[Math] Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)|\leq1 \forall x\in\mathbb R$

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Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $|f'(x)|\leq1 \forall x\in\mathbb R$. Then there exists $C$ in $\mathbb R $ such that

  1. $|f(x)|\leq C \sqrt |x|$ for all $ x$ with $|x|\geq 1$
  2. $|f(x)|\leq C |x|^2$ for all $ x$ with $|x|\geq 1$
  3. $f(x)=x+C$ for all $x \in \mathbb R $
  4. $f(x)=0$ for all $x \in \mathbb R $

If I take $f(x)=\frac{x}{2}$, then (4) is false, but I don't know how to prove or disprove others using the given conditions.Please help.

Thanks for your time.

Best Answer

  1. Does not hold, infact take $f(x)=\frac{x}{2} $. Suppose there exists $C$ $\in \mathbb{R}$ such that: $$ |f(x)|\leq C \sqrt |x| \text{ for all } x \text{ with } |x|\geq 1$$ Clearly from the inequality $C$ should be non negative. Then take $x=(2C+2)^2$, then $x \geq 1$, and so we get $$ \frac{(2C+2)^2}{2} \leq C \sqrt (2C+2)^2 = C(2C+2) $$
    Thus $$2{(C+1)^2} \leq C(2C+2) $$ i.e. $${(C+1)^2} \leq C(C+1) $$ and this is a contradiction.

  2. $f'$ is bounded above by $1$, so let $\sup |f'| = C \leq 1 $, then for any $x$ with $|x| \geq 1$, $f$ is differentiable on the interval $ ]0,x[$( or $]x,0[$ if $x\leq 0$) , so by mean value theorem we may write $$ |f(x)-f(0) | \leq C |x-0| $$ but $|x| \ geq 1$ so $|x| \leq |x|^2$ and $f(0)=0$ so $$ |f(x)|\leq C |x|^2 $$

  3. The same example you gave above can prove that 3. does not hold, with $f(x)=\frac{x}{2}$ .

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