How do I show that this set is dense in the function space?

Question

Consider the smooth manifold $\mathbb{S}^3$ embedded in $\mathbb{R}^4$, note that $$\widetilde{T}:= \frac{1}{\sqrt{2}}\mathbb{T}^2 = \left\{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4; \ x_1^2 + x_2^2 = x_3^2 + x_4^2 = \frac{1}{2}\right\} $$

is a two-dimensional smooth submanifold of $\mathbb{S}^3$.

I would like to know how to prove the following theorem

Theorem: Let $\mathcal{A}$ be the set $$\mathcal{A} := \{f: \mathbb{S}^3 \to \mathbb{R}; f \ \text{is a smooth function and} \ f(p) \neq 0, \ \forall \ p \in \widetilde{T}\},$$ then $\mathcal{A}$ is dense in the set $\mathcal{F(\mathbb{S}^3)} = \{f: \mathbb{S}^3 \to \mathbb{R}; f \ \text{is a smooth function}\}$.

I.e. if $f$ $\in$ $\mathcal{F}(\mathbb{S}^3)$ and $\varepsilon >0$, there exists $g$ $\in$ $\mathcal{A}$, such that

$$\sup_{x \in \mathbb{S}^3}|f(x) - g(x)|< \varepsilon \quad \text{and}\quad \sup_{x \in \mathbb{S}^3}\|\text{d}f_x - \text{d}g_x\|< \varepsilon .$$

Can anyone help me or give me some hints?

Answer 1

Where did you read this theorem? I think this is not correct. Let $f(x_1,\ldots,x_4) = x_1-\tfrac{1}{2\sqrt 2}$. This is definitely a smooth function on $S^3$ and $f(0,\tfrac 1{\sqrt 2},\tfrac 1 2,\tfrac 1 2) = -\tfrac 1{2\sqrt 2} < 0$, whereas $f(\tfrac 1{\sqrt 2},0,\tfrac 1 2,\tfrac 1 2) = \tfrac 1{2\sqrt 2} > 0$. Moreover, there is a path in $\widetilde T$ between these two points. How would you want to approximate $f$ by a continuous function $g$ that is non-zero on this path?

Answer 2

Your theorem is false: any function $f$ which takes both positive and negative values on $T$, for example $f(x) = x_1$, is an interior point in the complement of $A$.

Here's a rigorous argument: Note that $$A = A^+ \cup A^-$$ where $A^+ =\{f\in \mathcal{F}(S^3) ~\colon~ f > 0 ~\mathrm{ on }~ T \}$ and $A^- =\{f\in \mathcal{F}(S^3) ~\colon~ f < 0 ~\mathrm{ on }~ T \}$. This is because any $f \in A$ is continuous, and since $T$ is connected, $f$ cannot change sign unless it vanishes somewhere.

Consider $B^+ =\{f\in \mathcal{F}(S^3) ~\colon~ f \geqslant 0 ~\mathrm{ on }~ T \}$ and $B^- =\{f\in \mathcal{F}(S^3) ~\colon~ f \leqslant 0 ~\mathrm{ on }~ T \}$. Both $B^+$ and $B^-$ are closed for the pointwise convergence, and hence for the much stronger (but not helpful) topology you are considering on $\mathcal{F}(S^3)$. Therefore $B^+ \cup B^-$ is closed. It follows that the closure of $A$ is contained in $B^+ \cup B^-$, since $A = A^+ \cup A^- \subseteq B^+ \cup B^-$. Note that $B^+ \cup B^-$ is very far from everything, so your theorem is "very false" (a generic function in the complement of $A$ is not in its closure).