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Is the above statement true or is there a counter example which proves it wrong? I couldn’t find a prove.

(Before you close this thread: This question hasn’t been answered yet!)

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    $\begingroup$ That question you link is verrrrry different from your question (note that the body of that question is different from the title, unlike yours, which is what led to its closure). The hypothesis of bounded derivative in the body of that question makes all the difference in the world to uniform continuity. $\endgroup$ – Lee Mosher Jun 28 '17 at 13:59
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$f(x)=x^2$ is not uniformly continuous on $\Bbb R$, but is differentiable on that interval.

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False, consider $f:(0,\infty)\rightarrow \mathbb{R}$ with $f(x)=\frac{1}{x}$. This function is differentiable everywhere with $f'(x)=\frac{-1}{x^2}$ but this function is not uniformly continuous.

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It is false as others have given you counterexamples; but try to understand the differences between continuity, differentiability and uniform continuity.

Continuity of a function at a point is essentially a statement about the growth of function in the neighborhood of the point.

Differentiability of a function at a point is a statement about its smoothness at that point. Now you must observe that both these properties are local properties; i.e. you talk about them at a point. Even if you talk about them over an interval you define it as for any points in this interval we have so and so.

Whereas, uniform continuity of a function, is a global property; i.e. it gives you information about the growth of the function in any arbitrary interval you choose(definition). You should see yourself now that saying that a function is uniformly continuous carries a lot of information about it's growth; therefore you can not expect differentiability to imply uniform continuity.

Needless to say, this is not a proof or a rigorous argument. I just wrote it so that you can see why the examples are working i.e because they are growing too fast.

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You can say it is if the derivative is bounded and then use lipchitz condition through lagranges mean value to show it will be uniformly continuous. If it is given that derivative is bounded you can say so. Else it js not necessary.

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