A Proof of Weierstrass's Theorem
Author(s): Dunham Jackson
Reviewed work(s):
Source: The American Mathematical Monthly, Vol. 41, No. 5 (May, 1934), pp. 309-312
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2300993 .
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1934]
A PROOF OF WEIERSTRASS S THEOREM
309
A PROOF OF WEIERSTRASS'S THEOREM'
By DUNHAM JACKSON,University
of Minnesota
Weierstrass's theorem with regard to polynomial approximation can be
stated as follows:
If f(x) is a givencontinuousfunctionfora < x < b, and if e is an arbitrarypositivequantity,it is possible to constructan approximatingpolynomialP(x) such
that
1f(x) - P(X)I < E
for a ? x < b.
This theorem has been proved in a great variety of differentways. No
particularproofcan be designated once forall as the simplest,because simplicity depends in part on the preparationof the reader to whom the proofis addressed. A demonstrationwhich followsdirectlyfromknownfactsabout power
series or Fourierseries,forexample, is not so immediateif a derivationof those
facts has to be gone throughfirst.A proof commonlyregarded as among the
simplest and neatest is the one due to Landau2 in which an approximating
polynomialis given explicitlyby means of a certain type of "singular" integral.
The purpose of this note is to presenta modificationor modifiedformulationof
Landau's proofwhichis believed to possess furtheradvantages of simplicity,at
least fromsome points of view.
Let f(x) be a given continuous functionfora ? x ? b. Without essential loss
of generalityit can be supposed that 0 <a < b < 1, since any finiteintervalwhatever can be carried over into an interval contained in (0, 1) by a linear transformation,underwhichany continuousfunctionwill go into a continuousfunction and any polynomialinto a polynomialof the same degree. For convenience
in the writingof the formulas which enter into the proof, let the function
f(x), supposed given in the firstinstanceonly fora < x ? b, be definedoutside the
interval (a, b) as follows:
f(x) = 0
for
x
f(x)= -f(a)
a
f(x) =
f(x) =0
1-x
_bf(b)
x _ 0,
for0<x<a,
forb < x <1,
for
x>1.
Then f(x) is definedand continuousforall values of x. The question at issue is
that of approximatingit by means of a polynomialforvalues of x belongingto
I
Presentedto the MinnesotaSectionoftheAssociation,May 13,-1933.
E. Landau, Uberdie Approximation
einerstetigen
Funktion
durcheineganzerationaleFunktion,Rendiconti
del CircoloMatematicodi Palermo,vol. 25 (1908),pp. 337-345;see also CourantHilbert,Methoden
derMathematischen
Physik,vol. I, secondedition,Berlin,1931,pp 55-57.
2
310
A PROOF OF WEIERSTRASS'S
[May,
THEOREM
the interval (a, b).
Let J,ndenote the constant
(
in=
- u2)
du,
and let
Pn(x)
f(t) [
=-
-
(t -
x)2]ndt.
The integrand in the latter integral is a polynomial of degree 2n in x with
coefficients
whichare continuousfunctionsof t,and the integralis foreach value
of n a polynomialin x of degree 2n (at most) with constantcoefficients.
If 0 ? x _ 1, the value of the integralis not changed if the limitsare replaced
by -1 +x and 1+x, sincef(t) vanishes everywhereoutside the interval (0, 1),
and vanishes in particularfrom-1 +x to 0 and from1 to 1+x:
Pn(x) =
1
f
,1+Z
f(t) [1 -
(t-
x)2]ndt.
By the substitutiont- x = u this becomes
I1 r
nP(X) - +
(1)
U2)ndu.
f(x + u)(I-
If the equation
1 =-f?(1
-
u2)ndu
is multipliedby f(x), this factor,being independentof u, may be writtenunder
the integralsign:
(2)
f(x)
=-
I 1
ff(x)(
-
U2)ndu.
Hence, by subtractionof (2) from(1),
(3)
Pn(x) -f(x)
1 r'
= -J
[f(x + U)
- f(X)](I
-
U2)ndu.
The problem now is to show that the value of this expressionapproaches zero
as n becomes infinite.
Let e be any positive quantity. Since f(x) is (uniformly)continuous there
is a 6 >0 (independent of x) such that If(x+u) -f(x) I < e/2 for |u I 8. Let
M be the maximumof !f(x)I. Then lf(x+u) -f(x) I :52M forall values ofu.
For IUIu , 1
U2/82,and
If(x + u)
-
f(x) I _ 2MU2/2.
1934]
311
A PROOF OF WEIERSTRASS 'S THEOREM
For any value of u, one or the otherofthe quantitiese/2,2Mu2/82is greater
than or equal to |f(x+u) -f(x) I, and theirsum therefore
is certainlygreater
or
than equal to jf(x+u) -f(x) I:
If(x + u)
f(x)
-
< e/2 + 2Mu2/52
forall values of u. Consequently,for0 _ x < 1,
P.(x)
-f(x)I <
in
-1
-e/2
.r
(e/2)(1-u2)ndu 2M2+
+
Jn1
2M
a2
(1-2)ndu
2J
J -2(
U2)1du
Let thelastintegralbe denotedby J'n.By integration
by parts,
I
Jn =
IUU(
(I1-
U2nd
&(
)S"
1
r(I -u2)nrl
du=
+fu
J_.1 2(n+1)
-
2) n+1- 1
Jn+1
( +
2(n +1)
But Jn+1
< Jn,since1-u2 <1 throughout
theinterior
oftheintervalofintegra= (1 -u2) (1 -u2)" < (1 -u2)". So
tion and hence (1 -u2)+1
h
1~~~~~
2(ne+ 1)
J,,
2(n + 1)
It followsthat as soon as n is sufficiently
large
2MJ,'
52Jn
and consequently
IPn(x) - f(X)l
< el
for0? x< 1 and in particularfora ? x< b. This is thesubstanceof theconclusionto be proved.
The above proofhas something
in commonwiththatof S. Bernstein,3
the
fundamental
difference
beingthatLandau's integralis usedhereinsteadofthe
algebraicformula
fora binomialfrequency
distribution.
It was in factsuggested
to thewriter,not by consideration
of Bernstein'sproofas such,but by a conversationwithProfessor
W. L. Hart on the subjectof Bemoulli'stheorem.A
noteworthy
characteristic
of Bernstein'sproofis that it makesWeierstrass's
theoremin effect
a corollary
ofthatofBernoulli.
An alternative
organization
ofthepresentmethodofproofis as follows.Let
f(x) at firstbe not merelycontinuous,but subjectto the Lipschitzcondition
I See e.g. P61yaand Szego,Aufgaben
undLehrsittze
aus derAnalysis,vol. I, Berlin,
-1925,pp.
66, 230.
312
[May,
A PROOF OF WEIERSTRASS 'S THEOREM
If(x2) - f(Xl)I
(4)
< XX2
- Xl
If this conditionis satisfiedby f(x) as originallydefinedfora < x < b, it will be
satisfied,possiblywith a different
value of X, when the definitionis extended to
coverall valuesofx. Then,in (3), If(x+u) -f(x) I XuI, and
IPO(X)-f(X) l
Let
6 = 1/n/2,and
=
1
Xr'
yJ
U (I - U2)du
2Xr
U(U2)(1_
let
U(
-u2)ndu,
I2 =
u(
-
u2)ndu,
so that IPn(x)-f(x)I <2X(I1+I2)/Jn. In A1,u<1/n"l2, and
I, ?
n-12(l
-
u2)ndu<
n-12(l
< 1/nl2.
u2)ndu= 2n-"12Jn 211/Jn
-
In 12, 1/u_n"l2 and u=u2/u<nl?2u2, so that by application of the previous
reckoningwith Jn'
r1
*
?
12 < nhl2f
u2(1 - u2)ndu n1I2 u2(1 - u2)ndu = lnl'2J <knl"2Jn/(n+ 1),
212/J,
<
lnl"2/(n
+
1)
<
J
1/nl2.
So it appears not merelythat I P.(x) -f(x) is less than a quantity independent
of x which approaches zero as n becomes infinite,but also, more specifically,4
that it is less than a quantity of the orderof 1/n"12.This is proved, to be sure,
only fora functionsatisfying(4). But (4) is satisfiedin (a, b) by any continuous
functionwhose graph is a broken line made up of a finitenumber of straight
line segmentswith finiteslope, and as any functionwhateverthat is continuous
fora ? x _ b can be approximatedwithany desiredaccuracy by a functionof this
special type, the general conclusion of Weierstrass's theorem follows immediately.
The method can be applied equally well to the proof of Weierstrass's
theoremon the trigonometric
approximationof a periodiccontinuousfunction,
by the use of de la Vallee Poussin's integral'
1
rT
x\
/~~~~~~~~~~~~~~~~~~t-
f(t) Cos2n
)dt,
Hn =
irT
Cos2-du,
and to the proofof correspondingtheoremson the approximaterepresentation
of continuousfunctionsof morethan one variable.
I Cf. C. de la Vall6ePoussin,Sur I'
approximation
desfonctions
d' unevariablerMelle
etdeleurs
d6rivUes
par des polyndmes
et des suiteslimittesde Fourier,Bulletinsde la Classe des Sciences,
Acad6mieRoyalede Belgique,1908,pp. 193-254;pp. 221-224.
6 See de la Vall6ePoussin,loc. cit.,pp..228-230.