _ 1,\dots,y _ n such that K\subset J _ {y _ 1}\cup\dots\cup J _ {y _ n}. Put g _ x=\max(h _ {y _ 1},\dots,h _ {y _ n}). By step 2, g _ x\in\mathcal{B}, and the above relations show that g _ x has the other properties.

Step 4.\quad Given a real function f, continuous on K, and \varepsilon > 0, there exists a function h\in\mathcal{B} such that |h(x)-f(x)|<\varepsilon\, (x\in K).

Since \mathcal{B} is uniformly closed, this statement is equivalent to the conclusion of the theorem.

Let us consider the functions g _ x, for each x \in K, constructed in step 3. By the continuity of g _ x, there exist open sets V _ x containing x, such that g _ x(t)<f(t)+\varepsilon\, (t\in V _ x). Since K is compact, there exists a finite set of points x _ 1,\dots,x _ m such that K\subset V _ {x _ 1}\cup\dots\cup V _ {x _ m}. Put h=\min(g _ {x _ 1},\dots,g _ {x _ m}). By step 2, h\in\mathcal{B}, and we have h(t)>f(t)-\varepsilon\, (t\in K), h(t)<f(t)+\varepsilon\, (t\in K).

Theorem theorem775. Let \mathcal{A} _ \mathbb{R} be the set of all real functions on K which belong to \mathcal{A}.

If f\in\mathcal{A} and f=u+iv, with u,v real, then 2u=f+\bar f, and since \mathcal{A} is self-adjoint, we see that u\in\mathcal{A} _ \mathbb{R}. If x _ 1\neq x _ 2, there exists f\in\mathcal{A} such that f(x _ 1)=1, f(x _ 2)=0; hence 0=u(x _ 2)\neq u(x _ 1)=1, which shows that \mathcal{A} _ \mathbb{R} separates points on K. If x\in K, then g(x)\neq0 for some g\in\mathcal{A}, and there is a complex number \lambda such that \lambda g(x)>0; if f=\lambda g, f=u+iv, it follows that u(x)>0; hence \mathcal{A} _ \mathbb{R} vanishes at no points of K.

Thus \mathcal{A} _ \mathbb{R} satisfies the hypothesis of Theorem theorem774. It follows that every real continuous function on K lies in the uniform closure of \mathcal{A} _ \mathbb{R}, hence lies in \mathcal{B}. If f is a complex continuous function on K, f=u+iv, then u\in\mathcal{B}, v\in\mathcal{B}, hence f\in\mathcal{B}. This completes the proof.