Advanced book on Mathematics Olympiad

Real Analysis 525

f

(i+ 2

n

)

+f

(i

n

)

2

≥f

(

i+ 1

n

)

+

2

n

,i= 1 , 2 ,...,n,

or

f

(

i+ 2

n

)

−f

(

i+ 1

n

)

≥f

(

i+ 1

n

)

−f

(

i

n

)

+

4

n

,i= 1 , 2 ,...,n.

In other words, (^) i+ 1 ≥ (^) i+^4 n. Combining this fornconsecutive values ofigives
(^) i+n≥ (^) i+ 4.
Summing this inequality fori=0ton−1 and canceling terms yields
f( 2 )−f( 1 )≥f( 1 )−f( 0 )+ 4 n.
This cannot hold for alln≥1. Hence, there are no very convex functions.
Second solution: We show by induction onnthat the given inequality implies
f(x)+f(y)
2
−f

(

x+y

2

)

≥ 2 n|x−y|, forn≥ 0.

This will yield a contradiction, because for fixedxandythe right-hand side gets arbitrarily

large, while the left-hand side remains fixed.

The statement of the problem gives us the base casen=0. Now, if the inequality

holds for a givenn, then for two real numbersaandb,

f(a)+f(a+ 2 b)

2

≥f(a+b)+ 2 n+^1 |b|,

f(a+b)+f(a+ 3 b)≥ 2 (f (a+ 2 b)+ 2 n+^1 |b|),

and

f(a+ 2 b)+f(a+ 4 b)

2

≥f(a+ 3 b)+ 2 n+^1 |b|.

Adding these three inequalities and canceling terms yields

f(a)+f(a+ 4 b)

2

≥f(a+ 2 b)+ 2 n+^3 |b|.

Settingx=a,y=a+ 4 b, we obtain

f(x)+f(y)

2

≥f

(

x+y

2

)

+ 2 n+^1 |x−y|,