Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin

Question

I am looking for a continuous function $f: \mathbb{R}^2-\{0, 0\} \to \mathbb{R}$ satisfying the following condition: For any $g, h: \mathbb{R}\to\mathbb{R}$ continuous functions with $g(0)=h(0)=0$, we have $$ \lim_{t\to 0} f( g(t), h(t) ) = 0 $$ but $$ \lim_{(x, y)\to (0, 0)} f(x, y) $$ does not exist. In other words, the function $f$ is not continuous at the origin, but the limit of $f$ along any curve approaching the origin exists and is zero.

Motivation: In multivariable calculus, to prove that $f: \mathbb{R}^2\to\mathbb{R}$ is not continuous, we often find two curves approaching the origin on which $f$ has two different limits, which implies that $f$ is not continuous. So the converse would be false, if we can find a find an example of $f$ as described above.

Doesn't work if the domain is $\mathbb{R}^2$, you need more complicated spaces to have path-continuity not imply continuity. — Daniel Fischer, Feb 14 '16 at 22:41
@David: Ah good point. I have edited the question accordingly. — Prism, Feb 14 '16 at 22:46
@DanielFischer: Thanks! That answers the question, namely that such a function $f$ does not exist (because in some sense $\mathbb{R^2}$ is too nice). I would be happy to accept your answer if you could say a few words about the proof (for $X=\mathbb{R}^2$ or more generally for $X$ locally path connected + first countable). — Prism, Feb 14 '16 at 22:58
But even in $\mathbb{R}^2$, it's not sufficient to require the same limit along every straight line through the point to have continuity. I suppose you are aware of that, and probably have some example(s) for it? — Daniel Fischer, Feb 14 '16 at 22:58
@Daniel Would something like $\frac{x^2+y^2}{x}$ work? It is not continuous at $(0, 0)$ but the limit exists along every straight line. — Prism, Feb 14 '16 at 23:06
@Daniel: Right. I cannot come up with simple counter-examples. Is the counter-example you have mind a rational function of $x$ and $y$? — Prism, Feb 14 '16 at 23:12
No, I think rational functions don't work, you need something a bit more complicated. — Daniel Fischer, Feb 14 '16 at 23:14
@DanielFischer: Apparently, this rational function works. But I do wanna hear the counter-example you have in mind as well :) — Prism, Feb 14 '16 at 23:17
Ah, nice and simple. I had nothing specific in mind, just a general idea how to construct examples. — Daniel Fischer, Feb 14 '16 at 23:21

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

As mentioned in the related question, $\mathbb{R}^2$ is a too well-behaved space for such a function to exist.

If a topological space $X$ is locally path-connected and first countable, every function $f\colon X \to Y$, where $Y$ is any topological space, such that the composition $f\circ \alpha \colon I \to Y$ is continuous for all paths $\alpha \colon I \to X$ ($I \subset \mathbb{R}$ is some nondegenerate interval, open, half-open, or closed) is already continuous.

For if $f$ is discontinuous at $x_{\ast}\in X$, then by first countability there is a sequence $(x_n)_{n\in \mathbb{N}}$ in $X$ with $x_n \to x_{\ast}$ such that $f(x_n) \nrightarrow f(x_{\ast})$. By passing to a subsequence, we can assume that there is a neighbourhood $W$ of $f(x_{\ast})$ such that $f(x_n) \notin W$ for all $n$. Let $\{ V_k : k \in \mathbb{N}\}$ be a neighbourhood basis of $x_{\ast}$ consisting of path-connected sets, with $V_{k+1} \subset V_k$ for all $k$. By dropping some terms from the start of the sequence if necessary, we can assume that $x_n \in V_0$ for all $n$. We then define recursively $n_0 = 0$ and

$$n_{k+1} = \min \{ m > n_k : n \geqslant m \implies x_n \in V_{k+1}\}.$$

We can now replace $(x_n)$ by the subsequence $(x_{n_k})$, or equivalently assume that $n_k = k$ for all $k$.

Now we construct a path passing through the points $x_k$: Since $x_k$ and $x_{k+1}$ both lie in $V_k$, we can choose a path $\alpha_k \colon [2^{-k-1},2^{-k}] \to V_k$ with $\alpha_k(2^{-k-1}) = x_{k+1}$ and $\alpha_{k}(2^{-k}) = x_k$. Then define $\alpha \colon [0,1] \to X$ by

$$\alpha(t) = \begin{cases} x_{\ast} &, t = 0 \\ \alpha_k(t) &, 2^{-k-1} < t \leqslant 2^{-k}.\end{cases}$$

Since $\alpha_k(2^{-k-1}) = \alpha_{k+1}(2^{-k-1}) = x_{k+1}$, $\alpha$ is continuous on $(0,1]$, and since $\alpha([0,2^{-k}]) \subset V_k$, $\alpha$ is also continuous at $0$.

But $f(\alpha(2^{-k})) = f(x_k) \notin W$, so $f\circ \alpha$ is not continuous at $0$.

This is beautiful! Thanks so much Daniel – Prism Feb 14 '16 at 23:43 — Prism, Feb 14 '16 at 23:43

Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin

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