aa r X i v : . [ m a t h . HO ] A p r A Proof Of Cantor’s Theorem
S. WaltersA
BSTRACT . We present a short proof of Cantor’s Theorem (circa 1870s): if a n cos nx ` b n sin nx Ñ for each x in some (nonempty) open interval, where a n , b n are sequences of complex numbers, then a n and b n converge to 0.
1. P
ROOF O F C ANTOR ’ S T HEOREM
Cantor’s Theorem.
Let a n , b n be sequences of complex numbers such that lim n Ñ8 a n cos nx ` b n sin nx “ for each x in some open interval p c , d q . Then a n Ñ and b n Ñ .The proof presented here consists of reduction to the case C n sin nx Ñ , which is covered by Lemma B below and which we proceed to prove first. Lemma A.
Let δ ą be given and let B n be a bounded sequence of complexnumbers such that B n sin nt Ñ as n Ñ 8 for each ă t ă δ . Then B n Ñ . Proof.
Replacing t P p , δ q by t we have B n sin p n t q “ B n sin p nt q cos p nt q Ñ so the limit B n sin nx Ñ holds also for ă x ă δ . By repeating this doublingprocedure a finite number of times, we get B n sin nx Ñ for ď x ď π ď N δ for some fixed positive integer N . Therefore, f n p x q : “ B n sin nx Ñ for each x inthe closed interval r , π s . Further, since the sequence B n is bounded, thefunctions f n are bounded by an (integrable) constant on r , π s . Therefore, bythe Dominated Convergence Theorem ([3], Theorem 11.32) “ lim n π Z | f n p x q| dx “ lim n | B n | π Z | sin nx | dx “ lim n | B n | since ş π | sin nx | dx “ for each integer n ě . It follows that B n Ñ . (cid:3) The next lemma removes the boundedness condition in Lemma A.
Lemma B.
Let δ ą be given and let C n be a sequence of complex numberssuch that C n sin nt Ñ as n Ñ 8 for each ă t ă δ . Then C n Ñ . Date : March 30, 2020.2000
Mathematics Subject Classification.
Key words and phrases.
Trigonometric series, trigonometric polynomials, Can-tor/Lebesgue theorem, Fourier series.
Proof.
Consider the bounded sequence B n “ | C n | `| C n | , for which we have lim n B n sin nx “ lim n | C n | sin nx ` | C n | “ holds for each x in p , δ q . Hence B n Ñ by Lemma A, which in particularmeans that there is an integer M such that B n ă for n ą M . This says that | C n | ă (for n ą M ) so that C n is bounded. By Lemma A, applied again to C n , we see that C n Ñ . (cid:3) Proof of Cantor’s Theorem.
We are given that a n cos nx ` b n sin nx Ñ for each c ă x ă d . Let δ ą such that x ` δ ă d , so that c ă x ` t ă d holds foreach ď t ă δ . Then since the zero limit holds also for x ` t , we have a n cos n p x ` t q ` b n sin n p x ` t q “ a n cos p nx ` nt q ` b n sin p nx ` nt q“ a n “ cos p nx q cos p nt q ´ sin p nx q sin p nt q ‰ ` b n “ sin p nx q cos p nt q ` cos p nx q sin p nt q ‰ “ “ a n cos p nx q ` b n sin p nx q ‰ cos p nt q ` “ b n cos p nx q ´ a n sin p nx q ‰ sin p nt q . (1.1)Since this whole sum goes to 0, and since the first term in (1.1) goes to 0by hypothesis, it follows that the second term goes to 0, i.e. C n sin p nt q Ñ foreach ă t ă δ where C n : “ b n cos p nx q ´ a n sin p nx q is a fixed sequence (independentof t ) since here x is fixed. Lemma B above now applies to this and gives C n “ b n cos p nx q ´ a n sin p nx q Ñ as n Ñ 8 . Therefore taking the sum of absolute squares we get ˇˇ a n cos p nx q ` b n sin p nx q ˇˇ ` ˇˇ b n cos p nx q ´ a n sin p nx q ˇˇ “ | a n | ` | b n | which has zero limit, hence the result. (cid:3) R EFERENCES [1] G. Cantor,
Uber einen die trigonometrischen Reihen betreffenden Lehrsatz , Crelles Jour-nal f ¨ur Mathematik, Bd. (1870), 130-138.[2] R. L. Cooke, The Cantor-Lebesgue Theorem , American Mathematical Monthly , No. 7,558-565.[3] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, New York, Third Ed. 1976.D EPAR TMENT OF M ATHEMATICS & S
TATISTICS , U
NIVERSITY OF N OR THERN
B.C., P
RINCE G EORGE , B.C. V2N 4Z9, C
ANADA . E-mail address : [email protected] URL ::