I've come across two definitions for what it means for a function to be in $\mathfrak S$, the Schwartz space.
- A function $f \in \mathfrak S$ if $f \in C^\infty$ and for all $j, k \geq 0$ integers, $\displaystyle \lim_{\vert x \vert \to \infty} \bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert = 0$.
- A function $f \in \mathfrak S$ if $f \in C^\infty$ and for all $j, k \geq 0$ integers, there exists $C_{j, k}$ a constant, such that $\displaystyle \bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert \leq C_{j, k}$.
I've been able to show that $(1 \Rightarrow 2)$, but not the other way around.
To show that $(1 \Rightarrow 2)$, suppose that for all $j, k \geq 0$ integers with $\displaystyle \lim_{\vert x \vert \to \infty} \bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert = 0$. Then for all $\varepsilon > 0$, there exists $M$ such that if $\vert x \vert > M$, then $\displaystyle \bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert < \varepsilon$. In particular, $\displaystyle \bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert $ is continuous on $[-M, M]$ and hence achieves it's maximum, call it $C$. Define $$\displaystyle C_{j, k} = \max \bigg \{ \varepsilon = 1, \max_{x \in [-M, M]}\bigg \vert x^k \bigg( \frac{d}{dx} \bigg)^j f(x) \bigg \vert \bigg \}.$$
Can someone help me out with the converse? Is it even true?
Given $j$, $k$ positive integers there is a constant $C_{j,k+1}$ such that $$ \Bigl| x^{k+1} \bigl( \frac{d}{dx} \bigr)^j f(x) \bigr|\leq C_{j+1, k}\implies \Bigl| x^{k} \bigl( \frac{d}{dx} \bigr)^j f(x) \bigr|\leq \frac{C_{j, k+1}}{x}. $$