-1

Consider the bump function $f \in {\mathcal C}^\infty_c(]-1, 1[)$:

$$f(x) = \exp \left( \frac {1}{x^2-1} \right) $$

There is a canonical way to extend it so that the extension $\tilde f \in {\mathcal C}^\infty_c(\mathbb R)$.

In general,

the compactness of function support can always be preserved under extension.

Question. Why isn't this true also for the restriction?

Using the previous example, why isn't $ f \big\vert_{\left [-\frac 1 2 , \frac 1 2 \right ]} $ compactly supported? Isn't the interval $\left [- \frac 1 2, \frac 1 2 \right ]$ compact?

ric.san
  • 121
  • 1
    Yes it is. Any continuous function defined on a compact set has compact support. – Arctic Char Jul 05 '21 at 13:49
  • 2
    Possibly you're misinterpreting something. Say $f$ is the extension of $\exp(1/(x^2-1))$ that vanishes outside $[-1,1]$. Then it's certainly true that $f$ has compact support. But $f\ne C^\infty_c((-1,1))$ because the support of $f$ is not a compact subset of $(_1,1)$. – David C. Ullrich Jul 05 '21 at 14:03

1 Answers1

1

First of all, from what you quoted, it is not claimed that restriction would not preserves the property. Your example is one instance. However, it is uncommon, or moot, to talk about compact support when the domain is already compact: in this case every continuous functions defined on a compact set has compact support.

Back to your text. First, there's a mistake: your function $f$, defined on $(-1, 1)$, is not of compact support. But there is a canonical way to extend $f$ to a function on $\mathbb R$ which are both smooth and of compact support: just by extending by $0$. To show smoothness, some works is required, but once it is done, it's clear that the support is $[-1, 1]$ which is compact.

In general, if $U, V$ are two open sets in $\mathbb R^n$ and $U\subset V$, there is a canonical extension

$$ E : C^\infty_c(U) \to C^\infty_c(V)$$ just by extending by $0$. In this situation, the property of being compact support is preserved (proof is also easy, but is not applicable to the $f$ in your post).

Arctic Char
  • 16,007