TITLE: big-Oh Prove or disprove
QUESTION [1 upvotes]: Prove or disprove:
If $f(x) = O(h(x))$ and $ g(x) = O(k(x)) $ as $x \to \infty $ , then $f(x) - g(x) = O(h(x) - k(x))$ as $x \to \infty$.
Solution:
Counter-example to disprove the above assertion :
$f(x) = 4x^4 + 3x^3 \implies  f(x) = O(x^4). $
$g(x) = -x^5 + 4x^4 \implies g(x) = O(x^5).$
$f(x) - g(x) = x^5 + 3x^3 \implies f(x) - g(x) = O(x^5) \neq O(x^4 - x^5). $
I am not sure what $O(x^4 - x^5) $ means ? Is big- Oh linear operator ? If that is the case, then $O(x^4-x^5) = O(x^4) - O(x^5) = O(x^5) $ and my counter-example will not work. Any help appreciated.

REPLY [0 votes]: You can make it easier. Suppose $f(x)=g(x)=x$, then $(f-g)(x)=0$ and so we will have $O(0)=\{0\}$, set with only constantly zero function.
Meanwhile $2\sqrt{x}\in O(x)$ and $\sqrt{x}\in O(x)$, $x \to \infty$, but $2\sqrt{x}-\sqrt{x}=\sqrt{x}$ is not zero
For other questions. If we take that definition is $$O(f)=\{g: \exists C>0,\exists U_{\delta}(\infty), x \in U_{\delta}, |g(x)|\leqslant C|f(x)|\}$$ then $O(x^5-x^4)=O(x^4-x^5)=O(x^5),x \to \infty$ from one hand and $O(x^5)-O(x^4)=O(x^5)+O(x^4)=O(x^5),x \to \infty$ from another hand.

REPLY [0 votes]: It is very false, even for $h(x) = k(x)$. for simplicity, take $h(x)=x$, and $f(x)=2x,\;g(x)=x$. In this case it is very obvious that $f(x)-g(x)=x\neq O(0)$
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