Let \(W\) be the subset of the space \(C[0,1]\) of real-valued, continuous functions on \([0,1]\) satisfying the conditions: \[ |f(x) - f(y)| < |x - y| \qquad \int_0^1 f(x)^2\,dx = 1 \]
Prove that \(W\) is uniformly bounded, i.e., there exists \(M > 0\) such that \(|f(x)| \leq M\) for all \(x \in [0,1]\). Hint: Show first that \(|f(0)| \leq 2\) for all \(f \in W\).
Prove that \(W\) is a compact subset of \(C[0,1]\) under the sup norm \(\|f\|_\infty = \sup_{x \in [0,1]} |f(x)|\).
Erratum: with the strict inequality of the source, read for \(x \neq y\) (for \(x = y\) it would make \(W\) empty), part (B) is false because \(W\) is not closed. Let \(c = \frac{-1 + \sqrt{11/3}}{2}\), so that \(g(x) = x + c\) satisfies \(\int_0^1 g^2 = \frac13 + c + c^2 = 1\). For \(n \ge 2\) choose \(c_n\) with \(\int_0^1 \bigl((1 - \frac1n)x + c_n\bigr)^2\,dx = 1\) and \(c_n \to c\), and put \(f_n(x) = (1 - \frac1n)x + c_n\). Then \(|f_n(x) - f_n(y)| = (1 - \frac1n)|x - y| < |x - y|\) for \(x \neq y\), so \(f_n \in W\), and \(f_n \to g\) uniformly, but \(|g(1) - g(0)| = 1 = |1 - 0|\), so \(g \notin W\). Part (B) holds when the condition is \(|f(x) - f(y)| \le |x - y|\) for all \(x, y \in [0,1]\): that set is closed, and it is uniformly bounded by (A) and equicontinuous, hence compact by the Arzelà–Ascoli theorem.