Prove $f:\mathbb{R}^n\to\mathbb{R}$ is continuous iff for every curve, $\gamma:[a,b]\to\mathbb{R}^n: f\circ \gamma :\mathbb{R}\to\mathbb{R}$ is continuous.

$(\Rightarrow)$ is trivial.

$(\Leftarrow)$: Intuitively, $f$ is continuous at every subset on it's domain, so all in all $f$ is continuous.

How to formalize this? What should I rely on?


  • $\begingroup$ Are you familiar with $\epsilon-\delta$ proofs? $\endgroup$ – Tim Raczkowski Jan 13 '15 at 0:39
  • $\begingroup$ Yes I am @TimRaczkowski $\endgroup$ – AlonAlon Jan 13 '15 at 0:40
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    $\begingroup$ I think arguing by contradiction is a good way to go here. Suppose, there is a point $a\in\Bbb R^n$ and a curve $\gamma$ for which $f\circ\gamma$ is not continous at $a$. Formulate this in terms of $\epsilon$ and $\delta$ and show that $f$ is not continuous at $a$. $\endgroup$ – Tim Raczkowski Jan 13 '15 at 0:46
  • $\begingroup$ @TimRaczkowski: What you suggest is reproving the "trivial" direction. $\endgroup$ – Ted Shifrin Jan 13 '15 at 2:36
  • $\begingroup$ @TedShifrin yes, you are right. $\endgroup$ – Tim Raczkowski Jan 13 '15 at 2:37

Let's try this. Suppose $f$ is not continuous at $x=a$. Then there is an $\epsilon>0$ such that for every $n\in\Bbb Z_+$ there is a point $x_n$ such $|x_n-a|< \frac{1}{n} $ and $|f(x_n)-f(a)|\ge \epsilon$. Now, let $\gamma_n$ be a parametrized line from $x_{n}$ to $x_{n+1}$. Now, we can "string" the $\gamma_n$ together to create a path $\gamma$ such that $f\circ\gamma$ that is not continuous.

  • $\begingroup$ Right. It takes a tiny bit of work to get the stringing right and a continuous function on, say, $[0,1]$. $\endgroup$ – Ted Shifrin Jan 13 '15 at 4:06

Let $(p_i)_{i\ge 1}$ a sequence of points in $\mathbb{R}^n$ converging to $p$. Consider the following function $\gamma\colon [0,1] \to \mathbb{R}^n$: $\gamma(\frac{1}{i}) = p_i$, $\gamma$ is linear on each interval $[\frac{1}{i+1}, \frac{1}{i}]$, and $\gamma(0)=p$. One checks easily that $\gamma$ is continuous.

Assume that $f \circ \gamma$ is also continuous. Then $\lim_{t\to 0} f(\gamma(t)) = f(\gamma(0))$ and therefore $\lim_{ i \to \infty} f(\gamma(\frac{1}{i})) = f(\gamma(0))$ or $\lim_{ i \to \infty} f(p_i) = f(p)$.

It's easy from here.

This works more generally if instead of $\mathbb{R}^n$ we have a normed vector space.


This isn't the most elementary approach, but it's fun. Notice that the existence of the Hilbert curve means we can find an surjective continuous function $f:[-1,1]\rightarrow [-1,1]^n$. Let us, for convenience, assume that $f(0)=0$ and $\gamma(0)=0$ and prove continuity of $f$ at $0$. Then, using a construction similar to Hilbert's, choose some $\gamma$ which has that the image $$\gamma \left(\left[\frac{1}{4^n},\frac{2}{4^n}\right]\right)=\left[-\frac{1}{2^n},\frac{1}{2^n}\right]^n$$ and that the intervening intervals of the form $\left[\frac{2}{4^n},\frac{4}{4^n}\right]$ are interpolated arbitrarily. Now, if $f\circ \gamma$ is continuous then we know that, for any $\varepsilon>0$, there is an $\delta>0$ such that if $f(\gamma(t))<\varepsilon$ for all $|t|<\delta$. However, the image $\gamma((-\delta,\delta))$ always contains a neighborhood of $0$ due to the construction of $\gamma$. Therefore, we can say that $f(x)<\varepsilon$ for all $x$ in a neighborhood of $0$, ergo $f$ is continuous there.


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