506 Real Analysis
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Figure 663 f(m
3 n)
= 2 f(m− 1
3
3 n−^1)
+f(m+ 2
3
3 n−^1)
= 0 + 0 = 0.
Thusf( 3 mn)=0.
Finally, ifm≡ 2 (mod 3), then 2≤m≤ 3 n−1 and
3 f(m
3 n)
= 2 f(m+ 1
3
3 n−^1)
+f(m− 2
3
3 n−^1)
= 0 + 0 = 0.
Hencef( 3 mn)=0 in this case, too, finishing our induction.
Because the set{ 3 mn;m, n∈N}is dense in[ 0 , 1 ]andfis equal to zero on this set,
fis identically equal to zero.
(Vietnamese Mathematical Olympiad, 1999)
389.We argue by contradiction. Assume that there exista<bsuch thatf(a) =f(b),
say,f(a)>f(b).
Letg:R→R,g(x)=f(x)+λx, whereλ>0 is chosen very small such that
g(a)>g(b). We note that
lim
h→ 0 +g(x+ 2 h)−g(x+h)
h
=λ> 0 , for allx∈R.Sincegis a continuous function on a closed and bounded interval,ghas a maximum.
Letc∈[a, b]be the point wheregattains its maximum. It is important that this point is
notb, sinceg(a)>g(b). Fix 0<<λ. Then there existsδ=δ() >0 such that
0 <λ−<g(c+ 2 h)−g(c+h)
h<λ+, for all 0<h<δ.Fix 0<h 0 <min{δ,b− 2 c}. The above inequality written forh=h 0 ,h 20 ,h 40 , etc., yields