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couldnt find the answer for the following question. Maybe someone could help:

Let $C^{k}_{0}(\overline{D})$ denote the space of functions $f$ in $C^{k}(\overline{D})$ with $f=0$ on $\partial D$ . If $f \in C^{2}_{0}(\overline{D})$ then $|f| \leq C dist(x,\partial D)$. The same work for $|\nabla f|$. Why does this result hold? Does it work for function in $C^{1}_{0}(\overline{D})$? Thanks in advance!!

Micha
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1 Answers1

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Let $f \in C^1_0(\overline{D})$. I am assuming that $D$ is an open bounded subset of $\mathbb{R}^n$, so that there exists a constant $C>0$ such that $|\nabla f(x)| \leq C$ for every $x\in \overline{D}$.

Given $x\in D$, let $y\in\partial D$ be a projection of $x$ on $\partial D$ (i.e., a point such that $\text{dist}(x, \partial D) = |x-y|$).

Consider the restriction $\varphi(t) := f(y + t(x-y))$, $t\in [0,1]$. We have that $$ \varphi(0) = f(y) = 0, \quad \varphi(1) = f(x), \quad \varphi'(t) = \nabla f(y+t(x-y)) \cdot (x-y), $$ hence $$ |f(x)| = |\varphi(1) - \varphi(0)| = \left| \int_0^1 \varphi'(t)\, dt \right| \leq \int_0^1 \left|\nabla f(y+t(x-y)) \cdot (x-y)\right|\, dt \leq C |x-y| = C \, \text{dist}(x, \partial D). $$

A similar argument, if $f\in C^2_0$, gives the bound on $|\nabla f|$.

Rigel
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  • You believe you need to assume $D$ is convex for your argument to work in its current form. I think this is what the OP had in mind, anyways. – Nobody Aug 06 '17 at 14:56
  • I think that convexity is not needed. If $x$ and $y$ are as above, the segment $]y, x] := {y + t(x-y):\ t \in (0,1]}$ is contained in $D$. – Rigel Aug 06 '17 at 14:58
  • I stand corrected :-) – Nobody Aug 06 '17 at 15:15
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    The segment above is contained in $D$ even if $D$ is not convex (it follows from the definition of projection). – Rigel Aug 06 '17 at 15:42
  • Yes, I understand–the image must lie in $D$ for otherwise there exists a point with smaller distance to the boundary which is contradictory. As I mentioned, I stand corrected. – Nobody Aug 06 '17 at 15:43