# If a smooth function $f$ satisfies $\lvert f(x) \rvert \leq \lvert x \rvert$, all its derivatives are integrable with respect to Gaussian measure?

The question is as in the title.

Let $$f$$ be a smooth function on the real line satisfying $$\lvert f(x) \rvert \leq \lvert x \rvert$$.

Then, it is clear that $$f$$ itself is integrable with respect to the standard Gaussian measure: $$$$d\mu(x)=\frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx.$$$$

However, I am curious about its derivatives, since I would like to perform integration by parts for this function $$f$$. Could anyone please clarify for me?

Note that $$|f(x)| \leq |x|$$ doesn't really tell us much about $$f'(x)$$ anywhere except $$x=0$$, where we must have $$|f'(0)| \leq 1$$.
For instance, if we consider $$\phi_1, \phi_2$$ as a partition of unity subordinate to the ordered open cover $$(-2,2),\, \mathbb{R}\setminus [-1,1]$$, then $$f(x) = \phi_2(x)\sin(e^{x^2})$$ satisfies
• $$f \in C^\infty(\mathbb{R})$$
• $$|f(x)| = 0$$ when $$|x| \leq 1$$
• $$|f(x)| \leq 1$$
• $$|f'(x)|\,d\mu = Cxe^{x^2/2}\cdot|\cos(e^{x^2})|\,dx$$ when $$x > 2$$, which isn't integrable.
• Note that if you don't know about smooth partitions of unity, you could just consider $f(x) = x\sin(e^{x^2})$ to reach the same conclusion, just with a slightly messier derivative. Jun 23, 2023 at 0:18