Prove that , $e^{f(x)}-1$ is a schwartz class function when $f$ is a Schwartz function.

Question

Suppose $f$ is a Schwartz function, prove that $e^{f(x)}-1$ is also a Schwartz function.

In order to prove the result, we have to show that $|(e^{f(x)}-1)^{(n)}|\leq \frac{C_{m,n}}{|x|^{m}}$ for any nonnegative integer numbers $m$ and $n$.

Now the nth derivative for $(e^{f(x)}-1)^{(n)}$ is a very complicate formula involving derivatives of $f(x)$ and $e^{f(x)}$. $e^{f}$ can be expanded into a series but it's not so useful for our purpose.

Answer 1

For $n=0$ you can use that $|e^{f(x)}-1|\le C\,|f(x)|$ for $|x|$ large enough and a constant $C>1$. For $n\ge1$, it is easy to check by induction that $$ \bigl(e^{f(x)}\bigr)^{(n)}=e^{f(x)}\bigl(\text{polinomial}(f',f'',\dots,f^{(n)})\bigr) $$ Since $e^f$ is bounded and all the derivatives of $f$ are Schwarz functions, the desired inequalities follow.

Answer 2

Defintion and Proposition: $$h\in \mathcal S(\mathbb R)\Longleftrightarrow h\in C^\infty(\mathbb R)~~ and~~ \lim_{|x|\to\infty} x^m h^{(n)}(x) = 0$$

Let $$g(x)= e^{f(x)}-1$$

Show that for any $n\in\mathbb N $

$$g^{(n)} (x)=P[f,f',\cdots,f^{(n)}](x)e^{f(x)} = P[f,f',\cdots,f^{(n)}](x)g(x) +P[f,f',\cdots,f^{(n)}](x).$$ Where $P[f,f',\cdots,f^{(n)}]$ is some polynomial of $(f,f',\cdots,f^{(n)})-$variables.

Remark 1. $f,h\in\mathcal S(\mathbb R)\implies f+h, f\cdot h\in\mathcal S(\mathbb R)$ $$f\in \mathcal S(\mathbb R)\implies \forall k ~~f^{(k)})\in \mathcal S(\mathbb R)\implies P[f,f',\cdots,f^{(n)}] \in \mathcal S(\mathbb R)$$ 1. $$f\in \mathcal S(\mathbb R)\implies \lim_{|x|\to\infty} f(x)=0 \implies\lim_{|x|\to\infty} e^{f(x)} =1 \implies f~\text{is bounded and } C^\infty.$$ 2. $$e^{f}~\text{is bounded and} ~C^\infty ~~\text{and}~P[f,f',\cdots,f^{(n)}]\in \mathcal S(\mathbb R)\\ \implies \lim_{|x|\to\infty} x^mg^{(n)} =\lim_{|x|\to\infty} x^m P[f,f',\cdots,f^{(n)}](x)e^{f}=0$$ This show that $g = e^{f}-1 \in \mathcal S(\mathbb R)$ since it is $C^\infty$

3. List item

   $$\lim_{|x|\to\infty} x^mg^{(n)} =\lim_{|x|\to\infty}=0 ~~\text{and $g^{(n)}$ is continuous }~~\implies x^mg^{(n)} ~~\text{ is bounded. }$$ That is $$|(e^{f(x)}-1)^{(n)}|\leq \frac{C_{m,n}}{|x|^{m}}.$$