Continuity - "Bump Function"

Question

Question:

Fix $\delta$ > 0. If $H$ is a convex, compact subset of $\mathbb{R}^n$, show that $\partial H = H \setminus \operatorname{Int}(H)$ is compact and

$\varphi(x) = \left\lbrace \begin{array}{ll} \min\{\frac{1}{\delta}d(x,\partial H),1\}, & x \in H \\ 0, & x \not\in H \end{array}\right.$

is a continuous function that vanishes on $\partial H$.

What I did: The compactness of $\partial H$ is easily proven by showing that it's closed since it's trivially bounded as $H$ is bounded. The fact that $\phi$ vanishes at $\partial H$ comes from the definition of $\phi$ and its continuity.

What I have not been able to prove though is its continuity. I'm unsure how the convexity of $H$ will play a role in the proof; I think it will involve $\phi(tx+(1-t)y)=\min\{\frac{1}{\delta}d(tx+(1-t)y,\partial H),1\}$ for $t\in[0,1]$. The idea I have in my mind seems to not make use of convexity: If you take a cover of $H$ made of balls $\{B_\delta(x):x\in H\}$ and take $x,y\in H$ s.t $d(x,y)<\frac{\delta}{4}$, we can guarantee that the $x$ and $y$ are at least in neighbouring balls and therefore that $|d(x,\partial H)-d(y,\partial H)|\leq \delta$($*$).

But I'm unsure how to proceed from here and I would guess that $*$ doesn't actually follow from what we have.

Answer 1

Convexity does not play a role here: distance to a closed set is always (Lipschitz) continuous: here. Together with the fact that minimum of two continuous function is continuous (see here), one sees that

$$f(x) :=\min\{ \frac{1}{\delta } d(\cdot, \partial H), 1\}$$

is continuous on $\mathbb R^n$. Lastly, since $g(x) = 0$ is a continuous function on $\mathbb R^n\setminus \operatorname{Int} H$ and $g = f$ on $\partial H$, we see that $\varphi$ is continuous, say by the pasting lemma.

Answer 2

Convexity is not needed. The function $x \to d(x,\partial H)$ is a continuous function and minimum of two continuous functions is continuous. So the only question is what happens when a sequence $\{x_n\}$ in $H$ converges to a point $x$ not in $H$. Clearly, $x \in \overset {-}H$ so $x$ must be on the boundary. It follows that $\phi (x_n) \to 0 =\phi (x)$.