# Is this "continuous" function really continuous?

Let's assume $$n \ge 2.$$ Suppose $$f:\Bbb R^n \times \Bbb R \to \Bbb R^{n+1}$$ has the form $$f(x,t) = (\phi_t(x),t),$$ where $$\phi_{t_0}:\Bbb R^n \to \Bbb R^n$$ is a continuous function for each fixed $$t_0$$. Assuming also that there exists a dense set $$D\subset \Bbb R^n$$ such that for each $$x_0 \in D$$, the map $$f(x_0,\cdot)$$, which is the restriction of $$f$$ to the line $$\{x_0\}\times \Bbb R$$, is uniformly continuous.

Is it true that $$f$$ must be continuous?

I suspect that the answer is negative, but I cannot think of a counterexample yet. Of course, here we are talking about the standard Euclidean topology only.

No, this is not necessarily true. As an example, let $$n = 2$$, $$\phi_t: \mathbb{R}^2 \to \mathbb{R}^2$$ be defined by,

$$\phi_t(x, y) = \begin{cases} (t^{|x|}, y) &, \text{ if }t \in (0, 1], x \in [-1, 1]\\ (t, y) &, \text{ if }t \in (0, 1], x \notin [-1, 1]\\ (0, y) &, \text{ if }t \leq 0\\ (1, y) &, \text{ if }t > 1 \end{cases}$$

Then indeed $$\phi_t$$ is a continuous (in fact, uniformly continuous) function for each $$t$$. Let $$D = (\mathbb{R} \setminus \{0\}) \times \mathbb{R} \subset \mathbb{R}^2$$, which is dense. For any $$x_0 \in D$$, one easily verifies that $$f(x_0, \cdot)$$ is uniformly continuous. But $$f$$ is not continuous. In fact, $$f((0, 0), \cdot)$$ is not continuous. It has a jump discontinuity at $$t = 0$$.

Here is an even simpler example with the even stronger condition that $$D = \mathbb{R}^2$$:

$$\phi_t(x, y) = \begin{cases} (\frac{\min\{t^2, x^2\}}{t^2 + x^2}, y) &, \text{ if }(t, x) \neq (0, 0)\\ (0, y) &, \text{ if }(t, x) = (0, 0) \end{cases}$$

Then $$\phi_t$$ is uniformly continuous for any $$t$$ and $$f(x_0, \cdot)$$ is uniformly continuous for any $$x_0 \in \mathbb{R}^2$$. But $$f$$ is not continuous, since,

$$\lim_{x \to 0} f((x, x), x) = (\frac{1}{2}, 0, 0) \neq (0, 0, 0) = f((0, 0), 0)$$

• I should note that these are quite standard examples. I simply adapted them to the OP’s specific question. Commented Jul 12 at 15:16
• Thank you for the nice examples. I forgot to write in the question that I want $\phi_t$ to be injective on $D$ as well, but that is too late to edit it now since you have already perfectly answered the original question. Do you happen to know a good counterexample if we add this further condition? Commented Jul 12 at 16:18
• @BigbearZzz By $\phi_t$ being injective on $D$, do you mean the map $t \mapsto \phi_t(x_0)$ is injective for all $x_0 \in D$? In that case just change the second example to $\frac{t^3}{|t|^3 + |x|^3}$. Then $D = (\mathbb{R} \setminus \{0\}) \times \mathbb{R}$ works. Commented Jul 12 at 16:40
• Sorry I wasn't being clear. I meant that for each fixed $t$, the map $x\mapsto \phi_t(x)$ is injective on $D$ (but could potentially be non-injective on $\Bbb R^n$). Commented Jul 12 at 16:52
• @BigbearZzz In that case, let $\phi_t(x, y) = (\frac{x^2}{t^2 + x^2} + x^2, y)$ if $(t, x) \neq (0, 0)$ and $\phi_0(0, y) = (1, y)$. Let $A$ be the set of positive algebraic numbers and negative transcendental numbers. Then $D = A \times \mathbb{R}$ works. Commented Jul 12 at 18:07