Take the 2-minute tour ×
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.

This is motivated from an exercise in real analysis:

Prove that $C([0,1])$ is not dense in $L^\infty([0,1])$.

My first question is how $C([0,1])$ is identified as a subset of $L^\infty([0,1])$? (I think one would never say something like "$A$ is (not) dense in $B$" if $A$ is not even a subset of $B$. )

First of all, $L^\infty([0,1])$ is defined as a quotient space, but $C([0,1])$ is a set of functions: $$ C([0,1]):=\{f:[0,1]\to{\Bbb R}|f \ \text{is continuous}\}. \tag{1} $$

I think one should also take $C([0,1])$ as $$ C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \ \text{for some g where g is continuous on}\ [0,1]\} \tag{2} $$ where $f\sim g$ if only if $f=g$ almost everywhere. But I've never read any textbook (PDE, measure theory, or functional analysis, etc) that defines $C([0,1])$ (or more generally $C(X)$ where $X\subset{\Bbb R}$ is compact) in this way before. Second question: Could anyone come up with a reference with such definition?


[EDITED:]The original title doesn't reflect my point. I've changed it accordingly.

[EDITED:] Some thoughts after reading the comments and answers:

When one regards $C([0,1])$ as a subset of $L^\infty([0,1])$, (1) is not correct, and (2) would be not correct either. The final version I can come up with is $$ C([0,1]):=\{f:[0,1]\to{\Bbb R}|f\sim g \ \text{for some g where g is continuous on}\ [0,1]\}\big/\sim. \tag{3} $$

share|improve this question
    
Perhaps a topological, simple explanation would be that there exists at least one open non-empty subset $\,S\subset L^\infty[0,1]$ not containing any continuous function. –  DonAntonio Mar 29 '13 at 15:13
    
each continuous function is its only representative, so there's a natural injection –  suissidle Mar 29 '13 at 15:16
    
Verify that the map $C([0,1]) \to L^\infty([0,1])$ is isometric. In particular, the image of $C([0,1])$ in $L^\infty([0,1])$ is closed. Of course, the map is not onto. –  Martin Mar 29 '13 at 15:18
    
The original notation I used, "$C([0,1])\bigg/\sim$", is not correct I think. I've edited it. –  Jack Mar 29 '13 at 15:19
1  
Let me try again: Definition (1) is the correct one. There is an obvious bounded linear map $T\colon C[0,1] \to L^\infty[0,1]$ given by $Tf = [f]$, where $[f]$ is the equivalence class of $f$ modulo $\sim$. The image of $T$ is (3). The answers explain that $T$ is an isometric isomorphism from (1) onto (3), so you can actually identify them via $T$. –  Martin Mar 30 '13 at 17:59

3 Answers 3

You can actually identify $C([0,1])$ and $C([0,1])/\sim$ because, two continuous fonctions who agree almost everywhere are equal.

Indeed, let $f,g \in C([0,1])$ be such that $A = \{x\in [0,1]\mid f(x) \neq g(x)\}$ is negligible. Then $A$ must have an empty interior, so its complementary is dense in $[0,1]$. The function $h = f-g$ is continuous, hence $h([0,1]) = h(\overline{[0,1]\setminus A}) \subset \overline{h([0,1]\setminus A)} = \{0\}$. This proves that $f=g$.

If you want to be really rigorous, it would be better to say that the natural injection $C([0,1]) \hookrightarrow \mathcal{L}^\infty([0,1])$ factorizes with $\sim$, so that it induces an injection $C([0,1]) \hookrightarrow L^\infty([0,1])$. That way, $C([0,1])$ is identified with the image of this injection.

share|improve this answer
    
I think $C([0,1])/\sim$ is just $C([0,1])$ since for continuous function, $f=g$ if and only if $f=g$ almost everywhere. So the quotient space is not need. I edited accordingly in my original post. –  Jack Mar 29 '13 at 16:04
    
You have functions in $C([0,1])$ but you only have classes of equivalence of function in $C([0,1])/\sim \subset L^\infty$, so you cannot say that they are equal but you can still identify them (with a canonical bijection). –  Siméon Mar 29 '13 at 16:12
    
Thank you for your explanation. In your first sentence, (2) is used for $C([0,1])$, right? –  Jack Apr 20 '13 at 13:20

Actually, such identifications are done in the textbook about partial differential equations and Sobolev spaces. For example, we can see theorems like "$C_0^\infty(\Bbb R^d)$, the set of smooth functions with compact support, is dense in $W^{1,p}(\Bbb R^d)$ for all $1\leqslant p<\infty$". This means that we identify a test function $\phi$ to the class of functions which are almost everywhere equal to $\phi$, as what is done in the OP.

The characteristic function of $g=[1/2,1)$ cannot be approached in $L^\infty$ by such functions, because there would be an $f$ almost everywhere equal to a continuous function such that $\sup_{x\in [0,1]\setminus N}|g(x)-f(x)|<1/3$, where $N$ is a set of $0$ measure. In particular, $\sup_{x\in [0,1/2)\setminus N}|f(x)|<1/3$ and $\sup_{x\in [1/2,1)\setminus N}|1-f(x)|<1/3$.

share|improve this answer
    
As I understand from your answer, one should have the indicator function $1_{{\Bbb Q}\cap[0,1]}\in C([0,1])$, though $1_{{\Bbb Q}\cap[0,1]}$ is obviously not continuous on $[0,1]$, right? –  Jack Mar 29 '13 at 15:30
    
No, the characteristic function of $\Bbb Q$ is in the class of the null function, so it's not the problem. In the example I gave, the problem comes from a jump. –  Davide Giraudo Mar 29 '13 at 15:34
    
Oh, what my question in the comment mean is $1_{{\Bbb Q}\cap[0,1]}=0$ almost everywhere and since $0\in C([0,1])$, one should identify $1_{{\Bbb Q}\cap[0,1]}$ as an element in $C([0,1])$. –  Jack Mar 29 '13 at 15:39
1  
Regarding your last claim: The indicator function of $[1/2,1)$ (on the interval $[0,1]$) is continuous a.e. The set of points where it's discontinuous is just $\{\tfrac12,1\}$. I think you meant to write that it is not a.e. equal to a continuous function. –  kahen Mar 29 '13 at 15:54
    
@kahen Thanks, fixed now (I think). –  Davide Giraudo Mar 29 '13 at 17:01

This includes an answer to the original question posted, modified to answer the latest version of the question at time of writing. I have left the original answer because it includes an observation (and an example of how that observation is applied) that illustrates why people are a bit careless about differentiating between the equivalence classes and their representatives.

The norm on $L^\infty[0,1]$ is the essential supremum, so it 'ignores' changes on null sets. By $\|f\|$ below, I mean the essential supremum norm. I use $[f]$ to mean the equivalence class of $f$, the notation is potentially confusing, but context will disambiguate. By $f_1 \sim f_2$, I mean that $\{x | f_1(x) \neq f_2(x) \}$ is a null set.

One identifies $C[0,1]$ with a subset of $L^\infty[0,1]$ by taking equivalence classes, ie, we are really dealing with $\{[f] | f \in C[0,1] \}$, which is a subset of $L^\infty[0,1]$. (As an aside, continuity means that the identification $f \mapsto [f]$ is injective.)

(I use $m$ below to denote the Lebesgue measure, however the observation holds for any measure, of course. The subsequent demonstration of 'not being dense' does depend on the Lebesgue measure.)

Observation: Suppose $P$ is some property on $\mathbb{R}$ (or $\mathbb{C}$ as the case may be), and suppose $f_1 \sim f_2$. Then we have $m \{x | P(f_1(x)) \} = m \{x | P(f_2(x)) \}$. This follows since $f_1(x) = f_2(x)$ a.e. $x$. So even though $[f]$ is an equivalence class, we can think about $m \{x | P(\,[f]\,(x)) \}$ with the understanding that we really mean $m \{x | P(f(x)) \}$ for some representative $f \in [f]$. This is what allows us to be somewhat blasé about dealing with a function vs. its equivalence class.

The previous observation can be extended considerably, but loosely the idea is that the measure of the set of points that satisfies a 'nice' property is independent of the particular representations from the equivalence classes. By a 'nice' property, I mean a property whose truth value at $x$ depends only on the values of the representations at $x$.

Now consider $ [1_{[\frac{1}{2},1]}]$, and $[c]$ where $c \in C[0,1]$.

I claim $\|[1_{[\frac{1}{2},1]}]-[c] \| \ge \frac{1}{2}$, and since $c$ was arbitrary, we see that $C[0,1]/ \sim$ cannot be dense in $L^\infty[0,1]$.

To see why the claim is true, we will prove the statement for specific representatives of $[1_{[\frac{1}{2},1]}], [c]$ (ie, $1_{[\frac{1}{2},1]}, c$ respectively) and then invoke the above observation to conclude.

Let $\gamma =c(\frac{1}{2})$. We have $|\gamma-1|+|\gamma| \ge 1$, and hence $\max(|\gamma-1|, |\gamma|) \ge \frac{1}{2}$. Continuity of $c$ implies that for any $\epsilon>0$, there is a $\delta >0$ such that $|c(x)-\gamma| < \epsilon$ whenever $x \in B(\frac{1}{2}, \delta)$. Noting that if $x \in B(\frac{1}{2}, \delta)$, we have $1-x \in B(\frac{1}{2}, \delta)$, we get $\max(|c(1-x)|, |c(x)-1|) \ge \max(|\gamma|-\epsilon, |\gamma-1|-\epsilon) \ge \frac{1}{2}-\epsilon$. Hence for $x\in (\frac{1}{2},\frac{1}{2}+\delta)$, we have $\max(|c(1-x)-1_{[\frac{1}{2},1]}(1-x)|, |c(x)-1_{[\frac{1}{2},1]}(x)|) \ge \frac{1}{2}-\epsilon$.

In particular, $m \{x |\ |c(x)-1_{[\frac{1}{2},1]}(x)| \ge \frac{1}{2}-\epsilon \} \ge \delta >0$. The above observation shows that this is true for all $c'\in [c], f'\in [1_{[\frac{1}{2},1]}]$, so it follows from the definition of essential supremum that $\|[c]-[1_{[\frac{1}{2},1]}]\| \ge \frac{1}{2}$.

share|improve this answer

Your Answer

 
discard

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.