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If $f(x)$ define as: $f(x) = x\cos(\log(x))$, when $x > 0$. $f(x) = 0$, when $x = 0$. How we prove that $f$ is uniform continuous in $[0, +\infty[$?

Sebastiano
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LLLLLLLL
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  • Take a derivative. – Michael Nov 12 '20 at 22:29
  • please use MathJax for the future: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – person Nov 12 '20 at 22:34
  • You can prove that it is Lipschitz continuous by showing that it's continuous at $0$ and that its derivative is bounded on $(0,\infty)$. –  Nov 12 '20 at 22:34

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$ f $ is continuous at $ x=0$ since

$$(\forall x>0)\;|f(x)|\le |x|$$

$ f $ is differentiable at $ (0,+\infty) $ and

$$(\forall x>0)\;$$ $$f'(x)=\cos(\ln(x))-\sin(\ln(x))$$

thus $$(\forall x>0)\;\;|f'(x)|\le 2$$

So, By MVT, $$(\forall x,y\ge0)\;\;(\exists c>0) \;\;:$$ $$|f(x)-f(y)|=|x-y||f'(c)|$$ $$\le 2|x-y|$$

$ f $ is then UC at $ [0,+\infty)$. You just can take $ \delta=\frac{\epsilon}{2}$.

hamam_Abdallah
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