What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$?
Question. What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$?
Notes. This question might seem "obvious" to those experienced in analysis but can confuse beginners who interpret the statement "literally": an element in \( C([0,1]) \) is fundamentally different from an element in \( L^\infty([0,1]) \). The former is a function on $[0,1]$, while the latter is an equivalence class of functions on $[0,1]$. Given this difference, how can we say \( C([0,1]) \) is a subset of \( L^\infty([0,1]) \)?
I will provide my answer below. Answers offering different perspectives are all welcome.
2 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
This is actually an instance of a broader concept called identification. To understand the concept of identification, it is possibly better to first look at a simpler example, and only then see how this applies to $L^\infty([0,1])$ versus $C([0,1])$.
What is identification?
Example: $\mathbb N\subseteq\mathbb Z$
You may be aware of the standard definition of integers as equivalence classes of pairs of natural numbers under the equivalence relation $(a,b)\sim(c,d)\iff a+d=b+c$. For example, the number $-1$ is, in this definition, the equivalence class $\{(0,1), (1,2), (2,3), (3,4), \ldots\}$. Now obviously a set of pairs of natural numbers is clearly something else than a natural number. And yet, we insist that the natural numbers form a subset of the integers. How can that be?
Well, the answer is that we do not really care about the representation, but only on its properties. If you think about the number $-1$, you normally do not think about equivalence classes of pairs of integers. Similarly, natural numbers in modern mathematics are defined as certain sets, but you wouldn't normally think of them as sets either.
Now it turns out that a certain set of integers (the non-negative integers, when using the convention that $0\in\mathbb N$.) has exactly the same structure as the natural numbers. In this case it specifically means that the non-negative integers fulfil Peano's axioms.
Said in another way, we could, in principle, have started by defining the integers in some way not involving the natural numbers, and then simply declare the natural numbers to be the non-negative integers, and all the properties we care about would still hold the same way.
The general concept
Based on the example above, we can work out the general concept of identification.
The first thing we note is that for each natural number, there is exactly one well-defined integer which it corresponds to; more formally, the mapping from natural numbers is an injection. Indeed we can write down that injection explicitly: $n\mapsto [(n,0)]$, where the square brackets denote the equivalence class of the pair. In the following I'll call that injection $\iota$.
But that's not enough: The mapping $n\mapsto [(0,n+1)]$ (that is, $n\mapsto -1-n$) is also an injection, and yet we would not consider it valid to identify the negative integers with the natural numbers. Rather, the important part here is that not only the elements, but also the relevant properties map. In the case of natural numbers vs. integers, those properties are the order, the addition and the multiplication. That is, if $m\le n$, then $\iota(m)\le\iota(n)$. Note that the first inequality uses the definition of $\le$ in the natural numbers, and the second uses the definition of $\le$ in the integers. The common term for this is that $\iota$ respects the order.
Similarly, we demand that $\iota(m+n) = \iota(m) + \iota(n)$ and $\iota(m\cdot n) = \iota(m)\cdot \iota(n)$. In other words, $\iota$ repects addition and multiplication.
In other words, the condition we need is the following:
- Given two mathematical structures (sets with additional properties) $A$ and $B$, we can identify elements of $A$ with elements of $B$ if there exists an injection $\iota:A\to B$ that respects the relevant properties.
Note that for this, it is not necessary that all properties of $A$ are represented in $B$, only those that are meaningful in $B$. For example, take the identification of the real numbers with the complex numbers that have zero imaginary part (yes, that's also an identification; complex numbers are pairs of real numbers, and the pair $(r,0)$ clearly isn't the same object as the real number $r$). The real numbers are ordered, the complex numbers aren't. But that is not a problem; it just means we don't have to consider the order when identifying $\mathbb R$ with that subset of $\mathbb C$.
How does this apply to $C([0,1])\subseteq L^\infty([0,1])$?
This is indeed a particularly straightforward case of identification because when talking about $L^\infty([0,1])$ what we really do care about are the actual functions in the equivalence classes; the reason why we put them into equivalence classes is just that it makes the structure easier to work with mathematically. Basically, all properties of $L^\infty([0,1])$ are already given by properties of the functions in the equivalence classes; the valid properties on $L^\infty([0,1])$ are simply those which respect the equivalence relation.
As we have seen above, to make the identification of members of $C([0,1])$, we need an injection that respects all relevant properties.
A mapping $\iota:C([0,1])\to L^\infty([0,1])$ is easily found: Given that all continuous functions on closed intervals are bounded, the obvious choice is $\iota(f) = [f]$ where $[f]$ denotes the equivalence class from $L^\infty([0,1])$ to which $f$ belongs.
What remains to be shown is that $\iota$ is an injection, and that $\iota$ respects all relevant properties.
Showing that $\iota$ is injective means showing that if two continuous functions are equal almost everywhere, they are equal everywhere. This is something that can indeed be proven, but I don't see how this can immediately be seen.
Showing that it respects all relevant properties, on the other hand, is trivial: Since the properties of members of $L^\infty([0,1])$ are already defined through the properties of the underlying functions, it basically follows by definition that the relevant properties of $[f]$ are the same as those of $f$.
0 comment threads
Short answer. That means there exists an injection $\iota:C([0,1])\to L^\infty([0,1])$ and one can identify $C([0,1])$ with the image $\iota(C([0,1]))$, which is a subset of $L^\infty([0,1])$.
Now a long one.
We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.
Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $$L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim.$$ Here, we only care about the set structure, so the quotient is a "quotient set."
Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with $\iota(f)= [f]$ where $[f]$ denote an equivalent class in $L^\infty([0,1])$.
By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a bijection between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.
With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.
Such identification is so common in analysis that people just use the short version.
0 comment threads