In the proof of the Tietze theorem, how essential was the clever decision in Step 1 to divide the interval \([-r, r]\) into three equal pieces? Suppose instead that one divides this interval into the three intervals
\[ I_1 = [-r, -ar], \quad I_2 = [-ar, ar], \quad I_3 = [ar, r], \]
for some \(a\) with \(0 < a < 1\). For what values of \(a\) other than \(a = 1/3\) (if any) does the proof go through?
Solution
Suppose \(f:A\to[-r,r]\), and replace the thirds in Step 1 by \[ [-r,-ar],\qquad[-ar,ar],\qquad[ar,r], \] with \(0<a<1\). The same Urysohn construction produces a continuous \(g:X\to[-ar,ar]\) that equals \(-ar\) on \(f^{-1}([-r,-ar])\) and \(ar\) on \(f^{-1}([ar,r])\).
The error satisfies \[ |f-g|\le q r, \qquad q=\max\{1-a,\,2a\}. \] Indeed, on the two outer pieces the worst error is \(r-ar=(1-a)r\), while on the middle piece both \(f\) and \(g\) lie in \([-ar,ar]\), so the worst error is \(2ar\).
The iterative Tietze proof works precisely when this error factor is a strict contraction: \[ q<1. \] Since \(a>0\) already gives \(1-a<1\), the additional condition is \[ 2a<1. \] Thus the proof goes through for exactly \[ \boxed{0<a<1/2.} \] For such \(a\), the residual norms decrease geometrically by \(q<1\), and the successive correction functions have uniformly summable sup norms. At \(a=1/2\) the estimate gives \(q=1\), so the iterative argument no longer forces convergence; for \(a>1/2\) it is even worse.