ALG-B3-M02-P020 A function squeezed downward

Hint 1. First prove nonnegativity by comparing \(s\) with \(s+1\).

Hint 2. If \(H(t_0)>1\), move left through midpoints and get larger and larger values.

For any \(s\), since \(s0\). Fix \(r1+2^n a\) for all \(n\). For \(n=1\), \(H(r)\ge H(t_0)^2=(1+a)^2>1+2a\). For the induction step, take \(m=(r+t_0)/2\). It lies between \(r\) and \(t_0\); applying the same midpoint argument gives \(H(m)>1+2^n a\), hence \(H(r)\ge H(m)^2>1+2^{n+1}a\). Thus the fixed number \(H(r)\) exceeds \(1+2^n a\) for all \(n\), impossible. Therefore \(H(t)\le1\) for all \(t\), and with nonnegativity, \(0\le H(t)\le1\).

A. Source analysis. Main objects: a functional inequality for every pair of ordered arguments.

B. Insufficient first move. Testing individual values does not bound the function from above.

C. Hidden observation. If one value exceeds 1, moving left through midpoints produces unbounded growth.

D. Required move. First prove nonnegativity, then run an induction that doubles the excess.

E. Number of ideas. Two ideas: nonnegativity and explosive midpoint induction.

F. Difficulty justification. Final level 8: the method is hidden and requires constructing an infinite contradiction.

G. Why it is not one-step. It is not one-step: the upper bound appears only after nonnegativity and induction.