Thread: differential topology question (manifold with boundary)

Let $x\in\partial X$ be a boundary point. Show that there exists a smooth nonnegative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ if and only if $z\in\partial U$, and if $z\in\partial U$, then $df_z(\vec{n}(z))>0$.

Note that in this context $\vec{n}$ is the outward unit normal.

I was assigned this for homework, but I'm pretty sure it's not possible. In a basic example, this is saying that there is an $f:\R\to\R$ such that $f(x)>0$ for all $x\in (0,\infty)$ and $f(0)=0$, and $f'(0)<0$ (since the unit normal is -1). But this isn't possible, for that says that as $x\to 0^+$, that f(x) is increasing, which clearly can't be the case for nonnegative functions.

Is my logic wrong somewhere? Any help would be appreciated, thanks!

Originally Posted by MattMan

Let $x\in\partial X$ be a boundary point. Show that there exists a smooth nonnegative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ if and only if $z\in\partial U$, and if $z\in\partial U$, then $df_z(\vec{n}(z))>0$.

Note that in this context $\vec{n}$ is the outward unit normal.

I was assigned this for homework, but I'm pretty sure it's not possible. In a basic example, this is saying that there is an $f:\R\to\R$ such that $f(x)>0$ for all $x\in (0,\infty)$ and $f(0)=0$, and $f'(0)<0$ (since the unit normal is -1). But this isn't possible, for that says that as $x\to 0^+$, that f(x) is increasing, which clearly can't be the case for nonnegative functions.

Is my logic wrong somewhere? Any help would be appreciated, thanks!

Hey man, we can't help with homework problems.

So don't help and just tell me if the problem is even doable. In fact don't even read the homework question just look at the second part.

Originally Posted by MattMan

So don't help and just tell me if the problem is even doable. In fact don't even read the homework question just look at the second part.

I would agree with your second part. Since and for some right-neighborhood of we have that is non-positive.

Originally Posted by MattMan

Let $x\in\partial X$ be a boundary point. Show that there exists a smooth nonnegative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ if and only if $z\in\partial U$, and if $z\in\partial U$, then $df_z(\vec{n}(z))>0$.

Note that in this context $\vec{n}$ is the outward unit normal.

I was assigned this for homework, but I'm pretty sure it's not possible. In a basic example, this is saying that there is an $f:\R\to\R$ such that $f(x)>0$ for all $x\in (0,\infty)$ and $f(0)=0$, and $f'(0)<0$ (since the unit normal is -1). But this isn't possible, for that says that as $x\to 0^+$, that f(x) is increasing, which clearly can't be the case for nonnegative functions.

Is my logic wrong somewhere? Any help would be appreciated, thanks!

Just one question: Why are you assuming on ? The problem only states that if and only if , so it could be negative on the half-line.

Originally Posted by MattMan

Show that there exists a smooth nonnegative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ if and only if $z\in\partial U$

The function needs to be nonnegative in a neighborhood of 0, as well as 0 at 0. It could be identically 0, there but then you run into the same problem.

Originally Posted by MattMan

The function needs to be nonnegative in a neighborhood of 0, as well as 0 at 0. It could be identically 0, there but then you run into the same problem.

Yes, you're right, but I think it may be a typo, since if you ask non-positive instead the problem is easily solvable in a half-space from where it shouldn't be much trouble to jump to the general case.

Thanks for the confirmation of my suspicions.