Let $f : I\subset\mathbb{R} → \mathbb{C}$ be a continuous function on the closed interval $I$. A modulus of continuity of $f$ is any real-extended valued function $\omega: [0, ∞] → [0, ∞]$, vanishing at $0$ and continuous at $0$, that is $\lim_{t\to0}\omega(t)=\omega(0)=0.$

We say $f$ admits $\omega$ as modulus of continuity if and only if, $$\forall x,x'\in I: \|f(x)-f(x')\|\leq\omega(|x-x'|).$$ Here is my question: Does any function $f$ as above admit a modulus of continuity?

  • $\begingroup$ If you allow $w$ to take the value $+\infty$ then the function $w = +\infty$ everywhere should work no? $\endgroup$ – user37238 Sep 2 '13 at 8:13
  • $\begingroup$ I don't think so because we want $\omega$ to be continuous at $0$ and to be $0$ at $0$. $\endgroup$ – stroem Sep 2 '13 at 8:15
  • $\begingroup$ @user37238, but there is a condition at $0$. $\endgroup$ – njguliyev Sep 2 '13 at 8:17
  • $\begingroup$ Sorry, I didn't pay attention to the continuity of $w$. $\endgroup$ – user37238 Sep 2 '13 at 8:22
  • $\begingroup$ The complex-analysis tag seems to be misplaced. (The fact that the function is complex valued doesn't make it complex analysis.) $\endgroup$ – mrf Sep 2 '13 at 8:55

Yes, you can define it as $$\omega (t) := \sup_{0 < |x - x'|<t} \frac{\|f(x)-f(x')\|}{|x-x'|}$$ for $t>0$ and $\omega(0)=0$. Since a continuous function in the closed interval is uniformly continuous, $\omega$ will be continuous at $0$.

  • $\begingroup$ Thanks. But why holds $\lim_{t\rightarrow 0}\omega(t)=0$? $\endgroup$ – stroem Sep 2 '13 at 8:18
  • 1
    $\begingroup$ I explained it in my last sentence. Recall the definition of uniform continuity. $\endgroup$ – njguliyev Sep 2 '13 at 8:19
  • $\begingroup$ Sorry, my fault. Thanks a lot for your quick answer and time! $\endgroup$ – stroem Sep 2 '13 at 8:20

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