14.2 Uniformly continuous functions[2DQ]

It is easy to see that a function uniformly continuous is continuous at every point.

E355

[156]Prerequisites:355.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

\begin{equation} ∀ x,y∈ X_ 1,~ ~ d_ 2(f(x),f(y))≤ 𝜔(d_ 1(x,y))~ , \label{eq:modulo_ continuita} \end{equation}

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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

\[ ∀ x,y∈ X_ 1,~ ~ d_ 2(f(x),f(y))≤ L \, d_ 1(x,y) \]

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}

  357

  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 2, which shows that in this case the modulus \(𝜔\) defined in ?? is continuous and is finite).

Hidden solution: [UNACCESSIBLE UUID ’157’][UNACCESSIBLE UUID ’158’][UNACCESSIBLE UUID ’159’] [UNACCESSIBLE UUID ’15B’]

[15C]Let \((X,d)\) metric space and \(\mathcal F\) the set of uniformly continuous functions \(f:X→ℝ\), show that \(\mathcal F\) is a vector space.

This is more generally true if \(f:X→ X_ 2\) where \(X_ 2\) is a normed vector space (to which we associate the distance derived from the norm).

Hidden solution: [UNACCESSIBLE UUID ’15D’] [15F] Difficulty:*.Let \((X_ 1,d_ 1)\) and \((X_ 2,d_ 2)\) metric spaces, with \((X_ 2,d_ 2)\) complete. Let \(A⊂ X_ 1\) and \(f:A→ X_ 2\) be a uniformly continuous function. Show that there is a uniformly continuous function \(g:\overline A→ X_ 2\) extending \(f\); In addition, the extension \(g\) is unique.

Note that if \(𝜔\) is a continuity modulus for \(f\) then it is also a continuity modulus for \(g\). (We assume that \(𝜔\) is continuous, or, at least, that it is upper semicontinuous). Hidden solution: [UNACCESSIBLE UUID ’15G’][UNACCESSIBLE UUID ’15H’] [15J] Prerequisites:4.Let \(A⊂ ℝ^ n\) be bounded and \(f:A→ℝ\) a continuous function. Show that \(f\) is uniformly continuous if and only there exists a continuous function \(g:\overline A→ ℝ\) extending \(f\); In addition, the extension \(g\) is unique.

Hidden solution: [UNACCESSIBLE UUID ’15K’] [15M]Let \(f:(0,1]→ℝ\) be a continuous function. Prove that it is uniformly continuous, if and only if the limit \(\lim _{x→ 0+}f(x)\) exists and is finite. Hidden solution: [UNACCESSIBLE UUID ’15N’] [15P] Let \(f:[0,∞)→ℝ\) be a continuous function and such that the limit \(\lim _{x→∞} f(x)\) exists and is finite. Show that it is uniformly continuous. Hidden solution: [UNACCESSIBLE UUID ’15Q’] [15R]Let \(f:[0,∞)→ℝ\) be a continuous function, show that these two clauses are equivalent.

• There exists \(g:[0,∞)→ℝ\) uniformly continuous and such that the limit \(\lim _{x→∞} ( f(x)-g(x))\) exists and finite.

• \(f\) is uniformly continuous.

Hidden solution: [UNACCESSIBLE UUID ’15S’] [15T]Find an example of \(f:[0,∞)→ℝ\) continuous and bounded, but not uniformly continuous. Hidden solution: [UNACCESSIBLE UUID ’15V’] [15W] Let \(I⊆ℝ\) be an interval, and let \(f:I→ℝ\) be uniformly continuous. Let \(𝜔\) be the continuity modulus, defined by the eqz. ??, as in the exercise 1. Show that \(𝜔\) is subadditive i.e.

Knowing that \(\lim _{t→ 0+}𝜔(t)=0\) we conclude that \(𝜔\) is continuous. Hidden solution: [UNACCESSIBLE UUID ’15X’] [UNACCESSIBLE UUID ’15Y’] [15Z]Prerequisites:2. Let \(f:ℝ→ℝ\) be uniformly continuous; show that

or, equivalently, that there exists a constant \(C\) such that \(|f(x)|≤ C (1+|x|)\) for every \(x\). Hidden solution: [UNACCESSIBLE UUID ’160’] [161]Prerequisites:16. Let \((X_ 1,d_ 1)\), \((X_ 2,d_ 2)\) and \((Y,𝛿)\) be three metric spaces; consider the product \(X=X_ 1× X_ 2\) equipped with the distance \(d(x,y)=d_ 1(x_ 1,y_ 1)+d_ 2(x_ 2,y_ 2)\).   ² Let \(f:X→ Y \) be a function with the following properties:

• For each fixed \(x_ 1∈ X_ 1\) the function \(x_ 2↦ f(x_ 1,x_ 2)\) is continuous (as a function from \(X_ 2\) to \(Y\));

• There is a continuity modulus \(𝜔\) such that

  \[ ∀ x_ 1,\tilde x_ 1∈ X_ 2~ ~ ,~ ~ ∀ x_ 2∈ X_ 2~ ~ , 𝛿\big(f(x_ 1,x_ 2),f(\tilde x_ 1,x_ 2)\big) ≤ 𝜔\big(d_ 1(x_ 1,\tilde x_ 1)\big) \]

  (We could define this property by saying that the function \(x_ 1↦ f(x_ 1,x_ 2)\) is uniformly continuous, with constants independent of the choice of \(x_ 2\)).

Then show that \(f\) is continuous.

See also point 3 of the exercise 5.