# There exists differentiable curves arbitrarily close to the continuous ones

Let $$M$$ be a Riemannian manifold; if $$d$$ is the distance on $$M$$, we can consider the distance $$D$$ between any two continuous curves given by $$D(c, \gamma) = \max _{t \in [0,1]} d(c(t), \gamma(t))$$.

Let $$c:[0,1] \to M$$ a continuous curve. Is it true that for every $$\varepsilon > 0$$ there exists a differentiable (or even smooth) curve $$\gamma$$ such that $$D(c, \gamma) < \varepsilon$$?

I believe that this is true, and I also believe that I have proven it (the case of Peano-like curves make me doubt my reasoning, though). The idea is to consider sufficiently many points on $$c$$, join each two consecutive of them by a unique minimizing geodesic, join all these geodesic segments in a single zig-zag line and show that this line is at distance at most $$\varepsilon$$ from $$c$$. The problem is that its proof is 1.5 pages long (a lot of playing around with the injectivity radius), technical and boring. It would help me if, instead of including it in an article that I am writing, I could replace it with just a reference.

Assuming the result above is true, can you please provide a citable reference?

• Using Whitney embedding or Nash embedding, and tubular neighborhoods, it reduces to the same problem in Euclidean space. May 31 at 9:39
• @BenMcKay: Agreed, but using Nash is overkill. I am not sure about Whitney, because Whitney doesn't preserve distances. This means that the smooth curve may be "close" in the Euclidean space, but does it stay so when you pull it back on $M$? May 31 at 9:42
• @AlexM. but Nash has the same problem (you still have to compare the riemannian metric inside the imbedded manifold with the euclidean distance in $\mathbb{R}^N$). Whitney gives you the same as Nash in this case because you can choose a compact submanifold with boundary of $M$ containing the curve in its interior, and then use that any $2$ metrics on a compact Riemannian manifold are equivalent. May 31 at 13:26
• The problem of, given a compact submanifold $N\subseteq\mathbb{R}^N$, proving that the riemannian distance $d_N$ from the metric of $N$ and the restriction to $N$ of the Euclidean distance $d$ are equivalent can surely be solved by considering the function $F:N\times N\to\mathbb{R}$ given by $F(x,y)=\frac{d_N(x,y)}{d(x,y)}$ if $x\neq y$ and $F(x,x)=1\forall x$ and proving that it is continuous. But I don't know if that takes less time than constructing the smooth curve. May 31 at 14:54
• I don't think that it is necessary to embed the manifold into $\mathbb R^N$, and in fact, I tend to believe that the Riemannian structure is not essential by compactness. I imagine the following lemma: Fix $\epsilon>0$, and let $f\colon[0,1]\to\mathbb R^n$ be a continuous curve along with a smooth curve $g\colon[0,a)\to\mathbb R^n$ such that $\lVert f(t)-g(t)\rVert_{L^\infty([0,a))}<\epsilon$. Then there exists a smooth curve $h\colon[0,1]\to\mathbb R^n$ such that $\lVert f-h\rVert_{L^\infty([0,1])}<\epsilon$ and $b>0$ such that $g=h$ on $[0,b]$.
– Z. M
May 31 at 16:36

Theorem [Thm 3.3, Hirsch, Differential Topology] Let $$M$$ and $$N$$ be $$C^s$$-manifolds (with boundary), $$1\le s\le\infty$$. Then $$C^s(M,N)$$ is dense in $$C_S^r(M,N)$$ for $$0\le r, where $$C_S^r(M,N)$$ is equipped with the "strong" topology (when $$r=0$$, it becomes the compact-open topology).
In particular, when $$s=\infty$$, $$r=0$$, $$M$$ is compact and $$N$$ is equipped with a (Riemannian) metric, it leads to the result that you want.