lean_workbook_plus_72201

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Find the convolution $f\ast f$ of the function $f: [-1/2,1/2] \to \Bbb R$ given by $f(x) = 1$ for $|x| \le 1/4$ and $f(x) = 0$ otherwise.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_72201 (f : ℝ → ℝ) (hf : f = fun x => if abs x ≤ 1/4 then 1 else 0) : f * f = fun x => if abs x ≤ 1/2 then 1 else 0   :=  by sorry

Convolution became pointwise function multiplication, and the proposed convolution is a rectangle instead of the triangular overlap length.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Extend f=1_[-1/4,1/4] by0 onℝ and define(ff)(x)=∫f(t)f(x−t)dt. Prove(ff)(x)=max(1/2−|x|,0).

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.