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Honestly this is just a case of me not being great at constructing 'pathological' differentiable functions and I am not sure how to search for what I am looking for. This is related to an algebraic method of constructing cotangents that I have seen before, but it has been a long time so I don't remember the details.

Let $D$ be the set of functions $f$ on $\Bbb{R}$ which are differentiable at $0$ and have $f(0)=0$.

Suppose $f_0\in D$ has $f_0'(0)=0$.

Let $D\cdot D$ be the set of finite sums of products of elements of $D$. (i.e. the product ideal)

Is it the case that $f_0\in D\cdot D$? If not is there a counterexample?

It is easy to show that $f_0\in C\cdot D$ where $C$ is the set of continuous functions $f$ with $f(0)=0$ but I can't see if it is true in the restricted case.

Any help or references would be appreciated

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  • $\begingroup$ Not sure about sums of products of elements in $D$, but $x\mapsto x^{\frac{3}{2}}$ cannot be written as $g(x)h(x)$ with $g(0)=h(0)=g'(0)=h'(0)=0$. $\endgroup$
    – AnCar
    Apr 11, 2022 at 2:52
  • $\begingroup$ @AnCar Sorry, I probably wasn't clear enough. $g$ and $h$ can have non-zero derivative. In fact I have shown that if $f_0$ is twice differentiable then $f_0\in D\cdot D$. For example for $x^2$ we can write $x^2=x\cdot x$ $\endgroup$
    – Fishbane
    Apr 11, 2022 at 2:52
  • $\begingroup$ Sorry, $x^2$ was bad, corrected. $\endgroup$
    – AnCar
    Apr 11, 2022 at 2:55
  • $\begingroup$ Yes $x^{\frac{3}{2}}$ might be impossible, although it would be nice to have a proof. I'm not sure how to approach it though. $\endgroup$
    – Fishbane
    Apr 11, 2022 at 2:57
  • $\begingroup$ Well, fix some $\varepsilon>0$. On a nbh of $0$, $|g(x)|< \varepsilon |x|$ and the same for $h$ if $g'(0)=h'(0)=g(0)=h(0)=0$. If $x^{\frac{3}{2}}=g(x)h(x)$, then $x^{\frac{3}{2}}<\varepsilon^2 x^2$, contradiction. $\endgroup$
    – AnCar
    Apr 11, 2022 at 3:00

2 Answers 2

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Every function in $D$ satisfies $\limsup_{x \to 0} f(x)/x <\infty$, so every function $$h(x)=\sum_{i=1}^n f_i g_i$$ in $D \cdot D$ must satisfy $$\limsup_{x \to 0} h(x)/x^2 <\infty \,,$$ so in particular the function $x^{3/2}$ proposed by Ancar is not in $ D \cdot D$.

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  • $\begingroup$ That's a lovely clean argument. $\endgroup$
    – AnCar
    Apr 11, 2022 at 3:49
  • $\begingroup$ Thank you, this perfectly answers the question. Also it suggests that $f_0$ being twice differentiable might be a necessary and sufficient condition, although that is a completely different question. $\endgroup$
    – Fishbane
    Apr 11, 2022 at 4:04
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    $\begingroup$ Since all the individual limits exist, you can write $\lim_{x\to 0} \frac{h(x)}{x^2}=\sum_{i=1}^n \lim_{x\to 0}\frac{f_i(x)}{x}\lim_{x \to 0}\frac{g_i(x)}{x}$, meaning that $\lim_{x\to 0}\frac{h(x)}{x^2}$ does indeed exist. $\endgroup$
    – AnCar
    Apr 11, 2022 at 4:33
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    $\begingroup$ @AnCar Thanks, that looks correct. Thank you for all the help with this. $\endgroup$
    – Fishbane
    Apr 11, 2022 at 4:37
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    $\begingroup$ Indeed, limsups can be replaced by limits above. But note that $x^3 \sin(1/x^2)=x \cdot x^2 \sin(1/x^2)$ (defined to be zero at $0$) is in $D \cdot D$, but its derivative is not continuous at $0$. $\endgroup$ Apr 11, 2022 at 4:47
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Sorry about the confusion earlier. This should work assuming you are only interested in the question can $f_0$ be $gh$ with $g,h\in D$.

Take $f_0(x)=|x|\cdot x$. Clearly $f_0\in D$ and $\lim_{x\to 0} \frac{f_0(x)}{x}=\lim_{x\to 0}|x|=0$. Now assume $f_0=gh$ for some $g,h\in D$.

Fix $\varepsilon>0$ such that $2\varepsilon(|g'(0)|+|h'(0)|+2\varepsilon)<1$. Then on some small interval $(-\delta,\delta)$ we must have that $\frac{g(x)}{x}=g'(0)+\alpha(x)$ with $|\alpha(x)|<\varepsilon$ and similarly $\frac{h(x)}{x}=h'(0)+\beta(x)$ with $|\beta(x)|<\varepsilon$.

Multiplying, we get that $\frac{|x|}{x}=g'(0)h'(0)+\alpha(x)h'(0)+\beta(x)g'(0)+\alpha(x)\beta(x)$ for all $x\in(-\delta,\delta)$. But note that the LHS can take the values $\pm 1$ while the range of values on the RHS is less than $1$ by choice of $\varepsilon$, contradiction.

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