Page 236 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Proof: The proof is essentially based on a diagonalization argument. The simplest case is of real-
valued functions on a closed and bounded interval:
Let I = [a, b]  R be a closed and bounded interval. If F is an infinite set of functions ƒ : I  R which
is uniformly bounded and equicontinuous, then there is a sequence f of elements of F such that
n
f converges uniformly on I.
n
Fix an enumeration {x } of rational numbers in I. Since F is uniformly bounded, the set of
i i = 1,2,3,...
points {f(x )} is bounded, and hence by the Bolzano-Weierstrass theorem, there is a sequence
1 f ÎF
{f } of distinct functions in F such that {f (x )} converges. Repeating the same argument for the
n1 n1 1
sequence of points {f (x )}, there is a subsequence {f } of {f } such that {f (x )} converges.
n1 2 n2 n1 n2 2
By mathematical induction this process can be continued, and so there is a chain of subsequences
{f } {f } . . .
n1 n2
such that, for each k = 1, 2, 3,…, the subsequence {f } converges at x ,...,x . Now form the diagonal
nk 1 k
subsequence {f} whose mth term f is the mth term in the mth subsequence {f }. By construction,
m nm
f converges at every rational point of I.
m
Therefore, given any  > 0 and rational x in I, there is an integer N = N(, x ) such that
k k
|f (x ) – f (x )| < /3, n, m  N.
n k m k
Since the family F is equicontinuous, for this fixed å and for every x in I, there is an open interval
U containing x such that
x
|f(s) – f(t)| < /3

for all f Î F and all s, t in I such that s, t ÎU .
x
The collection of intervals U , x ÎI, forms an open cover of I. Since I is compact, this covering
x
admits a finite subcover U , ..., U . There exists an integer K such that each open interval U ,
1 J j
1  j  J, contains a rational x with 1  k  K. Finally, for any t Î I, there are j and k so that t and
k
x belong to the same interval U. For this choice of k,
k j
|f (t) – f (t)|  |f (t) – f (x )| + |f (x ) – f (x )| + |f (x ) – f (t)| < /3 + /3 + /3
n m n n k n k m k m k m
for all n, m > N = max{N(, x ), ..., N(, x )}. Consequently, the sequence {f } is uniformly Cauchy,
1 K n
and therefore converges to a continuous function, as claimed. This completes the proof.
Theorem 1: Weierstrass Approximation Theorem
Let f: [a, b]   be continuous. Then there is a sequence of polynomials {P } ¥ = such that p  f
n n 1
n
uniformly.

Notes It is important that [a, b] is a closed interval. If it was open, we could take (0, 1) and
f(x) = 1/x, which is unbounded. But every polynomial is bounded on (0, 1) and therefore
no sequence of polynomials could converge to f uniformly.

It will suffice to prove Weierstrass Approximation Theorem on [0, 1] from which the general
case can be easily obtained. Recall the notion of uniform continuity from Analysis 1.
Let I   and f be a real-valued function on I. We say that f is uniformly continuous on I if

"  > 0  > 0 " x, y Î I : |x – y| <   |f(x) – f(y)| < .
Also remind, that a continuous function on [a, b] is always uniformly continuous.

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