Let \(\mathcal F\) be a uniformly bounded, equicontinuous family of real-valued functions on a metric space \((X,d)\). Prove that \[ g(x)=\sup\{f(x):f\in\mathcal F\} \] is continuous.
Solution
1. The value \(g(x)\) is finite for every \(x\in X\).
Uniform boundedness gives a constant \(M>0\) such that \[ \abs{f(x)}\leq M \] for every \(f\in\mathcal F\) and every \(x\in X\). Hence \[ -M\leq f(x)\leq M \] for all \(f\in\mathcal F\), so the supremum defining \(g(x)\) is a finite real number.
2. Fix \(x_0\in X\) and \(\varepsilon>0\). There exists \(\delta>0\) suchthat, whenever \(d(x,x_0)<\delta\),\[ g(x)\leq g(x_0)+\frac{\varepsilon}{2}. \]
By equicontinuity at \(x_0\), there exists \(\delta>0\) such that \[ d(x,x_0)<\delta \quad\Longrightarrow\quad \abs{f(x)-f(x_0)}<\frac{\varepsilon}{2} \] for every \(f\in\mathcal F\). Thus, for such \(x\) and every \(f\in\mathcal F\), \[ f(x) < f(x_0)+\frac{\varepsilon}{2} \leq g(x_0)+\frac{\varepsilon}{2}. \] Taking the supremum over \(f\in\mathcal F\) gives \[ g(x)\leq g(x_0)+\frac{\varepsilon}{2}. \]
3. For the same \(\delta\), whenever \(d(x,x_0)<\delta\),\[ g(x_0)\leq g(x)+\frac{\varepsilon}{2}. \]
The equicontinuity estimate from step 2 also gives, for every \(f\in\mathcal F\), \[ f(x_0) < f(x)+\frac{\varepsilon}{2} \leq g(x)+\frac{\varepsilon}{2}. \] Taking the supremum over \(f\in\mathcal F\) yields the stated inequality.
4. The function \(g\) is continuous at \(x_0\).
By steps 2 and 3, if \(d(x,x_0)<\delta\), then \[ \abs{g(x)-g(x_0)} \leq \frac{\varepsilon}{2} < \varepsilon. \] This is continuity at \(x_0\).
5. Q.E.D.
The point \(x_0\in X\) was arbitrary, so step 4 proves that \(g\) is continuous on \(X\).