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Whether these two functions are uniformly continuous:

1)$f$ such that $|f(x)-f(y)|\le\sqrt{|x-y|}\ \forall x\in(0,1)$

2)$f$ such that $f(x)=x^{\frac1{2}}\sin(\frac1{x^3})\forall\ x\in(0,1)$

For 1), I think there exists a counterexample because $|x-y|\le\sqrt{|x-y|}\forall x\in(0,1)$. As for 2), I think this is similar to $x\sin \frac1{x}$. Any ideas. Thanks beforehand.

1 Answers 1

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1) If $f:[0,1]\to \mathbb{R}$ is such that :

$$\forall (x,y)\in[0,1]^2,\vert f(x)-f(y)\vert\le\sqrt{\vert x-y\vert}$$

Then $f$ is uniformly continuous because given any $\epsilon>0$, if we choose $\delta=\epsilon^2$, we readily see that :

$$\forall (x,y)\in[0,1]^2, \vert x-y\vert\le\delta\implies\vert f(x)-f(y)\vert\le\epsilon^2$$

2) If $f$ is defined by :

$$f:[0,1]\to\mathbb{R},x\mapsto\cases{\sqrt{x}\sin(\frac{1}{x^3})\quad\mathrm{if}\,0

then $f$ is continuous (because $\sin$ is bounded and therefore $\lim_{x\to0}f(x)=0=f(0)$) and hence uniformly continuous by Heine theorem

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    then, is $x\sin \frac1{x}$ uniformly continuous on (0,1)? – 2017-01-19
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    @vidyarthi Of course yes .. if you extent the map first into $F$, by setting $F(0)=0$ and $F(x)=x\sin\left(\frac{1}{x}\right)$ for $0$[0,1]$. Then $F$ is uniformly continuous by Heine-Cantor theorem, and a fortiori, its restriction to $(0,1)$ is also uniformly continuous. – 2017-01-19