Let \(K\) be continuous on \([0,1]\times[0,1]\). Let \(\mathcal F\) be the family of functions \[ f(x)=\int_0^1 g(y)K(x,y)\,dy, \] where \(g\) is continuous on \([0,1]\) and \(|g(y)|\le1\) everywhere. Prove that \(\mathcal F\) is equicontinuous.
Solution
1. For every \(\varepsilon>0\), there is \(\delta>0\) such that\[ \abs{x-x'}<\delta \quad\Longrightarrow\quad \abs{K(x,y)-K(x',y)}<\varepsilon \]for all \(x,x',y\in[0,1]\).
The function \(K\) is continuous on the compact set \([0,1]\times[0,1]\), so it is uniformly continuous. Apply uniform continuity to the two points \((x,y)\) and \((x',y)\), whose distance is \(\abs{x-x'}\).
2. With \(\delta\) as in step 1, every \(f\in\mathcal F\) satisfies\[ \abs{x-x'}<\delta \quad\Longrightarrow\quad \abs{f(x)-f(x')}<\varepsilon. \]
Let \[ f(x)=\int_0^1 g(y)K(x,y)\,dy \] with \(\abs{g(y)}\leq1\). If \(\abs{x-x'}<\delta\), then step 1 gives \[ \begin{aligned} \abs{f(x)-f(x')} &= \abs{\int_0^1 g(y)\bigl(K(x,y)-K(x',y)\bigr)\,dy}\\ &\leq \int_0^1 \abs{g(y)} \abs{K(x,y)-K(x',y)}\,dy\\ &< \int_0^1\varepsilon\,dy\\ &= \varepsilon. \end{aligned} \] The same \(\delta\) works for every admissible \(g\) and hence for every \(f\in\mathcal F\).
3. The family \(\mathcal F\) is equicontinuous.
Step 2 is precisely the uniform-in-\(f\) \(\varepsilon\)–\(\delta\) condition for equicontinuity on \([0,1]\).
4. Q.E.D.
Step 3 proves the claim.