Is there always a way up?

Question

I am trying to find a simple criterion for a real continuous function $f$ on a connected, open subset $U$ of $\mathbb R^n$ that would imply the following property (P)

For any $x, y \in U$ such that $f(x)<f(y)$, there exists a continuously differentiable path $\gamma:[0,1] \rightarrow U$ such that $ \gamma(0)=x, \gamma(1)=y$ and $f(\gamma(t))$ is increasing.

Intuitively, this "there is always a way up" property will hold for $f$ such that $f^{-1}((c, \infty))$ is connected for any $c \in \mathbb R$. However, if the set $f^{-1}(c)$ is highly irregular then my intuition might break down.

So the question is, does this criteria guarantee the property (P)? If not, is there a simple criterion for $f$ such that (P) is satisfied?

Added: Since the question is for satisfying my curiosity rather than solving a specific problem, feel free to modify any hypothesis of the question (changing 'connected' to 'path connected' or changing the regularity of $f$ for example). Any result, positive or negative, is welcomed.

Added: The criterion I proposed has a trivial counterexample: consider $f$ on $\mathbb R$ having a unique maximum. Anyway, the main question is to find a criterion for (P). I guess it should be that the inverse image of any interval is path connected.

Answer 1

I think the answer is no: $f^{-1}((c,\infty))$ being connected is not sufficient. My idea for a counterexample is to begin with a set that is connected but not path connected (the topologist's sine curve), and try to make some of the regions of the form $f^{-1}((c,\infty))$ resemble that set, so that they are also connected but not path connected.

Let $n=2$ and $U = (-1,1) \times (0,1)$. Let $G$ be the graph of the function $y = e^{-x}\sin(1/x)$ with $x \in (0,1/2)$. (Roughly, this is a copy of the topologist's sine curve limiting onto the line segment $\{0\} \times [0,1]$, but tapering in such a way that the sine-like part of $G$ is contained entirely in $U$.) Now define your function $f$ by taking $f(x,y) = 0$ if $(x,y) \in G$, $f(x,y) = -x$ if $x \leq 0$, and $f(x,y) = -\mathrm{dist}((x,y),G)$ if $x > 0$ and $(x,y) \notin G$. Roughly, this function is constantly $0$ on $G$, and steadily increasing to the left of the $Y$-axis, but it is negative everywhere to the right of the $Y$-axis except on $G$.

Now the point is you cannot "go upwards" from a point in $G$ to a point that is left of the $Y$-axis. The reason is that $\overline{G}$ ($G$ plus the line segment it limits to) is not path-connected. Any path beginning in $G$ must either leave $G$ while still to the right of the $Y$-axis (but then $f$ decreases), or it must stay in $G$ forever. Nonetheless, I'm pretty sure that $f^{-1}((c,\infty))$ is connected for all $c$.

Answer 2

Another counter-example in $\mathbb{R}^2$ is the Mount-St.Helens function $$ e^{-(x^2+y^2)} - \frac12\, e^{-9\,\left( \left(x-\frac15\right)^2 +y^2\right)},$$ on $U=(0,1)^2$, because one cannot go upwards e.g. from close to $(1,1)$ to $(\frac15,0)$. In view on these counter-examples one may try the criterion that for all $c\in \mathbb{R}$ both sets $$ f_{<c}~=~ f^{-1}\left( (-\infty,c)\right),~~~{\rm and}~~ f_{>c}~=~ f^{-1}\left( (c,\infty)\right)$$ must be connected. This would at least also fix the previous counterexamples by Will Brian and for e.g $e^{-x^2}$ in $\mathbb{R}$ by the OP.