Perturbing a path

Question

Suppose $\gamma : [0,1] \to \mathbb{R}^n$ is a smooth path, such that $\gamma([0,1]) \subseteq U$, where $U$ is an open subset of $\mathbb{R}^n$. Now I would like to consider the path $\gamma_\varepsilon : [0,1] \to \mathbb{R}^n$, where $$\gamma_\varepsilon^i(t) := \gamma^i(t) + \varepsilon f(t)$$ for some $i = 1,\dots,n$, $\varepsilon > 0$ and $f \in C^\infty_c(0,1)$ and the other components stay the same. I think now, that for $\varepsilon$ sufficiently small, it should be the case that $\gamma_\varepsilon$ is contained in $U$. Is that true?

I tried to prove it, however, I did not succeed. My idea was to use something like a neighbourhood of the form $$U_\delta := \{x \in \mathbb{R}^n : \mathrm{dist}(x,\gamma[0,1]) < \delta\}$$ for some $\delta > 0$. Since we know that $\gamma[0,1]$ is compact, I guess this should be possible. I tried a proof by contradiction.

Edit. If we can show the existence of such a $U_\delta$, I think the rest is easy: For any $t \in [0,1]$ we then have $$|\gamma_\varepsilon(t) - \gamma(t)| \leq \varepsilon \|f\|_\infty$$ Hence $$\mathrm{dist}(\gamma_\varepsilon(t),\gamma[0,1]) = \inf_{s \in [0,1]}|\gamma_\varepsilon(t) - \gamma(s)| \leq |\gamma_\varepsilon(t) - \gamma(t)| \leq \varepsilon \|f\|_\infty$$ for arbitrary $t \in [0,1]$ and hence for $\varepsilon$ sufficiently small we get that $\varepsilon \|f\|_\infty < \delta$.

Answer 1

Let's assume, that there is no such $U_\delta$. That means that $\forall\delta>0\ \exists t_\delta:B_\delta(\gamma(t_\delta))\not\subseteq U$. For the sequence of $\delta_n=1/n$, there is a sequence of $t_n$. Because $0\le t_n\le1$, we can choose a converging subsequence $t_k$. Let $\tau=\lim t_k$ and $M=\gamma(\tau)$.

Since smooth functions are locally Lipschitz continuous, the tail of sequence $\gamma(t_k)$ stays inside the neighbourhood of $M$ and $\lim\gamma(t_k) = M$.

Consider the radius $r$ of neighbourhood of $M$. $\exists N\forall m>N,\gamma(t_m)\in B_{r/2}(M)$. But if we take $m > 2/r$, then since $B_{\delta_m}(\gamma(t_m))\not\subseteq U$, $\exists P\not\in U:\mathrm{dist}(P,\gamma(t_m))<\delta_m<r/2$. Triangle inequality gives us: $$ \mathrm{dist}(P,M) \le \mathrm{dist}(P,\gamma(t_m)) + \mathrm{dist}(M,\gamma(t_m)) < r $$ But $B_r(M)\subset U$. We got a contradiction.

Answer 2

The assumption that $f$ is smooth and compactly supported in $(0,1)$ is unnecessary. We only need $f$ to be a bounded function on $[0,1]$, so let's say $|f|\le M$.

Since $\gamma$ is continuous, the image set $\gamma\big([0,1]\big)$ is compact. Since $\gamma\big([0,1]\big)$ is contained in the open set $U$, each point in the image set is contained in a ball contained in $U$. By compactness, take finitely many of these balls $B_1,\dots,B_k$ such that $\gamma\big([0,1]\big)\subset\bigcup_{j=1}^k B_j\subset U$.

Finally, if $0<\epsilon<\mathrm{dist}\Big(\gamma\big([0,1]\big),\Bbb R^n\smallsetminus\big(\bigcup B_j\big)\Big)\Big/M$, then the image set of $\gamma_\epsilon = \gamma + \epsilon(f,\dots,f)$ is wholly contained in $U$. The claim follows.

Answer 3

Lemma. Let $U \subseteq \mathbb{R}^n$ open and $A \subseteq U$ compact (or rather just closed). Then there exists $\delta > 0$ such that $$U_\delta := \{x \in \mathbb{R}^n : \mathrm{dist}(x,A) < \delta\} \subseteq U.$$

Proof. Towards a contradiction, suppose that no such $\delta$ exists. Hence we find a sequence $(x_n)$ in $\mathbb{R}^n$ such that $\mathrm{dist}(x_n,A) < \frac{1}{n}$ and $x_n \notin U$ holds for any $n \in \mathbb{N}$. Then $(x_n)$ is a Cauchy sequence. Indeed, for any $a \in A$ we have that $$|x_n - x_m| \leq |x_n - a| + |x_m - a|$$ Taking the infimum over $a \in A$ yields $$|x_n - x_m| \leq \mathrm{dist}(x_n,A) + \mathrm{dist}(x_m,A) < \frac{1}{n} + \frac{1}{m}$$ Since $\mathbb{R}^n$ is a Banach space, we find $x \in \mathbb{R}^n$ such that $x_n \to x$. We claim that $x \in A$. Indeed, since $\mathrm{dist}(-,A)$ is continuous, we get that $$\mathrm{dist}(x,A) = \lim_{n \to \infty}\mathrm{dist}(x_n,A) = 0.$$ Thus there exists some sequence in $A$ converging to $x$. But closedness of $A$ implies now that $x \in A$ (I think compactness is not needed at all). Since $U$ is open, we find a ball around $x$ contained in $U$. Contradiction.