tmua.me Extreme Set B Paper 2 · Question 9

A real-valued function $f$ , defined for every real number, satisfies

f(x+1)=f(x)+2x+1,\qquad f(2x)=4f(x)

for every real $x$ . Also, $0\leq f(x)\leq1$ whenever $0\leq x\leq1$ .

A student wishes to prove that $f(x)=x^2$ for every real $x$ . The student puts $g(x)=f(x)-x^2$ and correctly obtains

g(x+1)=g(x),\qquad g(2x)=4g(x),\qquad |g(x)|\leq1\ \text{when }0\leq x\leq1.

For a fixed real $x$ and each positive integer $n$ , the student chooses an integer $k_n$ such that

0\leq2^n x-k_n<1.

Which of the following validly completes the proof, without assuming that $f$ or $g$ is continuous?

ASince

g

has period

1

g(2^n x)=g(x)

. Combining this with

g(2^n x)=4^n g(x)

gives

g(x)=0

BThe bound on

f

gives

|4^n f(x)-k_n^2|\leq1

. Dividing by

4^n

and letting

n

increase gives

f(x)=x^2

COne integer

k

can be chosen so that

0\leq2^n x-k<1

for every

n

. Thus

g

is bounded on the sequence

2^n x

, forcing

g(x)=0

DRepeated scaling gives

g(x)=4^{-n}g(x/2^n)

. Since

x/2^n

approaches

0

, the bound on

g

gives

g(x)=0

EInteger translations give

g(2^n x-k_n)=g(2^n x)=4^n g(x)

. Hence

|g(x)|\leq4^{-n}

for every positive integer

n

, so

g(x)=0

FThe numbers

2^n x-k_n

lie between

0

and

1

, so they approach

0

n

increases. Therefore

g(2^n x-k_n)

approaches

g(0)=0

, forcing

g(x)=0

examday