EDB — 156

[156]Prerequisites:[155]↺↻.Let \(f:X_ 1→ X_ 2\) with \((X_ 1,d_ 1)\) and \((X_ 2,d_ 2)\) metric spaces.

A monotonic (weakly) increasing function \(𝜔:[0,∞)→ [0,∞]\), with \(𝜔(0)=0\) and \(\lim _{t→ 0+}𝜔(t)=0\), such that

is called continuity modulus for the function \(f\). (Note that \(f\) can have many continuity moduli).

For example, if the function \(f\) is Lipschitz, i.e. there exists \(L{\gt}0\) such that

then \(f\) satisfies the eqz. ?? by placing \(𝜔(t)=L t\).

We will now see that the existence of a continuity modulus is equivalent to the uniform continuity of \(f\).

• If \(f\) is uniformly continuous, show that the function

  \begin{equation} 𝜔_ f(t) = \sup \{ d_ 2(f(x), f(y)) ~ :~ x,y∈ X_ 1,d_ 1(x,y)≤ t \} \label{eq:modulo_ cont_ con_ sup} \end{equation}

  17

  is the smallest continuity modulus. ¹

• Note that the modulus defined in ?? may not be continuous, and may be infinite for \(t\) large — find examples of this behaviour.

• Also show that if \(f\) is uniformly continuous, there is a modulus that is continuous where it is finite.

• Conversely, it is easy to verify that if \(f\) has a continuity modulus, then it is uniformly continuous.

If you don’t know metric space theory, you can prove the previous results in case \(f:I→ ℝ\) with \(I⊆ℝ\). (See also the exercise [15W]↺↻, which shows that in this case the modulus \(𝜔\) defined in ?? is continuous and is finite).