508 Real Analysis
ma =f( 0 ), and hencefis constant on[ 0 ,a]. Passing to the limit witha→1, we
conclude thatfis constant on the interval[ 0 , 1 ]. Clearly, constant functions satisfy the
property, providing all solutions to the problem.
(Gazeta Matematica ̆(Mathematics Gazette, Bucharest), proposed by M. B ̆aluna) ̆
392.Letφ:[ 0 , 1 ]×[ 0 , 1 ]be a continuous surjection. Defineψto be the composition
[ 0 , 1 ]
φ
−→ [ 0 , 1 ]×[ 0 , 1 ]
φ×id
−→ [ 0 , 1 ]×[ 0 , 1 ]×[ 0 , 1 ]
pr 12
−→ [ 0 , 1 ]×[ 0 , 1 ],wherepr 12 :[ 0 , 1 ]×[ 0 , 1 ]×[ 0 , 1 ]→[ 0 , 1 ]×[ 0 , 1 ]is the projection of the cube onto
the bottom face. Each function in the above chain is continuous and surjective, so the
composition is continuous and surjective. Moreover, because the projection takes each
value infinitely many times, so doesψ. Therefore,ψprovides the desired example.
393.The first example of such a function was given by Weierstrass. The example we
present here, of a functionf :[ 0 , 1 ]→[ 0 , 1 ], was published by S. Marcus in the
Mathematics Gazette, Bucharest.
If 0≤x≤1 andx= 0 .a 1 a 2 a 3 ...is theternaryexpansion ofx, we let thebinary
representation off(x)be 0.b 1 b 2 b 3 ...,where the binary digitsb 1 ,b 2 ,b 3 ,...are uniquely
determined by the conditions
(i)b 1 =1 if and only ifa 1 =1,
(ii)bn+ 1 =bnif and only ifan+ 1 =an,n≥1.
It is not hard to see thatf(x)does not depend on which ternary representation you
choose forx. For example,
f( 0. 0222 ...)= 0. 0111 ··· = 0. 1000 ··· =f( 0. 1000 ...).Let us prove first that the function is continuous. Ifxis a number that has a unique
ternary expansion and(xn)nis a sequence converging tox, then the firstmdigits of
xnbecome equal to the firstmdigits ofxfornsufficiently large. It follows from the
definition offthat the firstmbinary digits off(xn)become equal to the firstmbinary
digits off(x)fornsufficiently large. Hencef(xn)converges tof(x),sofis continuous
atx.
Ifxis a number that has two possible ternary expansions, then in one expansion
xhas only finitely many nonzero digitsx = 0 .a 1 a 2 ...ak 00 ...,withak = 0. The
other expansion is 0.a 1 a 2 ...a′k 222 ...,witha′k=ak−1(=0 or 1). Given a sequence
(xn)nthat converges tox, for sufficiently largenthe firstk−1 digits ofxnare equal to
a 1 ,a 2 ,...,ak− 1 , while the nextm−k+1 are eitherak, 0 , 0 ,...,0, ora′k, 2 , 2 ,...,2.
Iff(x)=f( 0 .a 1 a 2 ...ak 00 ...)= 0 .b 1 b 2 b 3 ...,then fornsufficiently large, the first
k−1 digits off(xn)areb 1 ,b 2 ,...,bk− 1 , while the nextm−k+1 are eitherbk,bk+ 1 =
bk+ 2 = ··· =bm(the digits off(x))or1−bk, 1 −bk+ 1 = ··· = 1 −bm. The two
possible binary numbers are 0.b 1 b 2 ...bk− 10111 ...and 0.b 1 b 2 ...bk− 11000 ...;they