Let a power function $f \colon [0, \infty) \to [0, \infty)$ be defined by $ f(x) = x^{\alpha}$ with $\alpha >1$.

From my own understanding, this function $f(x) = x^{\alpha}$ is differentiable on $[ 0, \infty)$ with its first derivative $f'(x) = \alpha x^{\alpha-1}$ for all $x \in [ 0, \infty)$. Moreover, the derivative function $f'(x) = \alpha x^{\alpha-1}$ is continuous on $ [ 0, \infty)$. So, I think the function $f$ is continuously differentiable on $ [ 0, \infty)$.

Is it true?

On the other hand, I thought that if $f$ is continuously differentiable on $ [ 0, \infty)$, then $f$ should also be Lipschitz continuous on $[0, \infty)$. However, $f$ is not Lipschitz continuous on $[0, \infty)$, even for $(0,\infty)$. Thus, from this point of view, I thought $f$ might not be continuously differentiable on $[0, \infty)$.

Could anyone help explain this question? Also, when the function $f$ is Lipschitz continuous? And, when the function $f$ is continuously differentiable?

  • $\begingroup$ A function can be continuously differentiable without being Lipschitz continuous. Take for example $f(x)=x^2$. $\endgroup$ – TonyK Apr 3 '17 at 1:02
  • $\begingroup$ Having said that, I can't quite understand your problem: your $f$ is Lipschitz continuous on $[0,c]$ for every $c > 0$. $\endgroup$ – TonyK Apr 3 '17 at 1:07
  • $\begingroup$ Thanks @TonyK. You are right, since for Lipschitz continuous, we need extra condition that the derivative $f'(x)$ should be bounded. Apparently, $f'(x)$ is not bounded on $[0, \infty)$. $\endgroup$ – Paradiesvogel Apr 3 '17 at 1:10

Over a closed and bounded subset of $\mathbb{R}$, every continuously differentiable function is Lipschitz. In your case the domain is not bounded, so you cannot make this claim. That is, in your question $f$ is continuously differentiable but not Lipschitz, and this is not breaking maths per se.

Suppose $g:[0,5] \rightarrow [0, 5^a]$ is defined by $g(x) = x^a, a > 1$.

Then $g$ is continuously differentiable as you noted before. However, it is also Lipschitz. Can you prove this?

  • $\begingroup$ Yeah, you are right. I think I mess the definitions about "differentiable at $x$" implies "continuous at $x$" with the definitions about "differentiable on a set (interval) $I$ ". $\endgroup$ – Paradiesvogel Apr 3 '17 at 1:19

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