If you have a smooth function, the first order taylor remainder (with the polynomial part divided out ) is also smooth

Question

This is from the text "Distributions and Operators" by Gerb Grubb where the passage is taken from the section about the Cauchy Principal value:

The text claims $\varphi_1(x)$ is smooth but I'm having some problem seeing this. May I ask how it is shown that we have continuity and smoothness at $x=0$?

Also is it true that the higher order taylor remainder (without the polnomial part) are smooth functions?

Please do not just ask. Edit your post to include your attempts to prove this. (At the very least your final question...) — Anne Bauval, Apr 24 '23 at 06:23
The main question looks very much like a FAQ. Look for duplicates before asking (and before answering, @Jose27). — Anne Bauval, Apr 24 '23 at 06:25
Just asking a question is frowned upon: you need a Certificate Of Time Spent. Only suffering makes you worthy of answers here. — Mariano Suárez-Álvarez, Apr 24 '23 at 06:32
And of course one also cannot simply answer questions, either. You have to walk the walk first! Let no one take this place for one where people ask questions and answer them — hopefully someone sets up such a site, it would be even be fun. — Mariano Suárez-Álvarez, Apr 24 '23 at 06:33
This answers your question If $f(x)$ is smooth and odd, must $f(x)/x$ be smooth? easy-to-predict multiduplicate, found using approach0 https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24%5Cfrac%7Bf%5Cleft(x%5Cright)-f%5Cleft(0%5Cright)%7D%7Bx%7D%24&p=1 — Anne Bauval, Apr 24 '23 at 15:59
Of course I meant you main (FA)Q. As for your final additional question, I first understood it like the answerer (whether $f−P$ is smooth when $f$ and $P$ are), but you probably asked a less trivial question (not solved below although you accepted the answer), which is however also a duplicate: If $f$ is smooth and we write with Taylor's formula $f(x)=f(0)+f'(0)x+...+\frac{1}{k!}f^{(k)}(0)x^k+h(x)x^k$ is then also $h$ a smooth function? — Anne Bauval, Apr 24 '23 at 16:25
As for the rules for asking on this site, better look here: https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question than listen to sarcasms — Anne Bauval, Apr 24 '23 at 21:30

score 1 · Accepted Answer · answered Apr 24 '23 at 06:17

Continuity at 0 can be seen, for instance, from L'Hospital's rule and $\varphi_1(0)=\varphi'(0)$. As for differentiability, notice that $$ \varphi_1'(0):= \lim_{x\to 0} \dfrac{\varphi_1(x)-\varphi_1(0)}{x}= \lim_{x\to0} \dfrac{\varphi(x)-\varphi(0)-x\varphi'(0)}{x^2}, $$ and L'Hospital's applied twice (or Taylor approximation) give that $\varphi_1'(0)=\varphi''(0)/2$. From here, you can do an induction argument and use Taylor's theorem/L'Hospital's rule.

As for the Taylor remainder, notice that $f(x)=T_n(x)+R_n(x)$ for a smooth function $f$, where $T_n$ is the $n$-th order Taylor polynomial (centered at 0), and $R_n$ the remainder, then $R_n$ is the difference of smooth functions.

If you have a smooth function, the first order taylor remainder (with the polynomial part divided out ) is also smooth

1 Answers1