Is a smooth function with bounded partial derivatives defined on an open set in Euclidean space bounded?

Question

This question is a follow up to this one, which was answered essentially in the negative. So I am trying to find another way.

I have a connected bounded open set $X \subset R^n$; then I know it is also path-connected. I also have a smooth (say at least $C^2$) function $u$ on $X$ to the positive reals $P$. I also know that the absolute value of the partials $\partial u/\partial x_i,\; 1 \leq i \leq n$ are uniformly bounded on $X$. Finally, I have a limit point $z$ of $X$ and a sequence $\{z_n\} \subset X$ approaching $z$ . I would like to show that $\{u(z_n)\}$ is bounded (so I can extend it).

I don't know that $X$ is convex, so I can't use that. Instead, my idea is to integrate $u$ along paths $l(z_1,z_k) \subset X, \; k = 2,3,\ldots$, where $l(z_1,z_k)$ is of course the path in $X$ connecting $z_1$ and $z_k$. I would then show that this sequence of integrals is bounded by using the boundedness of the partials of $u$. I could then conclude that $u(z_n)$ itself is bounded. There's only one problem: While I believe, because $u$ is $C^2$, that $l(z_1,z_k)$ has finite length, I don't know that $\lim_{k \rightarrow \infty} l(z_1,z_k)< \infty$ (so I can't use the typical argument that the value of the integral is bounded by the bound on the partials times the length of the domain of integration). Is there any way to show that in this particular situation the limit of the length of the paths is finite? Perhaps by showing there is a neighborhood $U(z)$ of $z$ such that $u$ is bounded on the "last piece" $U(z) \bigcap X$? I am wary of drawing any conclusions in this situation because I know how crazy paths (even in $R^n$) can be.

Any ideas appreciated. Really.

Edit: Thanks to Paul Frost for suggesting edits (above in bold) to make things clearer. The partials being bounded uniformly means that there exist positive constants $K_i,\; 1 \leq i \leq n$ such that $\lvert \partial u(x)/\partial x_i(x) \rvert < K_i,\; \forall x \in X, 1 \leq i \leq n$.

Edit: Martin R seems to have provided a clever counter-example showing the answer to my question is no. I am still hopeful of getting the result I need as $X$ is actually the image, by an open map (a projection), of an $n$-dimensional integral manifold $L$. Given bounds on the partial derivatives of $L$ (and perhaps an additional restriction on the second derivative) I can use estimates on the minimum size of the image, in $X$, of any open "ball" in $L$ to show that, for every $x \in X$, if $U_x$ is a neighborhood of $x$ completely contained in $X$ then it must contain a ball of minimum radius (I have questions about these estimates, I will create another post to pose them). This obviously would get around counter examples such as the one provided by Martin R. My thanks to all who posted and tried to help.

Perhaps you want to ask whether $u$ is bounded on $U_\epsilon(z) \cap X$ if $z$ is a limit point of $X$? — Paul Frost, Sep 10 '22 at 12:11
@Paul Frost: $X$ is bounded (it's actually a subset of the $(n-1)$-dimensional simplex $S^{n-1}$. I mean by uniformly bounded that there is a constant $K > 0$ such that $\lvert \partial u(x)/\partial x_i(x) \rvert < K, \forall x \in X ; 1 \leq i \leq n$. My ultimate objective involves extending the graph of an integral manifold defined on $X$ to such a limit point. but an answer to your alternative question should suffice. I am just having a problem with the possibility of $u$ "winding around" an infinite path in $U_\epsilon (z)$. Thanks for taking the time to post. — user167131, Sep 10 '22 at 13:02
I think you should edit your question. At least you should say that $X$ is bounded. And a partial $\partial u/\partial x_i$ being bounded means that $\lvert \partial u/\partial x_i(x) \rvert \le K_i$ for all $x$, thus you get your common $K$ by taking the maximum. — Paul Frost, Sep 10 '22 at 13:29
You have not determined the range of $u$. Is it $\mathbb{R}$ or what? — , Sep 10 '22 at 14:08
@George Tsoutsinos: Thanks for posting, $u$ maps to the strictly positive reals (I call it $P$). The graph $\Gamma(u)$ of $u$ is actually an $n$-dimensional smooth integral manifold, I'm trying to show it can be extended to $z$. — user167131, Sep 10 '22 at 14:22

Martin R · Accepted Answer · 2022-09-10T21:12:49.257

4

The conjecture is wrong for $n=2$ (and therefore also in all higher dimensions). The idea of the following counterexample is to construct a domain $X$ where some boundary points can be reached only via a path of infinite length.

Let $X \subset \Bbb R^2$ be the open square $(0, 1)^2$, with the line segments $$ A_n = \left\{ (x, \frac 1n) \mid 0 \le x \le \frac 45 \right\} \, , n = 2, 4, 6, \ldots $$ and $$ B_n = \left\{ (x, \frac 1n) \mid \frac 15 \le x \le 1 \right\} \, , n = 3, 5, 7, \ldots $$ removed.

For $0 < x < 1/5$, define $f(x, y)$ as $1, 3, 5, \ldots$ in the squares between the line segments $A_n$, starting at the top.

For $4/5 < x < 1$, define $f(x, y)$ as $2, 4, 6, \ldots$ in the squares between the line segments $B_n$, starting at the top.

In each of the the remaining “stripes” let $f$ smoothly increase (or decrease) from $n$ to $n+1$, with bounded derivative in $x$ direction and constant in $y$-direction.

(One can for example set $$ f(x, y) = n + \frac{\phi(x-1/5)}{\phi(x-1/5)+\phi(4/5-x)} $$ in each stripe where $f$ increases from $n$ to $n+1$, where $\phi$ is the well-known smooth function $$ \phi(x) = \begin{cases} 0 & \text{ if } x \le 0 \, ,\\ \exp(-1/x)^2 & \text{ if } x > 0 \, . \end{cases} $$ Note that $\frac{\partial f}{\partial x}$ is bounded independently of $n$.)

The resulting function $f$ has the property that $\frac{\partial f}{\partial x}$ is bounded, and $\frac{\partial f}{\partial y} = 0$ everywhere. But for any sequence $(z_n)$ converging to a point $(x, 0)$, $0 \le x \le 1$, the sequence $f(z_n)$ is unbounded.

edited Sep 10 '22 at 21:12

answered Sep 10 '22 at 19:44

Martin R

128,226

Are you sure that $[0,1]^{2}$ is connected when you remove those line segments?? – Sep 10 '22 at 20:08
@GeorgeTsoutsinos: Yes, I am. Any two points in $X$ can be connected with a (polygonal) path. – Martin R Sep 10 '22 at 20:10
Are you sure that your set is open as the question says? I don't think so! – Sep 10 '22 at 20:27
OK ! Then find me an open neighborhood of $(\dfrac{1}{2},1)$ which is contained in the set! – Sep 10 '22 at 20:31
@GeorgeTsoutsinos: OK, that was a typo. One has to start with the open square. Thanks for letting me know. – Martin R Sep 10 '22 at 20:33
OK! But I would like a more rigorous definition of "In each of the remaining stripes let f smoothly increase from n to n+1 with bounded derivative in the x direction and constant in the y direction". Why don't produce a specific example? – Sep 10 '22 at 20:45
@GeorgeTsoutsinos: See updated answer. – Martin R Sep 10 '22 at 21:13
@Martin R: Thank you. Independent of whether the details are correct (no offense, trying to verify them would take me too far afield) your example is what I "feared". You constructed an infinite maze in which the path has to switch back and forth infinitely often. I will mark this as the answer because you did address my question. – user167131 Sep 10 '22 at 22:32

Is a smooth function with bounded partial derivatives defined on an open set in Euclidean space bounded?

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