Cauchy sequences converge

Question

Statement (taken from here):

Let ${f: {\bf N} \rightarrow {\bf C}}$, ${F: {\bf R}^+ \rightarrow {\bf C}}$ and ${g: {\bf R}^+ \rightarrow {\bf R}^+}$ be functions such that ${g(x) \rightarrow 0}$ as ${x \rightarrow \infty}$. Then the following are equivalent:

(i) One has $$\sum_{y \leq n < x} f(n) = F(x) - F(y) + O( g(x) + g(y) )$$ for all ${1 \leq y < x}$.

(ii) There exists a constant ${c \in {\bf C}}$ such that $$ \sum_{n < x} f(n) = c + F(x) + O( g(x) )$$

for all ${x \geq 1}$. In particular, ${c = -F(1) + O(g(1))}$.

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This exercise should be easy I think but for some reason I'm not grokking the use of the big Os.

My attempt:

i) $\rightarrow$ ii)

Set $y = 1$ in i) then you have

$$\sum_{1 \leq n < x} f(n) = F(x) - F(1) + O( g(x) + g(1) )$$

But I get stuck at trying to split the big O on the right side..

ii) $\rightarrow$ i)

Similar idea, but this time subtract up until $y$

$$\sum_{y \leq n < x} f(n) = \sum_{n < x} f(n) - \sum_{n < y} f(n) = $$

and applying ii) we have

$$ c + F(x) + O( g(x) ) - (c + F(y) + O(g(y))$$

Here $c$ cancels, and you get

$$ \sum_{y \leq n < x} f(n) = F(x) - F(y) + O(g(x)) - O(g(y))$$

and my problem becomes justifying joining the subtracting $O(g(y))$ part. I'm pretty sure that you can join $O(g) + O(h)$ into $O(g + h)$ by picking the max of the constants of the first expression but not the other way around

I'd prefer hints over solutions, thank you.

Answer 1

For i) -> ii), define the function

$$ H(x)=\sum_{n<x}f( n)-F(x). $$

Then use i) to conclude that $H(x)\to c$ for some $c$ as $x\to+\infty$. From i), it can also be concluded that

$$ H(x)=c+O(g(x)), $$

so we have

$$ \sum_{n<x}f(n)=F(x)+c+O(g(x)). $$

Answer 2

The second part is easy. $O(g)$ and $O(h)$ are just place holders for some unknown functions that satisfy some inequalities about their growth. It should be easy to show that more generally $C_1O(g)+C_2O(h)=O(g+h)$ for any constants $C_1, C_2$ and positive $g, h$.

The first part is slightly harder. First show that $\rho(x):=\sum_{n<x}f(n)-F(x)$ converges by showing it's Cauchy. Now we can determine $c$ in (ii) must be $\lim_{x\rightarrow\infty} \rho(x)$. Rewrite (i) as $\rho(x)-\rho(y)=O(g(x)+g(y))$, and take $x\rightarrow\infty$ to get (ii).

Note that $c=-F(1) + O(g(1))$ is by itself a trivial statement. I guess Tao wanted to say the constant from the big O in (i) can be used in (ii) to give a bound of $c$.