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How would you tackle the following expression:

$$\phi(x)^{(1+o(1))}=1$$

where $\phi:\mathbb{R}\rightarrow\mathbb{R}^+$ is continuous? $o(1)$ relates to $x\rightarrow\infty$. The specific function is known and I would like to prove the validity of the above expression for $\phi$. I am not sure how to proceed. What is a first step?

Might it be admissible to logarithmize the above Landau relation?

What is, after all, the correct interpretation for the expression $\phi(x)^{(1+o(1))}=1$?

user823
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  • Is it $o_{x \rightarrow 0}(1)$ or $o_{x \rightarrow \infty }(1) $or else? – Axel May 19 '21 at 14:30
  • It is $o_{x\rightarrow\infty}(1)$. – user823 May 19 '21 at 14:34
  • Do you have other information on $\phi$ is it positive? continuous? – Axel May 19 '21 at 14:43
  • It is continuous and strictly positive. – user823 May 19 '21 at 14:44
  • @vitamin d: there is a slight difference betweeen you and Axel. Axel says there is a minimal threshold for $x$ after which $\phi(x)=1$. – user823 May 20 '21 at 06:56
  • @Tintin I don't think both answers are correct. – Axel May 20 '21 at 14:14
  • On the interpretation of the expression $\phi(x)^{1+o(1)} = 1$, or more specifically, how to interpret an equation where big O/little o notation appears not just on the right hand side, here's a useful post with a few answers that explain it. – Elliot Yu May 20 '21 at 16:44
  • Note that (1) you need to quantify over all functions that satisfy the big O/little o's on the left hand side, but only require existence of some functions that satisfy the big O/little o's on the right hand side (of course in this particular case there's no big O/little o on the right); (2) the equal sign in this context is not symmetric, i.e. $\phi(x)^{1+o(1)} = 1$ isn't the same as $1 = \phi(x)^{1+o(1)}$. – Elliot Yu May 20 '21 at 16:52
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    Since for all $x$ sufficiently large, $|o(1)|\lt1$, an interpretation of the equation given above would be: for all $x$ sufficiently large, $\phi(x)=1$. – robjohn May 20 '21 at 19:08
  • @Tintin: An expression with Landau symbols on the left-hand side without any on the right-hand side looks weird and seems to be rather a typo. Could you add a reference? – Markus Scheuer Jun 04 '21 at 17:59

2 Answers2

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Two Interpretations

I discussed this with vitamin d in chat and mentioned that the little-o on the left-hand side of an "equation" was odd, but extrapolating from its meaning on the right, $f=o(g)$ would be more accurately be stated as $$ f\in o(g)=\left\{h:\lim_{x\to\infty}\frac{h(x)}{g(x)}=0\right\} $$

However, Elliot Yu commented below that in this answer it is said that when $o(g)$ appears on the left and $o(h)$ appears on the right, that $o(f)+o(g)=o(h)$ represents an inclusion of classes: $$ o(f)+o(g)\subset o(h) $$ that is, any element of $o(f)$ plus any element of $o(g)$ is an element of $o(h)$.

  1. However, in the case of this question, $o(1)$ appears only on one side of the "equation", so it is not clear whether this is meant to be a comparison of classes, or to say that $1\in\phi(x)^{1+o(1)}$ (which is what my answer below had assumed).

  2. If we do use the class comparison interpretation we would have $$ \phi(x)^{1+o(1)}\subset\{1\} $$ which would say that for any $u\in o(1)$, $$ \phi(x)^{1+u(x)}=1 $$ which implies that $\phi(x)=1$ for all $x\in\mathbb{R}$.

Interpretation

The proper interpretation of $$ \phi(x)^{1+o(1)}=1\tag1 $$ is that there exists a $u(x)$ so that $\lim\limits_{x\to\infty}u(x)=0$ and $$ \phi(x)^{1+u(x)}=1\tag2\\[6pt] $$

Conclusion

$\lim\limits_{x\to\infty}u(x)=0$ guarantees that there is an $x_0$ so that $$ \begin{align} x\gt x_0 &\implies|u(x)|\lt1\tag{3a}\\[6pt] &\implies1+u(x)\gt0\tag{3b} \end{align} $$ Therefore, $(2)$ and $\text{(3b)}$ say that $$ x\gt x_0\implies\phi(x)=1\tag4\\[6pt] $$ Thus, if for $x$ sufficiently large, $\phi(x)=1$, your function satisfies $(1)$.

robjohn
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  • $(+1)$ No worries. I am glad you posted also an answer. Thank you for providing a clear explanation. – Axel May 20 '21 at 21:13
  • (-1) I disagree with the interpretation here. According to this answer, as well as a few other answers under that same post, the interpretation of $\phi(x)^{1+o(1)} = 1$ should be that for every $u(x)$ such that $\lim_{x\to\infty} u(x) = 0$, we have the equality $\phi(x)^{1+u(x)} = 1$. – Elliot Yu May 20 '21 at 23:17
  • That doesn't work with my understanding: $f(x)=o(1)$ would then say that for every $u(x)$ such that $\lim\limits_{x\to\infty}u(x)=0$, we have the equality $f(x)=u(x)$. However, the actual meaning is there is a $u(x)$ such that $\lim\limits_{x\to\infty}u(x)=0$ and we have the equality $f(x)=u(x)$. – robjohn May 21 '21 at 00:05
  • The point is that big O and little o notation mean different things on the left vs on the right of the equal sign. On the right hand side, it implies that there exists a function that satisfies the notation, but on the left it should quantify over all such functions. Please take a look at the linked answer in my previous comment. Multiple answers there indicate the same understanding, so I'm reasonably confident that this is an accepted interpretation. – Elliot Yu May 21 '21 at 00:38
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It is how I understand your problem, I may be wrong so let me know if it is the case :

You can rewrite your $o(1)$ as a function $\varepsilon$ with $\varepsilon(x) \rightarrow0$ when $x\rightarrow \infty$.

It implies that you can find $x_0 \in \mathbb{R}$ such that :

$$\forall x\geq x_0, |\varepsilon(x)| \leq 1/2$$

Therefore,

$$\forall x\geq x_0, \quad 1/2 \leq 1+\varepsilon(x) \leq 3/2$$

Hence,

$$\forall x \geq x_0, \varphi(x)^{1+\varepsilon(x)} = 1$$

And as $\varphi$ is strictly positive then,

$$\forall x\geq x_0,(1+\varepsilon(x)) \ln \varphi(x) =0$$

But $\forall x \geq x_0, 1+\varepsilon(x)\geq1/2$ therefore:

$$\forall x\geq x_0, \ln \varphi(x) = 0 \quad \mathrm{i.e.} \quad \varphi(x) =1$$

To conclude, your expression for me means that for $x$ large enough, $\varphi(x) = 1$.

Axel
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  • Are you concluding that $\phi(x)=1$, so a constant, for $x\geq x_0$? – user823 May 19 '21 at 15:04
  • @Tintin Yes I am. – Axel May 19 '21 at 15:05
  • (+1) I hope my answer does not overlap yours too much, but I thought that certain points could benefit from a bit more emphasis. – robjohn May 20 '21 at 21:09
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    Note that this may not be the proper interpretation of $o(1)$ when on the left side of the equation. See the comments by Elliot Yu on my answer and the section I added on "Two Interpretations". – robjohn May 21 '21 at 08:49