a short proof of Bernstein's theorem on the analyticity of such functions. ... theorem [l] on "regu- ... By Rolle's theorem there exists a if on (t, x) such that. (4).
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 47, Number 2, February
1975
A PROOF OF BERNSTEIN'S THEOREM ON REGULARLY MONOTONICFUNCTIONS J. A. M. McHUGH ABSTRACT. class
(j
order
analyticity
This
paper
of such
a short
tially
case
proof)
theorem,
tion of class
C
if it is of
may depend
of Bernstein's
of Bernstein's
on the
theorem
That
we recall
that
on a real interval
subject
the reader
cited there.
theorem.
theorem
on this
we refer
the proof we present
Bernstein's
(which
proof
For background
original
of Bernstein's
motivated
a short
proof
paper by Boas 12], and the references
special
monotonie"
sign
functions.
functions.
of Bernstein's
"regularly
is of a fixed
We present
presents
monotonie"
tation
is called
derivative
of the derivative).
on the
larly
A function
and each
(and a presen-
to the brief
survey
The book [3] contains
proof for that
here.
[l] on "regu-
Before
a regularly
giving
monotonie
special
a proof of a
case
our proof function
(a, b) for which each derivative
par-
of is a func-
is of fixed sign
on (a, b). Theorem.
// F(x)
is regularly
monotonie
on (a, b), then
F(x)
is ana-
lytic on (a, b). Proof. the even
We assume
for convenience
part of F(x),
f(x),
for the odd part of F(x),
begin
by noting
following
that
a = - b < 0.
at x = 0.
and the analyticity
f(x) is itself
lemma is strategic
Lemma.
that
is analytic
of F(x)
regularly
We will prove
An analogous
thereby
monotonie
proof
follows.
on 0 < x < b.
that holds
We
The
for our proof.
// f{n)(x) < 0, f(n+l)(x) > 0, and /(n+2)W > 0, then for x £[0, b),
\R (x)\ > \R +,(x)|,
where
R (x) is the
mth remainder
in the Taylor
expansion
of f about x = 0. Proof
of Lemma.
Received
Write the remainder
by the editors
November
as
5, 1973-
AMS (MOS)subject classifications (1970). Primary 26A48, 26A90, 26A93; Secondary 26A24, 26A51, 26A84, 26A86Key words
and phrases.
C-infinity
functions,
analyticity,
regularly
monotonie
functions. Copyright © 1975. American Mathematical Society License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
358
A PROOF OF BERNSTEIN'S THEOREM
(1)
359
RM = —-— fV" >(;)(* - t)n-ldt. "
A sufficient
(n -
condition
1)! J 0
for the Lemma
(2)
to be true is that
-f0 n
for t on (0, x).
Since
ficient
for (2) to be true;
condition
/("+I)(r)
> 0, replacing
~
m by 1 in (2) yields
a suf-
namely
-f("Xt)-(x-t)f(n+1)(t)>o, or, if we write
g for /
,
(3)
-g(t)-(x-t)g'(t)>0.
By Rolle's
theorem
there
(4)
a if on (t, x) such that
-g(f)-(x-f)g'(0
Since
g" = /(n+2)>0, If instead
triples
(analytic
For, in this case,
is a polynomial. triples
or odd.
If /(n)
one has either
of the sign
is immediate.
Only for
+, -, + occur
is even,
/(n+1)U)
(perhaps
for /'*' after
odd.
/-»-/),
/ is even.
/("+1)(0)
> 0 for x > 0.
< 0 for x > 0. Thus /("+1)(x)
proof holds
attainable
when
say, then
> 0 for x > 0. Thus f(n+l)(x)
An analogous
is always
conclusion
can the triple
is even
/("+2)(x)
But also by supposition,
of the Lemma
the Lemma's
functions)
/(n)
0. By supposition,
= -g(x)>O.
(3) follows from (4) (Q.E.D. Lemma).
of the sign sequence
+, +, + or +, +, -,
polynomials
sign
exists
Since
= 0 =» / one of these
it follows
that
for
/ even
(5)
|Rn(x)| > |RB+1M|
for « = 0, 1, 2, ••• ,
for x on [O. b). Frequently = (-1)"
/"
and /"+1
==>/ is polynomial).
(6) then
are both > 0 (or both < 0) (else eventually Rewriting
r (*).,—£— "
for, say,
f*$
(»-«!
(1) as
Cf^Kxùd-ù^-^dt,
J 0
and /"¿ + U > 0 and
x > 0, one has
n ■
(7)
Q
