let-f-R-R-be-a-continuous-function-then-show-that-1-if-f-x-f-x-2-x-R-then-f-is-a-constant-function-2-if-f-x-f-2x-1-x-R-then-f-is-a-constant-function-

Question Number 207314 by universe last updated on 11/May/24

let f : R \to R be a continuous function then

show that

(1) if f (x) = f (x^{2}) \forall x \in R then f is a constant

function

(2) if f (x) = f (2 x + 1) \forall x \in R then f is a

constant function

Answered by Berbere last updated on 11/May/24

f (x) = f (x^{2}) \Rightarrow f i s c o n t i n u s

f (- x) = f (x) \Rightarrow w e s h o w n j s t f ∣ [0, \infty [i s c o n s t a n t e

i f x \in [0, 1]

l e t p \in [0, 1 [s u c h a t t h a t p x \in [0, p [

\Rightarrow f (p x) = f (p^{2} x^{2}) = (p^{2^{n}} x^{2^{n}}); \forall n \in N

f (p x) = \lim_{n \to \infty} f (p^{2^{n}} x^{2^{n}}) = f (0)

f (1 - \frac{1}{n}) = \lim_{n \to \infty} f (1 - \frac{1}{n}) = 0

= \lim_{n \to \infty} f (0) = f (0)

i f x > 1

f (\sqrt{x}) = f (x) \Rightarrow \forall n \in N^{*}

f (x^{\frac{1}{2^{n}}}) = f (x) \Rightarrow \lim_{n \to \infty} f (x^{\frac{1}{2^{n}}}) = f (\lim_{n \to \infty} e^{\frac{1}{2^{n}} l n (x)}) = f (1) = f (0)

\Rightarrow \forall x \in R f (x) = f (0)

2)

f (x) = f (2 x + 1)

x \overset{h}{\to} 2 x + 1

h o h \dots . h n t i m e s

= h^{n} (x) = a_{n} x + b_{n}

h^{0} (x) = x = f (x) = 2 x + 1

h^{n + 1} = 2 a_{n} x + 2 b_{n} + 1 = a_{n + 1} x + b_{n + 1}

b_{n + 1} = 2 b_{n} + 1; a_{n + 1} = 2 a_{n} \Rightarrow a_{n} = 2^{n}

b_{n} = 2^{n} - 1

h^{n} (x) = 2^{n} x + 2^{n} - 1

\Rightarrow y = 2^{n} x + 2^{n} - 1;

x_{n} = \frac{y}{2^{n}} - 1 + \frac{1}{2^{n}} \Rightarrow f (x) = f (y); \forall n y i s f i x e d

f (y) = \lim_{n \to \infty} f (\frac{y + 1}{2^{n}} - 1) = f (- 1)

f i s c o n s t a n t e

Commented by universe last updated on 11/May/24

t h a n k u s i r

Commented by Berbere last updated on 11/May/24

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