Let $\{U_i\}_{i\in I}$ be a locally finite collection of open subsets of $\mathbb{R}^n$, $K_i\subseteq U_i$ compact subsets, $\epsilon_i>0$ positive real numbers and a nonnegative natural number $k$.

Let now $K$ be another compact subset of $\mathbb{R}^n$. Since the collection is locally finite, $K$ intersects only finitely many of the $U_i$ nontrivially, let's say these are $U_1,...,U_n$.

Let $f\colon\mathbb{R}^n\rightarrow\mathbb{R}$ be a smooth function (all partial derivatives of all order exists and are differntiable) such that $\frac{\partial^{|\alpha|}}{\partial x^\alpha}f(x)<\epsilon_i$ for all $x\in K_i$, $i\in\{1,...,n\}$ and all multiindices $\alpha$ of order $|\alpha|\le k$, e.g. all partial derivatives of $f$ up to order $k$ are $\epsilon_i$-bounded in the compact subsets, which might intersect $K$.

Is there a smooth function $f'\colon\mathbb{R}^n\rightarrow\mathbb{R}$ for which the condition above does hold for all $i\in I$ and which agrees with $f$ on $K$?

My attempt was the following:

Choose a bump function $\delta\colon \mathbb{R}^n\rightarrow (0,\infty)$ which is $1$ on a neighborhood of $K$ and $0$ outside a small compact subset $L$ which includes $K$ and try $f'=\delta f$, but that seemed not to work out since a couldn't control the partial derivatives of the bump function in the area where it goes from $1$ to $0$.

  • $\begingroup$ I don't understand. Isn't $f$ already smooth? $\endgroup$ – Pedro Apr 7 '15 at 1:43
  • $\begingroup$ It is, but the conditions on the derivative holds only for $i\in\{1,..,n\}$ and for $f'$ it should hold for all $i\in I$ $\endgroup$ – Tom Apr 7 '15 at 12:08

No, this doesn't work in the stated generality. Let's take $n=1$ (one dimension) for simplicity, $k=2$, and $f(x)=x$. Consider these compact sets and corresponding $\epsilon$s:

  • $K_1=[0,1]$, $\epsilon_1=2$, $U_1=(-0.01,1.01)$
  • $K_2=[1,2]$, $\epsilon_2 =0.01$, $U_2=(0.99, 2.01)$.

Let $K=[0,0.98]$. This set intersects only $U_1$. Clearly, $f$, $f'$ and $f''$ are bounded by $\epsilon_1$ on $K_1$.

For the extension to satisfy the condition on $K_2$, it needs to have very small first derivative, among other things. But for the first derivative to drop from $1$ to $0.01$ within the interval $[0.98,1]$, the second derivative must be large, $\approx 50$. This would violate the restriction imposed by $\epsilon_1$ on $K_1$.


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.