How to show this inequality involving derivatives of a smooth function?

71 Views Asked by At

I must use the following:

Lemma: Let $f\in C^2([-1, 1], \mathbb R)$ and $\displaystyle C_j:=\max_{t\in [0, 1]}|f^{(j)}|$ with $j=0, 2$. Then $$|f^\prime(0)|\leq 4C_0(C_0+C_2).$$

to prove:

Proposition: Let $K_1$ and $K_2$ be two compact subsets of $\mathbb R^n$ such that $K_1\subset \textrm{int}(K_2)$. Let $W\subseteq \mathbb R^n$ be an open subset containing $K_2$ and let $f\in C^\infty(W)$. Then $$\displaystyle \left(\sup_{x\in K_1} \sum_{|\alpha|=1} |D^\alpha f(x)|\right)^2\leq C\sup_{x\in K_2} |f(x)|\left(\sup_{x\in K_2}|f(x)|+\sup_{x\in K_2} \sum_{|\alpha|=2} |D^\alpha f(x)|\right).$$

The author says it is an immediate consequence of the previous lemma but I can't see how. My questions are:

$(i)$ Is $W$ taken only for speaking of differentiablity of $f$, isn't it?

$(ii)$ What should I do for applying the lemma?

Any hints will be helpful.

Thanks.

Obs:

  1. The proof of the lemma follows from the mean value theorem.
1

There are 1 best solutions below

0
On

Yes, $W$ is mentioned only to state the differentiability assumption precisely.

Let $\delta=\operatorname{dist} (K_1,K_2^c)$; note that $\delta>0$. Fix $x\in K_1$ and $j\in \{1,\dots,n\}$. Let $g(t) = f(x+ t \delta e_j)$ where $e_j$ is the $j$th basis vector. Note that $g$ is defined and smooth on $[-1,1]$.

Applying the lemma to $g$ yields $$\delta |\partial_j f(x_0)|= |g'(0)| \leq 4\max_{[-1,1]} |g| (\max_{[-1,1]} |g|+\max_{[-1,1]} |g''|).$$ The right hand side is estimated by the analogous expression involving $f$, and the result follows.