3
$\begingroup$

Question. Let $\Omega$ be an open subset of $\mathbb R^3$, and let $f: \Omega \to \mathbb R$ be continuous. Then $f$ is

(a) Lipschitz, if it is $C^1$ and $\nabla f$ is bounded;

(b) uniformly continuous, if $\Omega$ is a disk;

(c) Lipschitz, if $\Omega = \mathbb R^3$;

(d) uniformly continuous, if it is differentiable, $\nabla f$ is bounded and $\Omega$ is a half-plane.

I know (a) is not justified, since we need further information about $\Omega$ (needs to be convex). Also the function $f: \mathbb R^3 \to \mathbb R$ such that $f(x,y,z) = x^n + y^n + z^n$ with $n \geq 2$ in the naturals seems to work as a counterexample to (c).

The main points I don't understand, though, are the requirements on $\Omega$ we find in (b) and (c): if $\Omega \subseteq \mathbb R^3$, how can it be open if it is a disk or a half-plane – in general, a "flat", 2D subset? Wouldn't all points in $\Omega$ be boundary points then?

At any rate, a disk without its boundary circumference (that's what I think they could be meaning when they say "open") is not compact, so Heine's theorem does not apply and (c) is not necessarily true, so I'm thinking (d) by exclusion, although I don't know which theorems would directly justify picking it as an answer. Of course, proving it is Lipschitz in those conditions would automatically prove it's uniformly continuous. Any input is appreciated!

$\endgroup$
  • 1
    $\begingroup$ I concur with your points about "disc" and "half-plane". I suspect there is a typo and that it was meant to be $\mathbb R^2,$ not $\mathbb R^3$. $\endgroup$ – DanielWainfleet Jun 14 '17 at 1:16
  • $\begingroup$ That's what I originally thought, but seeing $\mathbb R^3$ appears not once, but twice in the text, I'd started wondering whether something was eluding me. $\endgroup$ – giobrach Jun 14 '17 at 1:20
  • $\begingroup$ Typos tend to propagate. When you edit your own work and you have an x on one line and on a later line you have a y, and you know they should be the same, you're likely to change the y to x, even if the x was a typo. $\endgroup$ – DanielWainfleet Jun 14 '17 at 1:41
0
$\begingroup$

regarding (a) this is certainly true if the space is $R^3$ but not in general. Just take the mean value theorem along a line between two points $x$ and $y.$ $$ |f(x) - f(y)| \leq \sup |\nabla f| |x-y| $$

see https://math.stackexchange.com/a/1913493/332594 for discussion of the non-convex case.

regarding (b) If it's continuous on the closure then it's uniformly continuous. If only on the interior then it's not true.

(c) take $f(x) = e^{x_1}.$

(d) you can argue using the mean value theorem as above, since the half-place certainly is convex.

$\endgroup$
  • $\begingroup$ (a) If the line over which to apply Lagrange's theorem is not completely included within $\Omega$, then there are points on that line where the function $f$ is not defined, so we need convexity. But yes, we agree that we do need more hypotheses on the structure of $\Omega$. (b) Indeed, the question does not mention any property of the extension of $f$ to the closure of $\Omega$, so I marked it false. (d) Can you expand on this hint? $\endgroup$ – giobrach Jun 14 '17 at 0:13
  • $\begingroup$ See this answer for the convexity hypothesis: math.stackexchange.com/a/1913493/332594 $\endgroup$ – giobrach Jun 14 '17 at 0:15
  • $\begingroup$ Let $\Omega=\Bbb R^3\setminus\{x_1\ge 0, x_2=0\}$. Let $f(x_1,x_2,x_3)=0$ if $x_1<0$ and $=x_1$ if $x_1\ge 0, x_2>0$ and $=-x_1$ if $x_1\ge 0, x_2\le 0$. Then $f$ has "jumps" along the "fault plane" $\{x_1\ge 0,x_2=0\}$. $\endgroup$ – Vim Jun 14 '17 at 1:04
  • $\begingroup$ I have update the answer $\endgroup$ – Mark Joshi Jun 14 '17 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.