To clarify Remark 6.6: A family of functions \(\mathcal F\subset C(K,\mathbb R^m)\) mapping a set \(K\subset\mathbb R^n\) into \(\mathbb R^m\) is pointwise equicontinuous on \(K\) provided for every \(x\in K\) and \(\epsilon>0\) there exists some \(\delta>0\) (which may depend on \(x\)) such that \(\|f(x)-f(y)\|<\epsilon\) for all \(f\in\mathcal F\) and \(y\in K\) with \(\|x-y\|<\delta\). The family \(\mathcal F\) is uniformly equicontinuous if for every \(\epsilon>0\) there exists some \(\delta>0\) such that \(\|f(x)-f(y)\|<\epsilon\) for all \(f\in\mathcal F\) and \(x,y\in K\) with \(\|x-y\|<\delta\). Prove that these definitions are equivalent when \(K\) is compact.
Solution
Goal: Prove pointwise equicontinuity and uniform equicontinuity of a family \(\mathcal F \subseteq C(K;\mathbb R^m)\) are equivalent when \(K\) is compact.
1. Uniform equicontinuity implies pointwise equicontinuity.Proof: if \(\delta\) works for all \(x, y\) simultaneously, then in particular it works for all \(y\) with \(y\) near a fixed \(x\); so the pointwise condition holds with the same \(\delta\) (independent of \(x\), hence certainly allowed to depend on \(x\)).
2. Pointwise equicontinuity implies uniform equicontinuity.
2.1. Fix \(\varepsilon > 0\).For each \(x \in K\) choose \(\delta_x > 0\) such that \(\|x - y\| < \delta_x \Rightarrow \|f(x) - f(y)\| < \varepsilon/2\) for all \(f \in \mathcal F\).Proof: the pointwise equicontinuity hypothesis at \(x\) with \(\varepsilon/2\) in place of \(\varepsilon\).
2.2. The balls \(B(x, \delta_x/2)\), \(x \in K\), cover \(K\).Proof: \(x \in B(x, \delta_x/2)\) for each \(x\).
2.3. \(K\) is covered by finitely many of them: \(K \subseteq \bigcup_{i=1}^n B(x_i, \delta_{x_i}/2)\).Proof: compactness of \(K\) applied to the open cover of
2.2.
2.4. Set \(\delta := \min_i \delta_{x_i}/2 > 0\).If \(\|x - y\| < \delta\), then for every \(f \in \mathcal F\), \(\|f(x) - f(y)\| < \varepsilon\).Proof: \(x\) lies in some \(B(x_i, \delta_{x_i}/2)\) by 3, so \(\|x - x_i\| < \delta_{x_i}/2\); then \(\|y - x_i\| \le \|y - x\| + \|x - x_i\| < \delta + \delta_{x_i}/2 \le \delta_{x_i}/2 + \delta_{x_i}/2 = \delta_{x_i}\) since \(\delta \le \delta_{x_i}/2\).Both \(x\) and \(y\) are within \(\delta_{x_i}\) of \(x_i\), so \(\|f(x) - f(x_i)\| < \varepsilon/2\) and \(\|f(y) - f(x_i)\| < \varepsilon/2\), hence \(\|f(x) - f(y)\| \le \|f(x)-f(x_i)\| + \|f(y)-f(x_i)\| < \varepsilon\) by the triangle inequality.This holds for all \(f \in \mathcal F\) simultaneously, and \(\delta\) is independent of \(x, y, f\).
2.5. Q.E.D. Proof: 4 is uniform equicontinuity.