Which of the following sets are dense in $C[0,1]$

Question

Which of the following sets are dense in $C[0,1]$ with respect to sup-norm topology?

$1$. {$f$$\in$ $C[0,1]$ : $f$ is a polynomial }

$2$. {$f$$\in$ $C[0,1]$ :$f(0)$=$0$}

$3$. {$f$$\in$ $C[0,1]$ :$f(0)$$\neq$$0$}

$4$. {$f$$\in$$C[0,1]$ :$\int_{0}^{1}f(x)dx$$=$5}

I am thinking to apply Stone-weierstrass theorem but I don't know how to use it properly. Please help!

Answer 1

Answer: $1,\, 3$

Proof:

1. Stone-Weierstrass theorem states explicitly the dense property of the polynomials in $(C[0,1],\parallel\cdot\parallel_\infty)$

2. It is not dense. Because it is a proper closed subset of $(C[0,1],\parallel\cdot\parallel_\infty)$ as it is the inverse image of $\{0\}$ under the continuous map $f\mapsto f(0)$.

3. It is dense. It is obvious that every element of that set is a limit point of that set. So now, let $f(0)=0$. Then the sequence of functions $f_n(x)=f(x)+\frac{1}{n}$ converges to $f$ with respect to supnorm and $f_n(0)=\frac{1}{n}\neq 0$.

4. It is not dense. Because it is the inverse of the set $\{5\}$ under the continuous map $f\mapsto \int_0^1 f(x)dx$. And hence it is a proper closed set.

Answer 2

Hint :

1. Stone-weierstrass theorem
2. Think about $\{x,x^2,x^3\cdots\}$
3. Think about $\{a+x^i : a\in \mathbb{R}\backslash \{0\}; i\in \mathbb{N}\}$
4. I have no hint for this as of now but i believe this is too small to generate $C[0,1]$

Answer 3

(1) follows directly from the stone-weirstrass theorem, since $[0,1]$ is a closed interval.

(2) is not dense, show that $f(x)=1$ is not a limit point

(3) Here, the closure of this set is $C[0,1]$, try showing that any $g\in C[0,1]$ with $g(0)=0$ is a limit point.

(4) Not dense, again show that $f(x)=1$ is not a limit point.

Answer 4

Hint for 2 :

Let, $S=\{f\in C[0,1]:f(0)\not =0\}$.

Then $S=\{f\in C[0,1]:f(0)<0\}\cup \{f\in C[0,1]:f(0)>0\}$.

So, $\overline S=\{f\in C[0,1]:f(0)\le0\}\cup \{f\in C[0,1]:f(0)\ge0\}=C[0,1]$.

So, $S$ is dense in $C[0,1]$.

Again as $S$ is the union of two open sets so it is open subset of $C[0,1]$.

Hence its complement is nowhere dense.

That is, the set $\{f\in C[0,1]:f(0)=0\}$ is nowhere dense.

Similarly option 4 is nowhere dense.