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: \begin{equation} d\mu(x)=\frac{1}{\sqrt{2\pi}} e^{-x^2/2} dx. \end{equation}

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?


1 Answer 1


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.
  • $\begingroup$ 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. $\endgroup$ Jun 23, 2023 at 0:18

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