Define \(g : \mathbb {R}\to \mathbb {R}\) by

which is continuous (it is piecewise linear and the pieces agree at the breakpoints \(a, b\)). Let \(f(x) = 1 - \prod _{i=1}^{\ell }(1 - g(x[i]))\). By Lemma 5, \(f\) is continuous (a finite product and difference of continuous maps). If \(x[i] \ge b\) for all \(i\) then every \(g(x[i]) = 1\), so the product is \(0\) and \(f(x) = 1\). If \(x[i] {\lt} a\) for all \(i\) then every \(g(x[i]) = 0\), so the product is \(1\) and \(f(x) = 0\).

Define \(g : \mathbb {R}\to \mathbb {R}\) by \(g(x) = 1\) if \(x \le 1\), \(g(x) = 2 - x\) if \(1 {\lt} x \le 2\), and \(g(x) = 0\) if \(x {\gt} 2\), which is continuous. Let \(f_1(x) = (x, 1)\) and \(f_2(x) = (0, x)\) in \(\mathbb {R}^{\ell +1}\), both continuous. Set

By Lemma 5, \(f\) is continuous, and one checks directly that it equals \((x,1)\) when \(x[\ell ] \le 1\) and \((0,x)\) when \(x[\ell ] {\gt} 2\) (and interpolates continuously in between).

Let \(g\) be the same continuous function as in Lemma 54, and let \(f_1(x) = \left(\frac{x}{\lVert x \rVert _1}, 1\right)\) and \(f_2(x) = \left(0, \frac{x}{\lVert x \rVert _1}\right)\), both continuous on the relevant domain (nonzero inputs). Then

is continuous by Lemma 5 and satisfies the required piecewise definition.