Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the cardinality of set of all smooth functions belonging to $L^1$ or $L^2$ ? What is that of set of all integrable or square integrable functions ?

share|cite|improve this question
Can you tell us what ideas you've tried? It is quite strange that you know what the terms in the question mean and at the same time have no idea of how to proceed... – Mariano Suárez-Alvarez Nov 8 '10 at 6:36
Hint: a continuous function is determined by its value at all rational points. – Yuval Filmus Nov 8 '10 at 6:38
the direct procedure is to check if there is any injective or surjective or bijective mapping with a set with which it has to be compared. – Rajesh Dachiraju Nov 8 '10 at 6:41
For infinite cardinals $2c = c$. I suggest you start by reviewing your cardinal arithmetic. – Yuval Filmus Nov 8 '10 at 7:04
@a little don: Smooth functions aren't determined by their Taylor series. (Analytic functions are.) – Jonas Meyer Nov 8 '10 at 7:16
up vote 5 down vote accepted

A continuous function is determined by its value on the rational points, so there are at most $\aleph^{\aleph_0} = \aleph$ of them.

Conversely, it's not difficult to find $\aleph$ smooth (integrable) functions in $L_1 \cap L_2$, just take any such non-zero function $f$ and consider $\{rf : r \in \mathbb{R}\}$.

EDIT: For definiteness, $x \mapsto e^{-x}$ is $C^\infty$ and in $\bigcap_{p>0} L_p$.

share|cite|improve this answer
Which means cardinality of smooth integrable functions should be more than or equal to $\aleph$. But set of smooth functions is a subset of set of all continuous functions.Hence the cardinality of set of all smooth integrable functions is $aleph$. Please tell me whether my understanding is correct. Also please suggest any hint/reference for the approach to prove your HINT. (representation of a continuous function by its values at all rational points) – Rajesh Dachiraju Nov 8 '10 at 7:43
Your understanding is correct. Hint to the hint: the rational numbers are dense in the reals. – Yuval Filmus Nov 8 '10 at 7:49
given the values of a continuous function at all rational points, it is possible to deduce the value of the function at any irrational point. I am not able to prove this even though i know the fact that rationals are dense in real and the meaning of it. Please suggest a proof/hint. – Rajesh Dachiraju Nov 8 '10 at 9:53
@Yuval Filmus: By "$\aleph$" do you mean $c=2^{\aleph_0}$? – Arturo Magidin Nov 8 '10 at 15:42
@Yuval Filmus: I just have never seen it before. I suspect set theorists might object too... – Arturo Magidin Nov 8 '10 at 19:18

The second part of the question asks for the cardinality of the set of integrable or square-integrable functions. I will assume you actually mean "functions" rather than "equivalence classes of functions under the relation of equality almost everywhere".

Let $\beta$ be the cardinality of the set of all functions from $\mathbb{R}$ to $\mathbb{R}$; by standard set theory this is the same as the cardinality of the powerset of the real numbers: $\beta = 2^{|\mathbb{R}|} = 2^{2^{\aleph_0}}$.

Certainly the set of integrable functions, and the set of square integrable functions, can have cardinality no more than $\beta$. It turns out this is exactly the cardinality.

Let $E$ be a Cantor set; the key properties are that $|E| = |\mathbb{R}|$ and the measure of $E$ is $0$. Consider the set of all functions that are $0$ for every $x$ that is not in $E$. All of these functions are both integrable and square integrable, because the measure of $E$ is zero. The cardinality of this set is the cardinality of the set of functions from $E$ to $\mathbb{R}$, which is exactly $\beta$ because $|E| = |\mathbb{R}|$.

share|cite|improve this answer

You can also find the cardinality of smooth functions by first considering the set of constant functions, having cardnality c, and then determine the cardnality of all of the other smooth functions by knowing that, because it is not constant (by definition) that at each rational number, (rational only because they are dense in the reals and thus define any continuous function) there must be either positive or negative infinitesimal change in the function (not zero which would cause a non-differentiable "sharp corner"),the set of all such "decisions" defining each function having cardinality 2^(ℵ0) (since the rationals are countable).

This gives the set of smooth functions to have cardinality

c+2^(ℵ0) = c+c =c

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.