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A Schwartz function need not have compact support

Statement

The Schwartz function $g (x) = e^{- π ∣ x ∣^{2}}$ has support all of $R^{n}$ , so a Schwartz function need not have compact support. This example is choice-free.

Facts & Assumptions

Given: An integer $n \geq 1$ .

[F1]

Polynomial Gaussians with positive parameter are Schwartz (Polynomial Gaussians are Schwartz).

[F2]

Support is the closure of the nonzero set, and compact support defines $C_{c}^{\infty}$ (The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ ).

Counterexample

1.1

Taking $P = 1, t = 1$ in [F1] gives $g \in S$ . Positivity of the real exponential gives $g (x) > 0$ for every $x$ , so [F2] gives $supp g = \overline{R^{n}} = R^{n}$ .

F1F2

2.1

The increasing open balls $B (0, k)$ , $k = 1, 2, \dots$ , cover this support and have no finite subcover: a finite union lies in the largest of those balls and misses a point further along the first coordinate axis. Thus the support is not compact, while step 1.1 supplies the Schwartz hypothesis.

step 1.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Topics in Real and Functional Analysis (2017) (standard reference, not scraped)