Let f be a differentiable function from R to R such that [| f( x ) - f( y ) | 2| x - y |^3…

Question

Let f be a differentiable function from

R to R such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$

for all  $$x,y \in $$ R.

If   $$f\left( 0 \right) = 1$$

then    $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$  is equal to :

Accepted Answer

$$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$

$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$

$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$

$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$  $$ \Rightarrow f'\left( x \right) = 0$$

$$ \Rightarrow f\left( x \right) = $$ constant

as  $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$

$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$

Solutions · 1