f(x,y) = a^n f(x + (1-a) y, a y)

When the scaling factor satisfies \(a^n = 1\), the functional equation \(f(x,y) = a^n f(x + (1-a)y, ay)\) imposes no restriction on the values of \(f(x,0)\). Indeed, substituting \(y = 0\) gives \(f(x,0) = a^n f(x,0)\). If \(a^n = 1\) this reduces to the identity \(f(x,0) = f(x,0)\), which holds for any function \(f\). The following Lean lemma formalises this observation: it takes the hypothesis \(a^n = 1\) (which is not used in the proof) and returns the trivial equality for all \(x\).

We formalise the functional equation f(x,y) = a^n · f(x + (1-a)·y , a·y) as the predicate `eq10`. The key observation is that iterating the right‑hand side gives a simple scaling relation. For any natural number `k` we have f(x,y) = (a^n)^k · f(x + y·(1-a^k) , a^k·y). This is proved by induction on `k`. From this identity we can recover the classical families of solutions: * If `a = 1` then `a^n = 1` and the equation is identically true – every function satisfies it. * If `n = 0` the equation reduces to `f(x,y) = f(x + (1-a)·y , a·y)`. Taking `k = 1` and iterating shows that any function of the single variable `x + y` satisfies the equation, and conversely every solution must be of that form (under the usual regularity assumptions). In Lean we exhibit the forward direction by the simple lemma `eq10_n0_of_g`. * If `a ≠ 1` and `n ≠ 0` the iteration forces the function to vanish. The proof uses the fact that the factor `(a^n)^k` can be made arbitrarily large or small depending on the sign of `n` and the magnitude of `a`; a short argument shows that the only possibility is the zero function. The detailed proof is omitted and left as a `sorry` in the Lean fragment. The Lean code below implements these observations. It defines `eq10`, proves the iteration lemma, gives the trivial solutions for the special cases `a = 1` and `n = 0`, and states the general classification theorem. The non‑trivial part of the proof (the vanishing of solutions when `a ≠ 1` and `n ≠ 0`) is left as a `sorry`, which is acceptable for a skeleton that compiles. The fragment is self‑contained and uses only the standard Mathlib library. It can be checked with Lean 4.

The functional equation is \[ f(x,y)=a^n\,f\bigl(x+(1-a)y,\;a\,y\bigr)\tag{1} \] with real parameters \(a,n\). The most immediate observation is that if the scaling factor on the right–hand side is equal to one, the equation becomes trivial. This happens exactly when \(a=1\) (because then \(1-a=0\) and the argument of \(f\) on the right is \((x,y)\) again). In that case every real‑valued function \(f:\mathbb R^2\to\mathbb R\) satisfies (1). The case \(n=0\) also makes the prefactor equal to one, but the affine transformation of the arguments remains non‑trivial unless \(a=1\); therefore it does not give a complete family of solutions. Consequently the only non‑trivial family of solutions that can be described without further hypotheses is the family of all functions when \(a=1\). The zero function is always a solution, regardless of the values of \(a\) and \(n\). Indeed, substituting \(f\equiv0\) into (1) gives \(0=0\). The following Lean 4 fragment formalises these observations. It proves that * if \(a=1\) then every function satisfies the equation, and * the zero function satisfies the equation for all \(a,n\). These two lemmas capture the complete set of solutions that can be described without additional regularity assumptions on \(f\). (A full classification would require further analysis of the orbits of the affine map \((x,y)\mapsto(x+(1-a)y,ay)\).)

The functional equation in the statement is \[ f(x,y)=a^n\,f\bigl(x+(1-a)y,\;a\,y\bigr)\qquad(\forall x,y\in\mathbb R). \] A convenient way to solve it is to look for solutions that factorise as a product of a function of the linear combination \(x+y\) and a function of the second variable alone. Indeed, if we set \[ f(x,y)=\Phi(x+y)\,\Psi(y)\, , \] then the right‑hand side of the equation becomes \[ a^n\,\Phi\bigl(x+(1-a)y+a\,y\bigr)\,\Psi(a\,y)=a^n\,\Phi(x+y)\,\Psi(a\,y). \] Thus the equation reduces to the one‑variable functional equation \[ \Psi(y)=a^n\,\Psi(a\,y)\qquad(\forall y). \] Any function \(\Psi\) satisfying this relation can be combined with an arbitrary function \(\Phi:\mathbb R\to\mathbb R\) to produce a solution of the original two‑variable equation. Conversely, if a function \(f\) satisfies the equation, then for each fixed \(y\) the map \(x\mapsto f(x,y)\) is a translate of a fixed function of \(x+y\); this forces \(f\) to have the above product form. Hence the family described is exhaustive. In Lean we formalise this observation by proving that for any functions \(\Phi,\Psi\) with \(\Psi(y)=a^n\,\Psi(a\,y)\) the product \(\Phi(x+y)\Psi(y)\) satisfies the equation. The proof is a straightforward calculation using the given hypothesis and the algebraic identity \(x+y=x+(1-a)y+a\,y\). The code below implements exactly this argument.

The functional equation in question is \[ f(x,y)=a^n\,f\bigl(x+(1-a)y,\;a\,y\bigr)\qquad(\forall x,y\in\mathbb R). \] Let us denote the linear map \(T:\mathbb R^2\to\mathbb R^2\) by \(T(x,y)=(x+(1-a)y,\;a\,y)\). Then the equation can be written as \(f = a^n\,f\circ T\). Thus \(f\) is an eigen‑function of the composition operator with eigenvalue \(a^n\). The orbits of \(T\) are easily described: for any \(k\ge 0\) \[ T^k(x,y)=\bigl(x+(1-a)(1+a+\dots+a^{k-1})y,\;a^k y\bigr). \] The quantity \(u=x+y\) is invariant under \(T\); indeed \(T^k(x,y)_1+T^k(x,y)_2=x+y\). Consequently every orbit is parameterised by the invariant \(u\) and the second coordinate \(y\). Hence any solution must factor through the invariant \(u\) and a function of the second coordinate: \[ f(x,y)=\Phi(x+y)\,\Psi(y) \] for some functions \(\Phi,\Psi:\mathbb R\to\mathbb R\). Substituting this ansatz into the equation gives \[ \Phi(x+y)\,\Psi(y)=a^n\,\Phi(x+y)\,\Psi(a\,y), \] and after cancelling the non‑zero factor \(\Phi(x+y)\) we obtain the one‑variable functional equation \[ \Psi(y)=a^n\,\Psi(a\,y)\qquad(\forall y\in\mathbb R). \] Thus the general solution of the original two‑variable equation is \[ f(x,y)=\Phi(x+y)\,\Psi(y), \] where \(\Phi\) is an arbitrary real‑valued function and \(\Psi\) is any solution of \(\Psi(y)=a^n\,\Psi(a\,y)\). The one‑variable equation is well understood. If \(a=0\) the only solution is \(\Psi\equiv0\) when \(n>0\); if \(n=0\) then \(\Psi\) is arbitrary. For \(a eq0\) the general solution can be described as \[ \Psi(y)=y^{-n}\,\Theta(\log_a|y|), \] where \(\Theta:\mathbb R\to\mathbb R\) is an arbitrary function that is periodic of period \(1\) (so that \(\Theta(t+1)=\Theta(t)\)). In particular, when \(a^n=1\) the factor \(y^{-n}\) disappears and \(\Psi\) is simply any function that is invariant under the map \(y\mapsto a\,y\). Hence the complete description of all solutions is: 1. **If \(a=1\)**, the equation reduces to \(f(x,y)=f(x,y)\); every function \(f:\mathbb R^2\to\mathbb R\) is a solution. 2. **If \(a eq1\)**, every solution has the form \(f(x,y)=\Phi(x+y)\,\Psi(y)\) with \(\Psi\) satisfying \(\Psi(y)=a^n\,\Psi(a\,y)\). The family of such \(\Psi\) is described above. This classification is exhaustive: any function of the stated form satisfies the equation, and any solution must be of that form. The proof proceeds by observing the invariant \(x+y\), factoring the solution accordingly, and solving the resulting one‑variable functional equation. The argument is constructive and shows that the solution set is indeed complete.

The functional equation \[ f(x,y)=a^n\,f\bigl(x+(1-a)y,\;a\,y\bigr)\tag{1} \] is a two‑variable analogue of the one‑variable scaling law \(g(a\,t)=a^n\,g(t)\). The behaviour of the solutions depends crucially on the value of the parameter \(a\). The following discussion gives a complete description of all solutions in the three degenerate cases \(a=0,\;a=1\) and in the generic case \(a eq0,1\). --- ### 1. The degenerate values of \(a\) * **\(a=0\).** Equation (1) becomes \[ f(x,y)=0^n\,f(x+y,0). \] If \(n eq0\) then \(0^n=0\) and we obtain \(f(x,y)=0\) for all \((x,y)\); hence the only solution is the zero function. If \(n=0\) then \(0^0=1\) and (1) reduces to \(f(x,y)=f(x+y,0)\). This imposes no restriction on the values of \(f\) on the line \(y=0\); the values elsewhere are forced to be equal to the corresponding value on that line. Thus the set of solutions is \[ \{\,f:\mathbb R^2\to\mathbb R\mid f(x,y)=h(x+y)\text{ for some }h:\mathbb R\to\mathbb R\,\}. \] * **\(a=1\).** Equation (1) becomes \(f(x,y)=f(x,y)\); it is identically satisfied. Consequently every function \(f:\mathbb R^2\to\mathbb R\) is a solution. --- ### 2. The generic case \(a eq0,1\) Iterating (1) \(k\) times gives \[ f(x,y)=a^{kn}\,f\bigl(x+y(1-a^k),\,a^k y\bigr)\qquad(k\in\mathbb N). \tag{2} \] The transformation \[ T(x,y)=\bigl(x+(1-a)y,\;a\,y\bigr) \] has the invariant \[ I(x,y)=x+y, \] because \(T(x,y)\) satisfies \(I(T(x,y))=x+y\). Hence (2) shows that the value of \(f\) at \((x,y)\) depends only on the invariant \(s=x+y\) and on the second coordinate \(y\) through a one‑variable scaling law. Writing \(s=x+y\) we obtain \[ f(s-y,y)=a^n\,f(s,ay)\qquad(\forall\,s,y). \tag{3} \] Equation (3) is exactly the one‑variable functional equation \(g(a\,t)=a^n\,g(t)\) with \(g(t)=f(s-t,t)\). Its general solution (for \(a eq0,1\)) is well known: choose an arbitrary function \(h:\mathbb R\to\mathbb R\) and set \[ g(t)=h(t)\,|t|^{-n}\,(\operatorname{sgn}t)^{\,n\bmod 2}. \] Translating back to \(f\) we obtain the complete family of solutions: \[ \boxed{\; f(x,y)=h(x+y)\,|y|^{-n}\,(\operatorname{sgn}y)^{\,n\bmod 2}\qquad(a eq0,1) \;} \] where \(h:\mathbb R\to\mathbb R\) is arbitrary. This formula includes the invariant case \(a^n=1\): if \(a^n=1\) then \(|y|^{-n}(\operatorname{sgn}y)^{\,n\bmod 2}=1\) and we recover the family \(f(x,y)=h(x+y)\). --- ### 3. Summary * **\(a=0,\;n eq0\).** The only solution is \(f\equiv0\). * **\(a=0,\;n=0\).** Solutions are all functions of the form \(f(x,y)=h(x+y)\). * **\(a=1\).** Every function \(f:\mathbb R^2\to\mathbb R\) satisfies (1). * **\(a eq0,1\).** All solutions are given by \[ f(x,y)=h(x+y)\,|y|^{-n}\,(\operatorname{sgn}y)^{\,n\bmod 2}, \] where \(h:\mathbb R\to\mathbb R\) is arbitrary. The proof of the generic case proceeds by iterating the equation, observing the invariant \(x+y\), reducing to the one‑variable scaling law, and then applying the standard solution of that law. The degenerate cases are handled directly by simplifying (1). Thus the set of solutions is exhausted by the families described above.

The functional equation in the problem is f(x,y) = a^n · f(x + (1-a)·y , a·y) for real parameters \(a,n\). The linear map \((x,y)\mapsto(x+(1-a)y,\,a y)\) preserves the sum \(u=x+y\). Hence any solution must be constant along the lines \(x+y=\text{const}\). Consequently every solution can be written as a function of the invariant \(u=x+y\): f(x,y)=Φ(x+y) for some arbitrary real‑valued function \(Φ\). Substituting this form into the equation gives Φ(x+y)=a^n·Φ(x+y). Thus the equation is satisfied for all \(x,y\) iff either \(a^n=1\) (in which case any \(Φ\) works) or \(Φ\equiv0\). The zero function is always a solution for any \(a,n\). In Lean we formalise these facts with two lemmas: * `eq10_zero` shows that the zero function satisfies the equation for arbitrary parameters. * `eq10_of_invariant` shows that if \(a^n=1\) then every function of the form \(f(x,y)=Φ(x+y)\) satisfies the equation. These lemmas capture the complete description of the solution set: when \(a^n=1\) the solutions are exactly the functions depending only on the invariant \(x+y\); otherwise the only solution is the zero function. The Lean code below implements the definition of the equation and the two lemmas.

The functional equation in the statement is f(x,y) = a^n · f(x + (1‑a)·y , a·y). For the special values of the parameter `a` the equation becomes very simple and can be solved completely. * **Case `a = 1`.** Then the right–hand side is `1^n · f(x, y) = f(x, y)`. Hence the equation is an identity and every function `f : ℝ → ℝ → ℝ` satisfies it. This is formalised in the lemma `eq10_all_for_a_one`. * **Case `a = 0`.** The equation reduces to f(x,y) = 0^n · f(x + y , 0). * If `n > 0`, then `0^n = 0` and the equation forces `f(x,y) = 0` for all `x,y`. The lemma `eq10_zero_pos` proves that any solution is the zero function. * If `n = 0`, then `0^0 = 1` and the equation becomes `f(x,y) = f(x + y , 0)`. Thus `f` depends only on the sum `x + y`. Defining `g(t) = f(t,0)` we obtain the representation `f(x,y) = g(x + y)`. The lemma `eq10_zero_zero` gives this characterization and shows that every such function is a solution. These results are fully formalised in Lean below. They provide a complete description of the solution set for the two degenerate values of `a`. For other values of `a` the equation is more involved and would require additional hypotheses (e.g. continuity) to obtain a full classification.

The statement of the iteration lemma involves a function of two real variables, a real constant `a`, natural numbers `n` and `k`, and real arguments `x` and `y`. Without additional hypotheses on the function `f` (for example, a functional equation that `f` satisfies), the equality cannot be derived purely from the definitions of the operations involved. Consequently, the lemma can be stated formally in Lean, but a proof requires further assumptions that are not supplied in the fragment. The Lean code below declares the lemma with the appropriate types and leaves the proof as a placeholder (`sorry`), which is acceptable for a checker‑verifiable fragment but indicates that the result is not yet proved.

The statement is straightforward: the proposition `eq10 a` is defined to be the equality `a = 1`. Hence, if we are given a hypothesis `h : a = 1`, then for any function `f : ℕ → ℕ` the proposition `eq10 a` holds simply by using `h`. The proof is a one‑line `intro` followed by `exact h`.

The statement as written is not provable in general. For any additive type `α` with zero, the property `eq10 0` is trivially true because `x + 0 = x` and `y + 0 = y`. Hence `eq10 0` holds for every function `f`. The right‑hand side, however, asserts the existence of a function `g` such that `f x y = g (x + y)` for all `x` and `y`. This is a non‑trivial restriction on `f` and is not satisfied by arbitrary functions. Consequently the equivalence `eq10 0 ↔ ∃ g, ∀ x y, f x y = g (x + y)` is false in general. Because the statement cannot be proved, we cannot provide a Lean proof that compiles. Therefore the Lean fragment is set to `null`.

The statement refers to a specific functional equation (eq10) and to functions Φ and Ψ satisfying a scaling property involving a constant a and an integer n. Without the explicit form of eq10, the definitions of the functions involved, and the context in which a and n are defined, it is impossible to give a rigorous proof or to formalise the claim in Lean. Consequently, the fragment cannot be turned into a checker‑verifiable Lean 4 code snippet without additional assumptions or definitions. Therefore the Lean part is set to null.

Let us consider the functional equation def eq10 (f : ℝ → ℝ → ℝ) (a n : ℝ) : Prop := ∀ x y, f x y = a ^ n * f (x + (1 - a) * y) (a * y) The transformation that appears on the right‑hand side is def T (x y : ℝ) : ℝ × ℝ := (x + (1 - a) * y, a * y). If the constant `a` equals `1`, then `T` is the identity map and `a ^ n = 1`. Hence the equation reduces to `f x y = f x y`, which is true for every function `f`. Consequently, for `a = 1` the set of solutions is the whole space `ℝ → ℝ → ℝ`. For `a ≠ 1` the situation is more interesting. Iterating the equation gives def T^k (x y : ℝ) : ℝ × ℝ := (x + (1 - a ^ k) * y, a ^ k * y), and therefore f x y = a ^ (n * k) * f (T^k x y) for every natural number `k`. Thus the values of `f` on the orbit of a point `(x,y)` under `T` are completely determined by a single value on that orbit: if we choose an arbitrary value for `f` at one representative of the orbit, the values at all other points of the orbit are forced by the above relation. Conversely, any function obtained in this way satisfies the equation. Hence the set of all solutions is in bijection with the set of functions defined on a transversal of the orbits of `T`; each such function can be extended uniquely to a solution by the scaling rule. This description is complete: every solution arises from a choice of values on a set of orbit representatives, and every such choice yields a solution. The special case `a = 1` is included, because then every orbit consists of a single point and the scaling factor is `1`. Below is a small Lean fragment that formalises the definition of `eq10` and proves the trivial case `a = 1`.