Finite-Dimensional Translation Spaces and Lie Group Projections

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1. The Start

This note starts with the following problem.

Example 1.1. Let $𝑓 : ℝ \to ℝ$ be a continuous non-constant function, which satisfies

𝑓 (𝑥 + 𝑦) 𝑓 (𝑥 - 𝑦) = 𝑓 (𝑥) 2 - 𝑓 (𝑦) 2

𝑓 (𝑥 + 𝑦) 𝑓 (𝑥 - 𝑦) = 𝑓 (𝑥) 2 - 𝑓 (𝑦) 2

𝑓 (𝑥 + 𝑦) 𝑓 (𝑥 - 𝑦) = 𝑓 (𝑥) 2 - 𝑓 (𝑦) 2

. Find all such functions.

I have the cover linear theorem in the Lie section for it. But the conclusion of that theorem depends on the choice of $𝐺$ $𝐺$ , $𝜋$ $𝜋$ and $Φ$ $Φ$ , which weakens its power.

2. The Climax

Thus, I wanted to find another approach. And I got the idea from the standard solution: Let $𝑔 𝑎 (𝑥) ≔ 𝑓 (𝑥 + 𝑎) - 𝑓 (𝑥 - 𝑎) 2 𝑓 (𝑎)$ . Now, if we plug $𝑓 = s i n$ into this expression, after some calculations, we can get:

𝑔 = c o s

𝑔 = c o s

. The same can be done with $𝑓 = s i n h$ .

In the case of $𝑓 = s i n$ , $𝑔 = c o s$ , it gives us a circle in a 2-dimensional space. So, we can treat this equation as a projection of some 2-dimensional curve on $ℝ$ $ℝ$ . More generally, most equations can be regarded as a projection of a 1-dimensional manifold in a high dimensional space on $ℝ$ $ℝ$ , and in which the curve can be one-parameterized by $ℝ$ $ℝ$ .

3. The End

Finally, after chosing conditions which are strong enough to make conclusions, we have the following theorem.

Theorem 3.1 (Finite-Dimensional Translation Space). Let $𝑓 : ℝ \to ℝ$ be continuous and define its translation space $𝑉 𝑓 = s p a n ℝ {𝜏 𝑎 𝑓 : 𝑎 \in ℝ}$ , where $(𝜏 𝑎 𝑓) (𝑥) = 𝑓 (𝑥 + 𝑎)$ .
If $d i m 𝑉 𝑓 < \infty$ , then:
$𝑓$ $𝑓$ is real analytic.
There exist distinct $𝜆 1, \dots, 𝜆 𝑟 \in ℂ$ and nonzero polynomials $𝑃 1, \dots, 𝑃 𝑟$ $𝑃 1, \dots, 𝑃 𝑟$ such that
$𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,$ $𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,$ $𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,$
and moreover
$𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .$ $𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .$ $𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .$
In particular, $𝑟 \leq d i m 𝑉 𝑓$ .
$𝑓$ $𝑓$ is a matrix coefficient of a one-parameter subgroup of $G L (𝑉 𝑓)$ $G L (𝑉 𝑓)$ .

Proof. Let $𝑇 𝑎 : 𝑉 𝑓 \to 𝑉 𝑓$ be the restriction of the translation operator $𝜏 𝑎$ $𝜏 𝑎$ to $𝑉 𝑓$ $𝑉 𝑓$ . Since $𝑉 𝑓$ $𝑉 𝑓$ is invariant and finite-dimensional, $𝑇 𝑎 \in G L (𝑉 𝑓)$ .

The map

𝜌 : ℝ \to G L (𝑉 𝑓), 𝜌 (𝑎) = 𝑇 𝑎

𝜌 : ℝ \to G L (𝑉 𝑓), 𝜌 (𝑎) = 𝑇 𝑎

𝜌 : ℝ \to G L (𝑉 𝑓), 𝜌 (𝑎) = 𝑇 𝑎

is a group homomorphism:

𝜌 (𝑎 + 𝑏) = 𝜌 (𝑎) 𝜌 (𝑏), 𝜌 (0) = 𝐼 .

𝜌 (𝑎 + 𝑏) = 𝜌 (𝑎) 𝜌 (𝑏), 𝜌 (0) = 𝐼 .

𝜌 (𝑎 + 𝑏) = 𝜌 (𝑎) 𝜌 (𝑏), 𝜌 (0) = 𝐼 .

It is continuous because for every $ℎ \in 𝑉 𝑓$ , the map $𝑎 \mapsto 𝑇 𝑎 ℎ$ is continuous in the finite-dimensional topology of $𝑉 𝑓$ $𝑉 𝑓$ .

Hence there exists a unique $𝐵 \in E n d (𝑉 𝑓)$ such that

𝜌 (𝑥) = 𝑒 𝑥 𝐵 \forall 𝑥 \in ℝ .

𝜌 (𝑥) = 𝑒 𝑥 𝐵 \forall 𝑥 \in ℝ .

𝜌 (𝑥) = 𝑒 𝑥 𝐵 \forall 𝑥 \in ℝ .

Define $𝐿 : 𝑉 𝑓 \to ℝ$ by $𝐿 (ℎ) = ℎ (0)$ . Then

𝑓 (𝑥) = (𝜏 𝑥 𝑓) (0) = 𝐿 (𝜏 𝑥 𝑓) = 𝐿 (𝑒 𝑥 𝐵 𝑓) .

𝑓 (𝑥) = (𝜏 𝑥 𝑓) (0) = 𝐿 (𝜏 𝑥 𝑓) = 𝐿 (𝑒 𝑥 𝐵 𝑓) .

𝑓 (𝑥) = (𝜏 𝑥 𝑓) (0) = 𝐿 (𝜏 𝑥 𝑓) = 𝐿 (𝑒 𝑥 𝐵 𝑓) .

𝑓 (𝑥) = (𝜏 𝑥 𝑓) (0) = 𝐿 (𝜏 𝑥 𝑓) = 𝐿 (𝑒 𝑥 𝐵 𝑓) .

Now $𝑓$ $𝑓$ is cyclic for $𝐵$ $𝐵$ : since $𝑉 𝑓 = s p a n {𝑒 𝑥 𝐵 𝑓 : 𝑥 \in ℝ}$ , and $𝑥 \mapsto 𝑒 𝑥 𝐵 𝑓$ is real analytic, its Taylor coefficients $𝐵 𝑘 𝑓$ $𝐵 𝑘 𝑓$ span $𝑉 𝑓$ $𝑉 𝑓$ . Therefore $𝑉 𝑓 = s p a n {𝐵 𝑘 𝑓 : 𝑘 \geq 0}$ .

Complexifying $𝑉 𝑓$ $𝑉 𝑓$ , the cyclic property implies the minimal polynomial of $𝐵$ $𝐵$ equals its characteristic polynomial. Thus each distinct eigenvalue $𝜆 𝑗 \in ℂ$ corresponds to exactly one Jordan block of size $𝑠 𝑗$ $𝑠 𝑗$ , and

𝑟 \sum 𝑗 = 1 𝑠 𝑗 = d i m 𝑉 𝑓 .

𝑟 \sum 𝑗 = 1 𝑠 𝑗 = d i m 𝑉 𝑓 .

On the block for $𝜆 𝑗$ $𝜆 𝑗$ , the matrix exponential contributes

𝑒 𝜆 𝑗 𝑥, 𝑥 𝑒 𝜆 𝑗 𝑥, \dots, 𝑥 𝑠 𝑗 - 1 𝑒 𝜆 𝑗 𝑥 .

𝑒 𝜆 𝑗 𝑥, 𝑥 𝑒 𝜆 𝑗 𝑥, \dots, 𝑥 𝑠 𝑗 - 1 𝑒 𝜆 𝑗 𝑥 .

𝑒 𝜆 𝑗 𝑥, 𝑥 𝑒 𝜆 𝑗 𝑥, \dots, 𝑥 𝑠 𝑗 - 1 𝑒 𝜆 𝑗 𝑥 .

Applying $𝐿$ $𝐿$ gives

𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,

𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,

𝑓 (𝑥) = 𝑟 \sum 𝑗 = 1 𝑃 𝑗 (𝑥) 𝑒 𝜆 𝑗 𝑥,

where $d e g 𝑃 𝑗 = 𝑠 𝑗 - 1$ . Hence

𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .

𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .

𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓 .

Since $\forall d e g 𝑃 𝑗 + 1 \geq 1$ ,

𝑟 \leq 𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓

𝑟 \leq 𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓

𝑟 \leq 𝑟 \sum 𝑗 = 1 (d e g 𝑃 𝑗 + 1) = d i m 𝑉 𝑓

Finally, ${𝑒 𝑥 𝐵 : 𝑥 \in ℝ}$ is a one-parameter subgroup of $G L (𝑉 𝑓)$ $G L (𝑉 𝑓)$ , and $𝑓$ $𝑓$ is its matrix coefficient.

□

Finally, as a sweet treat after our hard work, let us apply this theorem to solve the initial problem.

Solution. Let $𝑔 (𝑥) ≔ 𝜏 𝑎 𝑓 (𝑥) - 𝜏 - 𝑎 𝑓 (𝑥) 2 𝑓 (𝑎)$ , then $𝑔 \in 𝑉 𝑓$ . Hence,

𝑓 (𝑥 + 𝑦) + 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑥) 𝑔 (𝑦)

𝑓 (𝑥 + 𝑦) + 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑥) 𝑔 (𝑦)

𝑓 (𝑥 + 𝑦) + 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑥) 𝑔 (𝑦)

𝑓 (𝑥 + 𝑦) - 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑦) 𝑔 (𝑥) .

𝑓 (𝑥 + 𝑦) - 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑦) 𝑔 (𝑥) .

𝑓 (𝑥 + 𝑦) - 𝑓 (𝑥 - 𝑦) = 2 𝑓 (𝑦) 𝑔 (𝑥) .

Add them up, we obtain

𝑓 (𝑥 + 𝑦) = 𝑓 (𝑥) 𝑔 (𝑦) + 𝑓 (𝑦) 𝑔 (𝑥)

𝑓 (𝑥 + 𝑦) = 𝑓 (𝑥) 𝑔 (𝑦) + 𝑓 (𝑦) 𝑔 (𝑥)

𝑓 (𝑥 + 𝑦) = 𝑓 (𝑥) 𝑔 (𝑦) + 𝑓 (𝑦) 𝑔 (𝑥)

, which is

𝜏 𝑦 𝑓 = 𝑔 (𝑦) 𝑓 + 𝑓 (𝑦) 𝑔

𝜏 𝑦 𝑓 = 𝑔 (𝑦) 𝑓 + 𝑓 (𝑦) 𝑔

𝜏 𝑦 𝑓 = 𝑔 (𝑦) 𝑓 + 𝑓 (𝑦) 𝑔

. Since $𝑔 \in 𝑉 𝑓$ , $𝜏 𝑦 𝑓 \in s p a n {𝑓, 𝑔} \subset 𝑉 𝑓$ . Hence, $d i m 𝑉 𝑓 \leq 2$ .

By applying Theorem 3.1, we obtain

𝑓 (𝑥) = 𝐴 𝑒 𝜆 𝑥 + 𝐵 𝑒 𝜇 𝑥 o r (𝐴 + 𝐵 𝑥) 𝑒 𝜆 𝑥 .

𝑓 (𝑥) = 𝐴 𝑒 𝜆 𝑥 + 𝐵 𝑒 𝜇 𝑥 o r (𝐴 + 𝐵 𝑥) 𝑒 𝜆 𝑥 .

𝑓 (𝑥) = 𝐴 𝑒 𝜆 𝑥 + 𝐵 𝑒 𝜇 𝑥 o r (𝐴 + 𝐵 𝑥) 𝑒 𝜆 𝑥 .

Substitute them into the original equation, we obtain

𝑓 (𝑥) = 𝑘 𝑥 o r 𝑓 (𝑥) = 𝐴 s i n (𝜔 𝑥) o r 𝑓 (𝑥) = 𝐴 s i n h (𝑘 𝑥)

𝑓 (𝑥) = 𝑘 𝑥 o r 𝑓 (𝑥) = 𝐴 s i n (𝜔 𝑥) o r 𝑓 (𝑥) = 𝐴 s i n h (𝑘 𝑥)

𝑓 (𝑥) = 𝑘 𝑥 o r 𝑓 (𝑥) = 𝐴 s i n (𝜔 𝑥) o r 𝑓 (𝑥) = 𝐴 s i n h (𝑘 𝑥)

𝑓 (𝑥) = 𝑘 𝑥 o r 𝑓 (𝑥) = 𝐴 s i n (𝜔 𝑥) o r 𝑓 (𝑥) = 𝐴 s i n h (𝑘 𝑥)

Note 3.2.

𝜏 𝑎 𝜏 𝑏 𝑓 ​ = 𝜏 𝑎 + 𝑏 𝑓

is actually an application of association, since it implies

𝑓 ((𝑥 ​ + 𝑎) ​ + 𝑏) ​ = 𝑓 (𝑥 ​ + (𝑎 ​ + 𝑏))

. Hence, for primary mathematics, we may apply association to functional equations to solve it.

View Typst source

#import "@preview/noteworthy:0.4.0": noteworthy
#import "/themes/site/notes.typ": *
#show: web-equations
#import "/typ/shared.typ": *

#show: tylenotes.with(
  title: "Finite-Dimensional Translation Spaces and Lie Group Projections",
  date: "2026-08-14",
  tags: ("real", "lean"),
  summary: "A new proof classifying continuous non-constant solutions of f(x+y)f(x−y)=f(x)²−f(y)² using finite-dimensional translation spaces and Lie group projections.",
)

= The Start

This note starts with the following problem.
#example[
  Let $f: RR -> RR$ be a continuous non-constant function,
  which satisfies
  $ f(x+y)f(x-y)=f(x)^2-f(y)^2 $.
  Find all such functions.
]

I have the cover linear theorem in the Lie section for it.
But the conclusion of that theorem depends on the choice of
$G$, $pi$ and $Phi$, which weakens its power.

= The Climax

Thus, I wanted to find another approach. And I got the idea from the
standard solution:
Let $g_a (x) := (f(x+a)-f(x-a))/(2f(a))$.
Now, if we plug $f = sin$ into this expression,
after some calculations, we can get:
$ g = cos $.
The same can be done with $f = sinh$.

In the case of $f = sin$, $g = cos$, it gives us a circle
in a 2-dimensional space.
So, we can treat this equation as a projection of some
2-dimensional curve on $RR$.
More generally, most equations can be regarded as
a projection of a 1-dimensional manifold in a high dimensional space
on $RR$, and in which the curve can be one-parameterized by $RR$.

= The End
Finally, after chosing conditions which are strong enough to make conclusions,
we have the following theorem.

#theorem(title: "Finite-Dimensional Translation Space")[
  Let $f : RR -> RR$ be continuous and define its translation space
  $V_f = "span"_RR {tau_a f : a in RR}$, where $(tau_a f)(x) = f(x+a)$.

  If $dim V_f < infinity$, then:

  + $f$ is real analytic.
  + There exist distinct $lambda_1, dots, lambda_r in CC$ and nonzero polynomials $P_1, dots, P_r$ such that
    $ f(x) = sum_(j=1)^r P_j(x) e^(lambda_j x), $
    and moreover
    $ sum_(j=1)^r (deg P_j + 1) = dim V_f. $
    In particular, $r <= dim V_f$.
  + $f$ is a matrix coefficient of a one-parameter subgroup of $"GL"(V_f)$.
]

#proof[
  Let $T_a : V_f -> V_f$ be the restriction of the translation operator $tau_a$ to $V_f$.
  Since $V_f$ is invariant and finite-dimensional, $T_a in "GL"(V_f)$.

  The map
  $ rho : RR -> "GL"(V_f), quad rho(a) = T_a $
  is a group homomorphism:
  $ rho(a+b) = rho(a) rho(b), quad rho(0) = I. $

  It is continuous because for every $h in V_f$, the map $a |-> T_a h$ is continuous in the finite-dimensional topology of $V_f$.

  Hence there exists a unique $B in "End"(V_f)$ such that
  $ rho(x) = e^(x B) quad forall x in RR. $

  Define $L : V_f -> RR$ by $L(h) = h(0)$. Then
  $ f(x) = (tau_x f)(0) = L(tau_x f) = L(e^(x B) f). $

  Now $f$ is cyclic for $B$: since $V_f = "span"{e^(x B) f : x in RR}$, and $x |-> e^(x B) f$ is real analytic, its Taylor coefficients $B^k f$ span $V_f$.
  Therefore $V_f = "span"{B^k f : k >= 0}$.

  Complexifying $V_f$, the cyclic property implies the minimal polynomial of $B$ equals its characteristic polynomial.
  Thus each distinct eigenvalue $lambda_j in CC$ corresponds to exactly one Jordan block of size $s_j$, and
  $ sum_(j=1)^r s_j = dim V_f. $

  On the block for $lambda_j$, the matrix exponential contributes
  $ e^(lambda_j x), x e^(lambda_j x), dots, x^(s_j - 1) e^(lambda_j x). $

  Applying $L$ gives
  $ f(x) = sum_(j=1)^r P_j(x) e^(lambda_j x), $
  where $deg P_j = s_j - 1$. Hence
  $ sum_(j=1)^r (deg P_j + 1) = dim V_f. $
  Since $forall deg P_j + 1 >= 1$,
  $ r <= sum^r_(j=1) (deg P_j + 1) = dim V_f $

  Finally, $\{e^(x B) : x in RR\}$ is a one-parameter subgroup of $"GL"(V_f)$, and $f$ is its matrix coefficient.
]

Finally, as a sweet treat after our hard work, let us apply this theorem to solve the initial problem.

#solution[
  Let $g(x) := (tau_a f(x) - tau_(-a) f(x)) / (2 f(a))$, then $g in V_f$.
  Hence, $ f(x+y) + f(x-y) = 2f(x)g(y) $
  $ f(x+y) - f(x-y) = 2f(y)g(x). $
  Add them up, we obtain $ f(x+y) = f(x)g(y)+f(y)g(x) $,
  which is $ tau_y f = g(y) f + f(y) g $.
  Since $g in V_f$, $tau_y f in "span"{f,g} subset V_f$.
  Hence, $dim V_f <= 2$.

  By applying *Theorem 3.1*, we obtain
  $ f(x) = A e^(lambda x) + B e^(mu x) "or" (A+B x) e ^ (lambda x). $
  Substitute them into the original equation, we obtain
  $ f(x) = k x "or" f(x) = A sin (omega x) "or" f(x) = A sinh (k x) $
]

#note[
  $tau_a tau_b f = tau_(a+b) f$ is actually an application
  of association, since it implies 
  $f((x+a)+b) = f(x+(a+b))$. Hence, for primary mathematics,
  we may apply association to functional equations to solve it.
]