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Let $g(x, y, z)$ be a polynomial in the three variables $x, y, z$ that take values in $\mathbb{N}$. Prove that $g(x, y, z) = \mathcal{O}(x^k y^l z^m)$ for some $k, l, m \in \mathbb{N}$

I am not sure how to approach the big $O$ method for multiple variables, the following post did not help me that much further Formal definition of big-O when multiple variables are involved?. So far we have only dealt with single-variable polynomials. Also, am I right in assuming that what is meant here is that the maximal degrees of $x$, $y$ and $z$ in the polynomial are the numbers $k, l$ and $m$?

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    You can use $f \in \mathcal{O}(g) \Leftrightarrow \limsup_{x \to a} \left|\frac{f(x)}{g(x)}\right| < \infty$; (If $\lim$ exists, you can use it instead of $\limsup$) In your case we'd have $x\in \mathbb{R}^3$ and $a = \infty$. Though I got no idea what exactly $x\to \infty$ means for $x\in\mathbb{R}^3$. – Sudix Dec 06 '18 at 01:34

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Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:

$$ g(x,y,z)= a_0 +a_1 x + \dots + a_k x^k + b_0 + b_1 y + \dots +b_l y^l + c_0 + c_1 z + \dots c_m z^m $$

Now we let $x,y,z \geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term. $$g(x,y,z) \leq a_0 x^k y^l z^m + \dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + \dots +b_l y^l + c_0 x^k y^l z^m + \dots c_m x^k y^l z^m $$ So we get that:

$$ g(x,y,z) \leq \left( \sum_{i=0}^k a_i+ \sum_{i=0}^l b_i + \sum_{i=0}^m c_i \right) x^k y^l z^m$$ We can now complete the proof:

Let $x, y, z \geq 1$ and choose $M=\left( \sum_{i=0}^k a_i+ \sum_{i=0}^k b_i + \sum_{i=0}^k c_i \right)$, we then have that: $$ g(x,y,z) \leq M \cdot x^k y^l z^m$$

So $g(x,y,z) = \mathcal{O}(x^k y^l z^m)$. $\square$

Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form: $$g(x,y,z)= \sum_{p=0}^m \sum_{j=0}^l \sum_{i=0}^k a_{ijp}x^i y^j z^p.$$ We can apply the same trick. We let $x,y,z, \geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p \leq x^k y^l z^m $$ We can now approximate the sum by: $$ g(x,y,z) \leq \sum_{p=0}^m \sum_{j=0}^l \sum_{i=0}^k a_{ijp}x^k y^l z^m = \left( \sum_{p=0}^m \sum_{j=0}^l \sum_{i=0}^k a_{ijp} \right) x^k y^l z^m $$

Now we let $M=\left( \sum_{p=0}^m \sum_{j=0}^l \sum_{i=0}^k a_{ijp} \right)$: $$ g(x,y,z) \leq M x^k y^l z^m $$ And this is our desired result $\square$