# Real Analytic Maps with Dense Image

Let $$f$$ be a real analytic map from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ where $$n. Can we say image of $$f$$ is not dense in $$\mathbb{R}^m$$? I know this is not true for $$C^0$$ or $$C^1$$ maps.

The answer is no. Below I construct an analytic function $$f:\mathbb R \rightarrow \mathbb R^2$$ with dense image. Note, that this does not contradict Sard's theorem, since it only implies that the image has zero measure and does not say anything about density.

I was not able to come up with an example that is given by a simple formula, but it should be noted that if we replace real vector spaces with arbitrary analytic manifolds $$N$$ and $$M$$ with dimensions $$n < m$$ and ask the same question, then there are well-known classical examples. For instance, almost all geodesics on a torus are dense.

We start from a real analytic function $$\phi:\mathbb R \rightarrow [-1,1]^2$$ with dense image. It can be easily constructed as $$\phi(t) = (\phi_1(t), \phi_2(t)) = (\sin(\alpha t), \sin(\beta t)),$$ where $$\alpha, \beta \in \mathbb R\setminus \{0\}$$ and $$\alpha / \beta$$ is irrational.

Now let $$S \subset \mathbb R^3$$ denote the unit sphere. Using $$\phi$$ we define a real analytic function $$\psi:\mathbb R \rightarrow S$$ with dense image. It just maps $$t$$ to a point on a sphere that has spherical coordinates $$\pi \phi_1(t)$$ and $$\pi(\phi_2(t) + 1)/2$$. From the formulas that map spherical coordinates onto $$S$$ it is clear that $$\psi$$ is real analytic.

Finally, we note that, since $$\psi$$ is analytic, its image (by Sard's theorem, for instance) does not coincide with $$S$$. Moreover, if $$x \in S$$, then there is a real analytic diffeomorphism $$p_x$$ from $$S \setminus \{x\}$$ to $$\mathbb R^2$$ given by the stereographic projection. The function in question now can be constructed as $$p_x \circ \psi$$, where $$x \in S$$ is taken outside the image of $$\psi$$.

EDIT. A deleted answer to this question also gave me another insight into this problem. Namely, given any map $$c:\mathbb Z \rightarrow \mathbb R$$ there exists a real analytic function $$f:\mathbb R \rightarrow \mathbb R$$ that extends $$c$$. Using this and arbitrary dense sequence in $$\mathbb R^n$$ you can construct a real analytic function $$f:\mathbb R\rightarrow \mathbb R^n$$ such that $$f(\mathbb Z)$$ is dense.

I suppose, there are many ways to prove the statement above. I prefer complex analytic arguments that allow to construct a function $$f$$ with prescribed values on integers even with the additional constraint that f is an entire complex analytic function. The proof goes along the lines of the Mittag-Leffler theorem. See this post, for instance: Entire function with prescribed values

No. Let's consider the case $$n=1$$. Let $$x_j$$ be a sequence that is dense in $$\mathbb R^n$$. We can assume $$|x_j| < j$$ for all $$j$$. Then there is a real-analytic map $$f: \mathbb R \to \mathbb R^n$$ such that $$f(n) = x_n$$ for all positive integers $$n$$. In fact we can assume $$f$$ is analytic on the strip $$U = \{z \in \mathbb C: |\text{Im}(x)| < 1\}$$. We can define it as

$$f(x) = \sum_{n=1}^\infty x_n g_n(x-n)$$ where $$g_n(z) = \frac{\sin(\pi z)}{\pi z} \exp(-n z^2)$$ for $$z \ne 0$$, $$g_n(0) = 1$$. Note that $$g_n(z)$$ is an entire function (the singularity at $$0$$ being removable), and for $$|\text{Im }(z)|<1$$ we have $$|g_n(z)| < 4 \exp(n - n (\text{Re }z)^2))$$ This is enough to conclude that the series $$\sum_{n=1}^\infty x_n g_n(z-n)$$ converges uniformly on compact subsets of $$U$$, so that $$f$$ is analytic there.