-1

Let $f(x):=x$ and $g(x):=x^{2}$, then we say that $f=\mathcal{O}(g)$ if there exists $\alpha,x_{0}>0$ such that: $$ |f(x)|\leqslant\alpha|g(x)| $$ for all $x>x_{0}$

My question is if now $f,g\colon \mathbb{R}^{n}\to\mathbb{R}^{n}$, defined by $f(\mathbf{x}):=\mathbf{x}$ and $g(\mathbf{x}):=\mathbf{x}^{2}$ (element-wise squaring) then can we say that:

$f=\mathcal{O}(g)$ if there exists $\alpha>0$ and $\mathbf{x}_{0}$ such that: $$\|f\|_{p}\leqslant \alpha\|g\|_{p}$$ for all $\|\mathbf{x}\|_{p}>\|\mathbf{x}_{0}\|_{p}$?

Smylic
  • 8,098
SPARSE
  • 492
  • You can say $f=\mathcal{O}(|g|_p)$. – Kavi Rama Murthy May 24 '22 at 07:26
  • 1
    I don't understand the question. You are defining a new non-standard variant of the $,f=\mathcal{O}(g),$ notation which is now defined for multivariable functions. What do you want to know about it? – Somos May 27 '22 at 22:20
  • Seems like a duplicate of https://math.stackexchange.com/q/96809/620957. See in particluar https://en.wikipedia.org/wiki/Big_O_notation#Multiple_variables and https://people.cis.ksu.edu/~rhowell/asymptotic.pdf. – Milten May 30 '22 at 10:01

1 Answers1

0

You extend big O notation for multivariable functions. It doesn't conflict with the existing notation. However it is non-standart, so you should give the definition every time you use it (once per article or homework).

Smylic
  • 8,098