Is the above statement true or is there a counter example which proves it wrong? I couldn’t find a prove.
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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!)
On
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.
On
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.
$f(x)=x^2$ is not uniformly continuous on $\Bbb R$, but is differentiable on that interval.