The Schwartz space $\mathcal{S}$ is the space of smooth, infinitely differentiable and rapidly decreasing functions on $\mathbb{R^n.}$ Using a set builder we can rewrite $\mathcal{S}$ as: $$ \mathcal{S}(\mathbb{R^n})=\{f\in C^{\infty}(\mathbb{R^n}):||f||_{\alpha,\beta}<\infty \,, \forall \alpha, \beta\} \tag{1} $$ Where $\alpha,\beta$ are multi-indices and $$ ||f||_{\alpha,\beta}=\sup_{x\in \mathbb{R^n}} \left|x^\alpha D^\beta f(x)\right| \tag{2} $$


  1. How can one interpret intuitively the definition of the norm of $f$ given in $(2)?$ And how does it imply that such $f$'s (and all its derivatives) should be decaying?
  2. By "rapidly decreasing functions", is it hinted that the decay rate of such functions is exponential? (otherwise, "rapid" with respect to what?)
  3. Finally, how can one explain that any smooth function $f$ with compact support is in $\mathcal{S}(\mathbb{R^n})?$

1 Answer 1

  1. From (2) it follows that $$ |D^\beta f(x)|\le\frac{\|f\|_{\alpha,\beta}}{|x|^\alpha}. $$ $f$ and all its derivatives decay at $\infty$ faster than any negative power.

  2. The decay may not be exponential. Consider for instance $e^{-(1+x^2)^a}$ with $0<a<1/2$.

  3. If the support of $f$ is $K$, then for all multi-indices $\alpha,\beta$ $$ \sup_{x\in K}\bigl|x^\alpha D^\beta f(x)\bigr|<\infty. $$

  • $\begingroup$ What does $|x|^{\alpha}$ mean? $\endgroup$
    – Pedro
    Aug 18, 2016 at 16:58
  • $\begingroup$ $|x|$ is the euclidian norm of $x$. $\endgroup$ Aug 19, 2016 at 1:18

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