Modulus of continuity

Remarks

The function $f$ has a modulus of continuity if and only if $f$ is uniformly continuous, and then $\omega_f$ is the smallest one. The same definitions work for a map between metric spaces, with distances in place of $\abs{x - y}$ and $\abs{f(x) - f(y)}$.

Example: let $f(x) = \sqrt x$ on $D = [0, \infty)$. For $0 \le y \le x$, one has $\sqrt x - \sqrt y \le \sqrt{x - y}$, since squaring gives $x - 2\sqrt{xy} + y \le x - y$, that is $y \le \sqrt{xy}$. Equality holds for $y = 0$, so $\omega_f(\delta) = \sqrt \delta$, and this tends to $0$.

Non-example: let $f(x) = x^2$ on $\R$. For $\delta > 0$, one has $f(x + \delta) - f(x) = 2x\delta + \delta^2$, which is unbounded in $x$. So $\omega_f(\delta) = \infty$ for each $\delta > 0$, and $\omega_f$ does not tend to $0$.

The modulus $\omega(\delta) = L\delta$ gives the Lipschitz condition with constant $L$. The modulus $\omega(\delta) = C\delta^\alpha$ with $0 < \alpha \le 1$ gives the Hölder condition of exponent $\alpha$.