Question

Interpret and prove:
(i) o+o=o
(ii) o+O=O
(iii) O+O=O
(iv) o(O)=o[(iv) and (v) represent compositions of functions]
(v) o x O=o

Answer 1

Question #8076 Interpret and prove: 1) $o + o = o, 2)o + O = O, 3)O + O = O, 4)o(O) = o, 5)o \cdot O = o$. Solution. Through the solution for simplicity of notation I will use $f_{i} = f_{i}(x)$, $g = g(x)$.

1) If $f_{1} = o(g), f_{2} = o(g), x \to x_{0}$, then $f_{1} + f_{2} = o(g), x \to x_{0}$. The condition implies that $f_{i}(x) / g(x) \to 0, x \to x_{0}, i = 1, 2$, hence $\frac{f_{1} + f_{2}}{g} \to 0, t \to x_{0}$, thus $f_{1} + f_{2} = o(g), x \to x_{0}$.

2) If $f_{1} = o(g), f_{2} = O(g), x \to x_{0}$, then $f_{1} + f_{2} = O(g), x \to x_{0}$. It is obvious that if $f_{1} = o(g), x \to x_{0}$ entails $f_{1} = O(g)$, hence the statement of 2) will follow from 3).

3) $f_{i} = O(g), x \to x_{0}, i = 1,2$ then $f_{1} + f_{2} = O(g), x \to x_{0}$. The condition implies exist $\delta_{i}$ such that for all $x \in B_{\delta_i}(x_0)$ $|f_i(x)| < K_i|g(x)|, i = 1,2$. Take $\delta := \min \{\delta_1, \delta_2\}$, then those two inequalities hold. Hence for all $x \in B_{\delta}(x)$ $|f_1 + f_2| \leq (K_1 + K_2)|g(x)|$, thus $f_1 + f_2 = O(g)$.

4) If $f_{1} = O(f_{2}), g = o(f_{1}), x \to x_{0}$, then $g = o(f_{2}), x \to x_{0}$. For all $\epsilon$ exists such $\delta_{1}$ that $|g(x)| < \epsilon |f_{1}(x)|$, $x \in B_{\delta_1}(x_0)$. Moreover, exists $\delta_{2}$ and $K > 0$ such that for all $x \in B_{\delta_2}(x_0)$ $|f_{1}| \leq K|f_{2}|$, hence for all $x \in B_{\min\{\delta_1, \delta_2\}}(x_0)$ $|g| < K\epsilon |f_2|$ and we are done.

5) Can be proven in the same manner as 4).