2

I am not sure of my answers to this question. I would appreciate if someone can verify if my logic is correct, Thanks! So $C[0,1]$ is the vector space of all continuous function on the interval [0,1]. (over the real numbers) with the norm $||f||= \underset{x\in\left[0,1\right]}{sup}\left|f(x)\right|$ .

I am given two subsets of the vector space ,$ A=\left\{ g\in C\left[0,1\right]\,|\, g(1)=-g(0)\right\} , B=\left\{ g\in C\left[0,1\right]\,|\, g(1)=g(0)\right\} $ So the questions are , these two sets dense in $C[0,1]$ .

So, my answer was that neither A nor B is a dense subset of our vector space. Let there be an $\epsilon>0$ Let's take for example, $f(x)=x$ in both cases. $f(1)=1$ and $f(0)=0$ . And we are looking for functions $g,h$ such that $g(1)=-g(0)$ and $h(1)=h(0)$ , that satisfy $||f-g||<\epsilon $ and $||f-h|||<\epsilon $

Let's take $g$ for a second and $\epsilon =\frac{1}{4}$. So let's assume for a second that there exists a function like this. So for every $x \in [0,1]$ we get $|f(x)-g(x)|<\epsilon$ . So we get $|f(0)-g(0)|<\frac{1}{4}$ , therefore, $-\frac{1}{4}<g(0)<\frac{1}{4}$ . On the other hand, $|f(1)-g(1)|<\frac{1}{4}$ therefore, $\frac{3}{4}<g(1)<\frac{5}{4}$ . On the other hand we have $g(0)=-g(1)$, therefore we get we get $\frac{1}{4}<g(0)<\frac{1}{4}$ and , $-\frac{5}{4}<f(0)<-\frac{3}{4}$ . A Contradiction

In the same way, let's assume that there is a function $h$ that satisfies the conditions above. So for every $x\in[0,1]$ we get $|f(x)-h(x)|<\epsilon$ and therefore $|f(0)-h(0)|<\frac{1}{4}$, and therefore $-\frac{1}{4}<h(0)<\frac{1}{4}$ . And we have $|f(1)-h(1)|<\frac{1}{4}$ and therefore, $\frac{3}{4}<h(1)<\frac{5}{4}$, and since $h(1)=h(0)$ we get, $-\frac{1}{4}<h(0)<\frac{1}{4}$ and $\frac{3}{4}<h(0)<\frac{5}{4}$ , a contradiction.

Can someone verify this? Thank you.

Charles Carmichael
  • 595
  • You have the right overall idea but there are some technical errors in arriving at your contradictions. For instance $|f(1)-g(1)|<\epsilon$ doesn't imply $g(1) < f(1)-\epsilon$ (in fact it guarantees this to be false). Similarly, the lower bound $g(0) > -1+\epsilon$ does nothing to show that $|f(0)-g(0)|$ is greater than $1$. Your example will work, but you need to be precise with your inequalities. – Erick Wong Feb 15 '15 at 18:08
  • @ErickWong Thanks for replying! I had the picture in my had but didn't know how to formalize it, I will work on it and make it more precise. Glad to hear that the idea is correct, Thanks! – Charles Carmichael Feb 15 '15 at 18:10

2 Answers2

1

I agree that both subsets aren't dense in $C[0,1]$, and considering the function $f(x)=x$ will do the trick. However, your proof isn't quite correct. In particular $|f(1)-g(1)|<\epsilon$ does not imply $g(1)<f(1)-\epsilon$ (I suppose you mean $g(1)<f(1)+\epsilon$). I would say that $g(1)\in (1-\epsilon,1+\epsilon)$, hence $g(0)\in(-1-\epsilon,-1+\epsilon)$. So $g(0)<-1+\epsilon$. We also see that $|g(0)-f(0)|<\epsilon$. So $g(0)>0-\epsilon$. Now $0-\epsilon<g(0)<-1+\epsilon$ yields a contradiction for $\epsilon=\frac{1}{2}$. For the set $B$ we use the same argument with the final contradiction $1-\epsilon<g(0)<0+\epsilon$.

Uncountable
  • 3,520
1

You have a small mistake (with inequalities) in your work. Eventually, you'll need to give $\epsilon$ a value to create a contradiction.

Assume that $A$ is dense and let $f(x) = x$.

Then, $\forall \ \epsilon > 0$, $B_{||.||}(f,\epsilon) \bigcap A \neq \phi$.

In particular, for $\epsilon = \frac12$, we can find $g \in A$ such that $||f - g|| < \frac12$.

But:

$$||f - g|| < \frac12 \implies \sup_{x \in [0,1]}|f(x) - g(x)| < \frac12 \implies f(1) - g(1) < \frac12 \implies g(1) > f(1) - \frac12 \implies g(1) > \frac12 \implies -g(0) > \frac12 \implies f(0) - g(0) > \frac12 \implies \sup_{x \in [0,1]}|f(x) - g(x)| \ge \frac12 \implies ||f - g|| \ge \frac12$$

A contradiction.