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How to show that given a function $f(x)$ in the Schwartz space, $\left| x \right|f\left( x \right)$ is bounded

Edit: we are operating in the space $\mathbb R^n$

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    Can you write the definition of Schwartz space;this should follow almost immediately from the definitions – peek-a-boo Oct 23 '21 at 17:00
  • $S\left( {{\mathbb R^n}} \right) = \left{ {f \in {C^\infty }\left( {{R^n}} \right);\left| {{x^i}{\partial ^j}f\left( x \right)} \right|{\rm{ bounded , for , all , i}}{\rm{,j}}} \right}$ –  Oct 23 '21 at 17:05

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The Schwartz space is the space of rapidly decreasing functions, then if $f$ belong to Schwartz space, $$\displaystyle\lim_{|x|\to\infty}|x|^kD^\beta f(x)< \infty \mbox{ for all } k\in \mathbb N \mbox{ and } \beta\in\mathbb{N}^n.$$ In particular, if $k=1$ and $\beta = (0,0, \cdots,0)$, $$\displaystyle\lim_{|x|\to\infty}|x|f(x)=L< \infty.$$ This means that, given $\varepsilon>0$, exist $A>0$ such that $|x|>A$, $|x||f(x)|<\varepsilon+L$. It remains to be seen what happens to $|x|f(x)$ when $|x|\leq A$, but of course this remains bounded as it is a continuous function in the compact $|x|\leq A.$

Ilovemath
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  • I think instead of limit the supremum should be used. –  Oct 23 '21 at 17:11
  • see this proposition https://math.stackexchange.com/questions/989703/formal-proof-that-schwartz-space-is-space-of-rapidly-decreasing-functions, actually there are some equivalences – Ilovemath Oct 23 '21 at 17:12
  • is the definition of ${D^\beta }f\left( x \right) = \frac{{{\partial ^{{\beta _1}}}}}{{\partial x_1^{{\beta _1}}}} \cdots \frac{{{\partial ^{{\beta _n}}}}}{{\partial x_n^{{\beta _n}}}}f\left( x \right)$ correct, given that $\beta = \left[ {{\beta _1} \cdots {\beta _n}} \right]$ –  Oct 23 '21 at 17:32
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    yes, this is the definition. – Ilovemath Oct 23 '21 at 23:21