This is a question related with the regularity of harmonic maps.
Let $N\geq 1$ and $f:\mathbb{R}^N\to \mathbb{S}^2$, where $\mathbb{S}^2=\{x\in \mathbb{R}^3 : \|x\|=1\}$. Assume that the BMO semi-norm of $f$ is small, let say $[f]_{BMO}\leq \varepsilon$.
I would like to prove that the image of $f$ does not cover the sphere, more precisely that $\overline{\operatorname{Im}(f)}\neq \mathbb{S}^2$ (the closure w.r.t. the $\mathbb{R}^3$ topology), if $\varepsilon$ is small enough, but I am not sure that this is true.
Can someone give a proof or a counterexample? If it is false, maybe someone can give some further regularity assumptions on $f$ such that the result is true.
Remark: Here I used the euclidean norm $\|\cdot\| $of $\mathbb{R}^3$ for BMO, i.e. $$ [f]_{BMO}=\sup_{x\in\mathbb{R}^N,r>0} \frac{1}{|{B_r(x)}|}\int_{B_r(x)}\|{f(y)-f_{x,r}}\|\,dy, $$ where $f_{x,r}$ is the average $$f_{x,r}=\frac{1}{|{B_r(x)}|} \int_{B_r(x)} f(y)\,dy.$$
As pointed out in a comment, the question doesn't quite make sense because $f\in BMO$ is only defined almost everywhere. But it seems to me that in fact there does exist a continuous $f\in BMO(\mathbb R)$ with small norm mapping $\mathbb R$ onto $S$. I'm going to write $||f||_*$ for the BMO seminorm.
First a reduction to something a little simpler:
Lemma: Suppose $f$ is measurable and for every ball $B$ there exists a constant $\alpha_B$ with $$\frac1{|B|}\int_B|f-\alpha_B|\le c.$$Then $f\in BMO$, with $$||f||_*\le 2c.$$
Corollary: If $f\in BMO$ and $|\phi(x)-\phi(y)|\le M|x-y|$ then $$||\phi\circ f||_*\le 2M||f||_*.$$
Now, there certainly exists a Lipschitz map $\phi:\mathbb R^2\to S$ such that $\phi(B(0,1))=S$. So we need only show that there exists $f:\mathbb R\to\mathbb R^2$ such that $||f||_*<\epsilon$ and $$B(0,1)\subset f(\mathbb R).$$
Now recall someone's inequality:
Lemma There exists $c$ such that if $f\in H^1(\mathbb T)$ and $n_{j+1}\ge 2n_j$ then $$\sum_j|\hat f(n_j)|^2\le c||f||_{H^1(\mathbb T)}^2.$$ By duality:
Corollary There exists $c$ such that if $n_{j+1}\ge2n_j$, $\sum|c_j|^2<\infty$, and $$f(t)=\sum_jc_je^{in_jt}$$ then $$||f||_{BMO(\mathbb T)}^2\le c\sum_j|c_j|^2.$$
Now let $$c_j=c\lambda^j,$$where $3/4\le\lambda<1$. If you choose $c$ and $\lambda$ appropriately you have $$\sum|c_j|=2$$and $$\sum|c_j|^2\le\epsilon^2.$$We're going to let $$f(t)=\sum_jc_je^{1n_jt}$$for some sequence $n_j$ with $n_{j+1}\ge 2n_j$. Note that $f$ is continuous, and for any choice of $n_j$ satisfying that inequality we have $$||f||_*\le c\epsilon.$$
And now the fuzzy part: Given $c_j$ as above, if $n_j\to\infty$ fast enough we have $B(0,1)\subset f(\mathbb R)$.
That's a fact. A proof might take some space; if you're familiar with constructions of space-filling curves it should be somewhere between plausible and clear...
That gives $f\in BMO(\mathbb T)$, of course. You could argue that $BMO(\mathbb T)\subset BMO(\mathbb R)$, or you could multiply by a smooth cutoff function (equal to $1$ on some period of $f$) to get $BMO(\mathbb R)$.