Mathematicshard

What Is the Cardinality of the Set of Continuous Functions Where f(f(x)) = exp(x)?

Question

What is the cardinality of the set of continuous functions

f: \mathbb{R} \to \mathbb{R}

such that

f(f(x)) = e^x

for all

x

A deep dive into functional iteration and the square root of the exponential.

Cardinality of Solution Set

𝔠

The cardinality of the continuum (2^{ℵ₀}) — uncountably many solutions

1Understand the Functional Equation

We seek continuous functions

f: \mathbb{R} \to \mathbb{R}

satisfying

f(f(x)) = e^x

for all

x

. This means

f

is a **functional square root** of the exponential function:

f^{\circ 2} = \exp

Functional Iteration

If we write

f^{\circ n}

for the

n

-fold composition of

f

, then

f^{\circ 2} = f \circ f = \exp

and

f^{\circ 4} = \exp \circ \exp

. We're looking for a function that sits "halfway" between the identity and exp.

2Properties of f

Since

e^x

is strictly increasing and maps

\mathbb{R}

onto

(0, \infty)

, the function

f

must be strictly increasing (if

f

were decreasing,

f \circ f

would be increasing, which is consistent, but further analysis eliminates this case).

Property	exp(x)	f(x)	Reason
Monotonicity	Strictly increasing	Strictly increasing	Composition rule
Range	ℝ → (0,∞)	ℝ → ?	Must be compatible
Fixed points	None (e^x > x)	None	Forced by exp
Continuity	Yes	Yes (given)	Assumption

3Fixed Point Analysis

The exponential function has no real fixed points (

e^x > x

for all

x \in \mathbb{R}

). If

f

had a fixed point

p

where

f(p) = p

, then

f(f(p)) = f(p) = p

, but

e^p \neq p

. So

f

has **no fixed points** either.

No Fixed Points

Since

f

is continuous, strictly increasing, and has no fixed points, either

f(x) > x

for all

x

f(x) < x

for all

x

. In fact,

f(x) > x

(since

f(f(x)) = e^x > x

and both

f

applications must move in the same direction).

4The Construction via Fundamental Domain

The key technique uses **Schröder's equation** and fundamental domains. Since

f(x) > x

for all

x

, we can choose any interval

[a, f(a))

as a fundamental domain. Defining

f

freely (but continuously and increasing) on this interval determines

f

everywhere via the relation

f(f(x)) = e^x

  The Real Line Partitioned by f:

  ... |─── D₀ ───|─── D₁ ───|─── D₂ ───| ...
       a       f(a)      f²(a)     f³(a)

  D₀ = [a, f(a))  ← fundamental domain (free choice)
  D₁ = [f(a), f²(a))  ← determined by f on D₀
  D₂ = [f²(a), f³(a))  ← determined by exp on D₀

5The Cardinality Result

The freedom in choosing

f

on the fundamental domain

[a, f(a))

propagates to the entire real line. Since there are

\mathfrak{c} = 2^{\aleph_0}

continuous increasing functions on an interval, the solution set has cardinality

\mathfrak{c}

Solution typeContinuous, strictly increasing

FreedomChoice of f on one interval

Each choice extends uniquelyYes (via f∘f = exp)

CARDINALITY𝔠 = 2^{ℵ₀}

6Key Concepts

Schröder's Equation

The technique of "linearizing" a functional equation near a fixed point (or on a fundamental domain) is called Schröder's method. It reduces the problem of finding

f^{\circ 2} = g

to finding

f

on a single fundamental domain.

Cardinality of Function Spaces

The set of continuous functions

\mathbb{R} \to \mathbb{R}

has cardinality

\mathfrak{c}

. Our solution set is a proper subset but still has cardinality

\mathfrak{c}

because the fundamental domain freedom gives a continuum of choices.

Quiz

Test your understanding with these questions.

Does

f

have any fixed points if

f(f(x)) = e^x

What is the cardinality of the set of continuous solutions?