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I am looking for a smooth function $h:\mathbb{R}^2\to\mathbb{R}$ such that,

  1. $Supp(h)\subset B(0,1)$
  2. $|h(0)| \ge 1$
  3. $|\partial_x h| < \epsilon$ everywhere, where $\epsilon > 0$ is a given constant

I have seen this and this, where the answer is negative for bounded derivatives. But note that I don't require any control over the partial derivative $\partial_y h$.

So can such a function $h$ exist? Or is the situation still the same and we cannot expect any kind of derivative bound at all?

If it helps, geometrically I am trying to think of climbing a hill of height $1$ and base $B(0,1)$, by using a road with many hairpin bends : as I am traversing the road the inclination (i.e $\partial_x h$) doesn't grow too much; but as I am crossing a bend, the $\partial_y h$ value gets arbitrarily large. I apologize if this makes no sense at all!

Any help is appreciated. Cheers!

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For any ${\bf a}, {\bf b} \in \mathbb R^2$, and any polygonal path $\Gamma$ from $\bf a$ to $\bf b$, $$h({\bf b}) - h({\bf a}) = \int_\Gamma (\nabla h)({\bf r}) \cdot d{\bf r}$$
In particular, take ${\bf a} = (-1,0)$ and ${\bf b} = (0,0)$, and $\Gamma$ the straight line segment from $\bf a$ to $\bf b$. We get

$$1 \le h({\bf b}) - h({\bf a}) = \int_{-1}^0 \partial_x h(t,0)\; dt \le \max_{0 \le t \le 1} \partial_x h(t,0)$$

i.e. there must be some $t \in [0,1]$ with $\partial_x h(t,0) \ge 1$.

The point is that the first equation is true for every polygonal path, not just for some. You don't get to choose the road!

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