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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 ?

Asaf Karagila
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Rajesh D
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    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-Álvarez Nov 08 '10 at 06:36
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    Hint: a continuous function is determined by its value at all rational points. – Yuval Filmus Nov 08 '10 at 06:38
  • 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 D Nov 08 '10 at 06:41
  • @Yuval: Thanks! this Hint is new to me....let me try for a while – Rajesh D Nov 08 '10 at 06:43
  • Is there any similar hint for smooth functions ? – Rajesh D Nov 08 '10 at 06:46
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    For infinite cardinals $2c = c$. I suggest you start by reviewing your cardinal arithmetic. – Yuval Filmus Nov 08 '10 at 07:04
  • So the cardinality of set of all continuous functions is $\aleph^(\aleph_0^2)$ – Rajesh D Nov 08 '10 at 07:07
  • smooth functions have Taylor series... – futurebird Nov 08 '10 at 07:12
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    @a little don: Smooth functions aren't determined by their Taylor series. (Analytic functions are.) – Jonas Meyer Nov 08 '10 at 07:16
  • My hint only gives an upper bound, but note that $\aleph^{\aleph_0} = \aleph$. – Yuval Filmus Nov 08 '10 at 07:21
  • Another law of cardinal arithmetic: if $c \geq \aleph_0$ then $c^2 = c$ (for example, the rationals have the same cardinality as the naturals). – Yuval Filmus Nov 08 '10 at 07:22
  • @Yuval: Thanks again. So its $\aleph^(\aleph_0)$ for set of all continuous functions. – Rajesh D Nov 08 '10 at 07:25
  • See my note above: $\aleph^{\aleph_0} = \aleph$. – Yuval Filmus Nov 08 '10 at 07:26
  • So its (of set of all continuous functions) equal to the cardinality of set of all Real numbers – Rajesh D Nov 08 '10 at 07:27
  • By "smooth" do you actually mean "smooth", or do you mean "can be obtained by changing the values of a smooth function on a set of measure 0"? Many people would mean the latter when they say "smooth", but that will change the answer. – Carl Mummert Nov 08 '10 at 12:22
  • @Mariano: "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" is it a regular pun is it really strange....i am asking bcoz i am not from a math background..(also i did not attack any excersice problems from books...just got to know some definitions and theorems. – Rajesh D Nov 09 '10 at 06:42
  • I have an intuition but not sharp enough...in strictly mathematical sense – Rajesh D Nov 09 '10 at 06:44

3 Answers3

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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 $\bigcap_{p>0} L_p$.

Yuval Filmus
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  • 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 D Nov 08 '10 at 07:43
  • Your understanding is correct. Hint to the hint: the rational numbers are dense in the reals. – Yuval Filmus Nov 08 '10 at 07:49
  • 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 D Nov 08 '10 at 09:53
  • @Rajesh D: it's a basic exercise in real analysis. Use the sequential definition of continuity. – Carl Mummert Nov 08 '10 at 12:22
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    @Yuval Filmus: By "$\aleph$" do you mean $c=2^{\aleph_0}$? – Arturo Magidin Nov 08 '10 at 15:42
  • @Arturo: Yes, is that deprecated? – Yuval Filmus Nov 08 '10 at 19:17
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    @Yuval Filmus: I just have never seen it before. I suspect set theorists might object too... – Arturo Magidin Nov 08 '10 at 19:18
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    @Arturo: It is a matter of convention, in introductory courses $\aleph$ is usually a common notation for the continuum, later on you either stop using it altogether or just start using $2^{\aleph_0}$. I find it best to always clarify the notation when there is no consensus about it. – Asaf Karagila Jul 17 '11 at 06:06
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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 $|E| = |\mathbb{R}|$.

Carl Mummert
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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

Chris Henson
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