I'm a little bit unsure if I use the Big-O-notation in the following context correctly:

Consider a function $\varphi \in C^{\infty}(\overline M)$ on a compact manifold with boundary and a boundary defining function $x$, so we have $\partial M = \{ x=0\}$ and $x>0$ on the interior $M$. Assume that we can use coordinates $(x,y)$ to describe points on $\overline M$.

Now we're interested in the behaviour of $\varphi(x)$ for $x \rightarrow 0$.

Assume that $\varphi \in o(x)$ for $x \rightarrow 0$. By definition this means, that $\lim \limits_{x\rightarrow 0} |\frac{\varphi(x)}{x}| = 0$. This means, that if we write $\varphi(x) = x \cdot g(x)$ that $g$ still is zero when $x$ is zero. Is this correct so far?

If this is correct, then one could write (could one, asymptotically?) $g(x) = x \cdot h(x)$ where $h(0)$ may is zero but we don't know.

The crucial thing for me now is the following:

Can one then say that $\varphi \in O(x^2)$ as $x \rightarrow 0$ because

\begin{equation} \lim\limits_{x\rightarrow 0} |\frac{\varphi(x)}{x^2}| = |h(x)| \Rightarrow \lim\sup\limits_{x\rightarrow 0} |\frac{\varphi(x)}{x^2}| = |h(x)| < \infty \end{equation} And, in general: Is it correct to treat a smooth function $\varphi$ that satisfies $\varphi \in O(x^k)$ asymptotically like $x^k\cdot g(x)$ for another smooth function $g$? (for $x \rightarrow 0$)

Thanks for any help!


I may formulated this issue a little bit confusing. The main question above is: Does $\varphi \in o(x)$ imply that $\varphi \in O(x^2)$ when $x \rightarrow 0$?


Mathematicians usually write $\varphi(x) = o(x)$ as $x \to 0$ [and not some strange thing like $\varphi \in o(x)$].

For your question. If $\varphi(x) = O(x^2)$ then $\varphi(x) = x^2 h(x)$ with $\limsup |h(x)| < \infty$ . But there is no reason that $h(0)$ is even defined. Certainly it need not have a limit at $0$.

In your last question, what do you mean by "smooth"? Again, $g(0)$ need not be defined so if "smooth" refers to its behavior at $0$, then $g$ isn't.

added Nov 9

Question: $\varphi \in o(x)$ implies $\varphi \in O(x^2)$.

No, not in general. $\varphi(x) = x^{3/2}$ satisfies $\varphi(x) = o(x)$ but not $\varphi(x) = O(x^2)$.

If we are in a situation to apply Taylor's theorem for $\varphi$ at $0$, then maybe you can get the conclusion you want, however. This works as follows: Under the right conditions*, we have $$ \varphi(x) = \varphi(0)+\varphi'(0)\;x + \frac{\varphi'(0)}{2}\;x^2 + o(x^2) , $$ and from $\varphi(x) = o(x)$ we conclude $\varphi(0)=\varphi'(0)=0$, and therefore $\varphi(x) = O(x^2)$.

*The "right conditions" are the hypotheses of an appropriate case of Taylor's Theorem.

  • 2
    $\begingroup$ It's fairly common to write $\varphi\in o(x)$, especially if you dislike the non-commutative nature of $=$ in the notation $\varphi = o(x)$. $\endgroup$ – Rick Decker Nov 9 '12 at 3:31
  • $\begingroup$ Thank you for your answer. It is definitely correct to write $\varphi \in O(x^2)$, but think it as equality if you like it more. What is with my main Question: Does $\varphi \in o(x)$ imply $\varphi \in O(x^2)$? I did not say that the limit h(0) is well defined. I said "we don't know" but it may has more zeroes. Smooth refers to being an infinitely differentiable function on $\overline M$, so all points $p = (0, y) \in \partial M$ are included. $\endgroup$ – Robin Neumann Nov 9 '12 at 8:35
  • $\begingroup$ I wouldn't write $\varphi = o(x)$, either, but I would write $\varphi(x) = o(x)$. Or, more generally, I want to write things like $\sin x = x + o(x^2)$. $\endgroup$ – GEdgar Nov 9 '12 at 19:34
  • $\begingroup$ Thanks a lot for your edited answer, GEdgar! Taylor's theorem seems to be the right tool. In fact I believe it is valid in my case, I'm going to check this. And if one likes to express the error term with big-O-notation like in $f(x) = g(x) + O(x^2)$ I also would use it that way. But in general I slightly prefer to underline that $O(g(x))$ is related to a full set of functions. Thanks again! $\endgroup$ – Robin Neumann Nov 9 '12 at 19:46

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