Landau symbol definition

In Uni, we learned about the Landau-symbol $$\mathcal{O}$$. But the definition confuses me.

$$f=\mathcal{O}(g)$$ for $$x\rightarrow x_0$$ if and only if there exists a constant $$c>0$$ and a neighbourhood $$U(x_0)$$, so that for every $$x \in U(x_o): |f(x)| \le c\cdot|g(x)|$$

Now I was thinking that this counts for almost every function. E.g. let's take $$f(x) = x^2$$ and $$f(x)=x$$.

Let's also choose a neighbourhood: $$U(x_0) = [-10,10]$$ for $$x_0=0$$. Obviously for any $$x\in U(x_0)$$, I could say that $$|x^2| \le c\cdot|x|$$ with $$c = 10$$. My problem is that for any neighbourhood, I could choose a constant c that is high enough to satisfy the criterion.

Isn't the criterion always satisfied simply if c is large enough?

We cannot have $$c$$ such that $$|x| \leq c|x^{2}|$$ for $$x$$ in some interval around $$0$$. This is because the inequality fails whenever $$0<|x| <\frac 1 c$$.

• Could you explain why it fails? Isn't $|x|≤c|x^2|$ true for a c big enough? – Finn Eggers Mar 29 '19 at 12:07
• @FinnEggers What is required is a $c$ and a $\delta$ such that the inequality holds for every $x$ with $|x| <\delta$. The constant $c$ is not allowed to vary with $x$; it is a constant. So you cannot increase $c$ based on what $x$ you pick. – Kavi Rama Murthy Mar 29 '19 at 12:12
• That makes sense – Finn Eggers Mar 29 '19 at 12:37
• But can c change with Delta? – Finn Eggers Mar 29 '19 at 12:39
• Yes, but whatever $c$ you choose there is always an $x$ for which the inequality is violated. – Kavi Rama Murthy Mar 29 '19 at 13:13

No. Let $$f(x)=x, g(x)=x^2$$ and $$x_0=0.$$ Suppose that there is $$c>0$$ and a neigborhood $$U$$ of $$x_0$$ such that $$|f(x)| \le c\cdot|g(x)|$$ for all $$x \in U$$. This would imply that

$$1 \le c|x|$$ for all $$x \in U$$ with $$x \ne 0.$$

But this is absurd.

• Could you explain with this is being implied? – Finn Eggers Mar 29 '19 at 12:07