Finite-Dimensional Translation Spaces and Lie Group Projections

Real analysis/All notes14 Aug 2026

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1. The Start

This note starts with the following problem.

Example

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

$$𝑓\left(𝑥+𝑦\right)𝑓\left(𝑥-𝑦\right)={𝑓\left(𝑥\right)}^2-{𝑓\left(𝑦\right)}^2$$

. 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 $𝐺$, $𝜋$ and $\mathrm{\Phi }$, which weakens its power.

2. The Climax

Thus, I wanted to find another approach. And I got the idea from the standard solution: Let ${𝑔}_{𝑎}\left(𝑥\right)≔\frac{𝑓\left(𝑥+𝑎\right)-𝑓\left(𝑥-𝑎\right)}{2𝑓\left(𝑎\right)}$. 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 $ℝ$. More generally, most equations can be regarded as a projection of a 1-dimensional manifold in a high dimensional space on $ℝ$, and in which the curve can be one-parameterized by $ℝ$.

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 ${𝑉}_{𝑓}={\text{span}}_{ℝ}\left\{{𝜏}_{𝑎}𝑓:𝑎\in ℝ\right\}$, where $\left({𝜏}_{𝑎}𝑓\right)\left(𝑥\right)=𝑓\left(𝑥+𝑎\right)$.

If $dim{𝑉}_{𝑓}<\infty$, then:

1. $𝑓$ is real analytic.

2. There exist distinct ${𝜆}_1,\dots ,{𝜆}_{𝑟}\in ℂ$ and nonzero polynomials ${𝑃}_1,\dots ,{𝑃}_{𝑟}$ such that

   $$𝑓\left(𝑥\right)=\sum _{𝑗=1}^{𝑟}{𝑃}_{𝑗\left(𝑥\right)}{𝑒}^{{𝜆}_{𝑗}𝑥},$$

   and moreover

   $$\sum _{𝑗=1}^{𝑟}\left(deg{𝑃}_{𝑗}+1\right)=dim{𝑉}_{𝑓}.$$

   In particular, $𝑟\le dim{𝑉}_{𝑓}$.

3. $𝑓$ is a matrix coefficient of a one-parameter subgroup of $\text{GL}\left({𝑉}_{𝑓}\right)$.

Proof

Let ${𝑇}_{𝑎}:{𝑉}_{𝑓}\to {𝑉}_{𝑓}$ be the restriction of the translation operator ${𝜏}_{𝑎}$ to ${𝑉}_{𝑓}$. Since ${𝑉}_{𝑓}$ is invariant and finite-dimensional, ${𝑇}_{𝑎}\in \text{GL}\left({𝑉}_{𝑓}\right)$.

The map

$$𝜌:ℝ\to \text{GL}\left({𝑉}_{𝑓}\right),𝜌\left(𝑎\right)={𝑇}_{𝑎}$$

is a group homomorphism:

$$𝜌\left(𝑎+𝑏\right)=𝜌\left(𝑎\right)𝜌\left(𝑏\right),𝜌\left(0\right)=𝐼.$$

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 ${𝑉}_{𝑓}$.

Hence there exists a unique $𝐵\in \text{End}\left({𝑉}_{𝑓}\right)$ such that

$$𝜌\left(𝑥\right)={𝑒}^{𝑥𝐵}\forall 𝑥\in ℝ.$$

Define $𝐿:{𝑉}_{𝑓}\to ℝ$ by $𝐿\left(ℎ\right)=ℎ\left(0\right)$. Then

$$𝑓\left(𝑥\right)=\left({𝜏}_{𝑥}𝑓\right)\left(0\right)=𝐿\left({𝜏}_{𝑥}𝑓\right)=𝐿\left({𝑒}^{𝑥𝐵}𝑓\right).$$

Now $𝑓$ is cyclic for $𝐵$: since ${𝑉}_{𝑓}=\text{span}\left\{{𝑒}^{𝑥𝐵}𝑓:𝑥\in ℝ\right\}$, and $𝑥\mapsto {𝑒}^{𝑥𝐵}𝑓$ is real analytic, its Taylor coefficients ${𝐵}^{𝑘}𝑓$ span ${𝑉}_{𝑓}$. Therefore ${𝑉}_{𝑓}=\text{span}\left\{{𝐵}^{𝑘}𝑓:𝑘\ge 0\right\}$.

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

$$\sum _{𝑗=1}^{𝑟}{𝑠}_{𝑗}=dim{𝑉}_{𝑓}.$$

On the block for ${𝜆}_{𝑗}$, the matrix exponential contributes

$${𝑒}^{{𝜆}_{𝑗}𝑥},𝑥{𝑒}^{{𝜆}_{𝑗}𝑥},\dots ,{𝑥}^{{𝑠}_{𝑗}-1}{𝑒}^{{𝜆}_{𝑗}𝑥}.$$

Applying $𝐿$ gives

$$𝑓\left(𝑥\right)=\sum _{𝑗=1}^{𝑟}{𝑃}_{𝑗\left(𝑥\right)}{𝑒}^{{𝜆}_{𝑗}𝑥},$$

where $deg{𝑃}_{𝑗}={𝑠}_{𝑗}-1$. Hence

$$\sum _{𝑗=1}^{𝑟}\left(deg{𝑃}_{𝑗}+1\right)=dim{𝑉}_{𝑓}.$$

Since $\forall deg{𝑃}_{𝑗}+1\ge 1$,

$$𝑟\le \sum _{𝑗=1}^{𝑟}\left(deg{𝑃}_{𝑗}+1\right)=dim{𝑉}_{𝑓}$$

Finally, $\{{𝑒}^{𝑥𝐵}:𝑥\in ℝ\}$ is a one-parameter subgroup of $\text{GL}\left({𝑉}_{𝑓}\right)$, 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 $𝑔\left(𝑥\right)≔\frac{{𝜏}_{𝑎}𝑓\left(𝑥\right)-{𝜏}_{-𝑎}𝑓\left(𝑥\right)}{2𝑓\left(𝑎\right)}$, then $𝑔\in {𝑉}_{𝑓}$. Hence,

$$𝑓\left(𝑥+𝑦\right)+𝑓\left(𝑥-𝑦\right)=2𝑓\left(𝑥\right)𝑔\left(𝑦\right)$$$$𝑓\left(𝑥+𝑦\right)-𝑓\left(𝑥-𝑦\right)=2𝑓\left(𝑦\right)𝑔\left(𝑥\right).$$

Add them up, we obtain

$$𝑓\left(𝑥+𝑦\right)=𝑓\left(𝑥\right)𝑔\left(𝑦\right)+𝑓\left(𝑦\right)𝑔\left(𝑥\right)$$

, which is

$${𝜏}_{𝑦}𝑓=𝑔\left(𝑦\right)𝑓+𝑓\left(𝑦\right)𝑔$$

. Since $𝑔\in {𝑉}_{𝑓}$, ${𝜏}_{𝑦}𝑓\in \text{span}\left\{𝑓,𝑔\right\}\subset {𝑉}_{𝑓}$. Hence, $dim{𝑉}_{𝑓}\le 2$.

By applying Theorem 3.1, we obtain

$$𝑓\left(𝑥\right)=𝐴{𝑒}^{𝜆𝑥}+𝐵{𝑒}^{𝜇𝑥}\text{or}\left(𝐴+𝐵𝑥\right){𝑒}^{𝜆𝑥}.$$

Substitute them into the original equation, we obtain

$$𝑓\left(𝑥\right)=𝑘𝑥\text{or}𝑓\left(𝑥\right)=𝐴sin\left(𝜔𝑥\right)\text{or}𝑓\left(𝑥\right)=𝐴sinh\left(𝑘𝑥\right)$$

View Typst source

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#import "@preview/noteworthy:0.4.0": *
#import "/themes/site/notes.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) $
]