Let \(K\in C([0,1]\times[0,1])\) and define \[ Tf(x)=\int_0^1K(x,y)f(y)\,dy \] for \(f\in C([0,1])\). Prove that \(Tf\in C([0,1])\). Moreover, prove that \[ \Omega=\{Tf:\|f\|_\infty\le1\} \] is precompact in \(C([0,1])\) with the supremum norm.
Solution
Fix \(f\in C([0,1])\). Since \(K\) is uniformly continuous on the compact square, if \(x_n\to x\), then \[ \sup_{y\in[0,1]}|K(x_n,y)-K(x,y)|\longrightarrow0. \] Hence \[ |Tf(x_n)-Tf(x)| \le\|f\|_\infty\sup_y|K(x_n,y)-K(x,y)|\longrightarrow0. \] Thus \(Tf\) is continuous.
Now let \(\|f\|_\infty\le1\). Since \(K\) is bounded, \[ |Tf(x)|\le\int_0^1|K(x,y)|\,dy\le\|K\|_\infty, \] so the family \(\Omega\) is uniformly bounded. Also, for \(x,x'\in[0,1]\), \[ |Tf(x)-Tf(x')| \le\int_0^1|K(x,y)-K(x',y)|\,dy \le\sup_y|K(x,y)-K(x',y)|. \] The right-hand side tends to \(0\) as \(x'\to x\), uniformly in all \(f\) with \(\|f\|_\infty\le1\), again by uniform continuity of \(K\). Thus \(\Omega\) is equicontinuous.
By the Arzelà–Ascoli theorem, every sequence in \(\Omega\) has a uniformly convergent subsequence whose limit lies in \(C([0,1])\). Therefore \(\Omega\) is precompact in the supremum norm.