Will partial summation work for this problem? Define $$G(x)=\sum_{n \leq x} T\left(\frac{x}{n}\right)$$ and $G,T: [1,\infty) \to \mathbb R$
And function T satisfies the following conditions:
1) $T(x)=O(x)$
2) $T(x) \sim cx (x \to \infty)$
How to show that $G(x) \sim cx\log{x} (x \to \infty)$? I have tried to use the partial summation, but feel that might not work.
 A: Actually a little more is needed: $T$ must be bounded on bounded subsets of $[1,\infty)$.  Otherwise a counterexample is $T(t) = t + 1/(t-1)^2$ for $t > 1$, $T(1) = 0$.  
Hint: If 
$c_1 x < T(x) < c_2 x$ for $x > N$, and $|T(x)| <= B$ for $x \le N$, what can you say about $G(x)$?
A: Since your condition $2$ implies your condition $1$, perhaps you meant that for $x\ge1$
$$
|T(x)|\le c_1x
$$
I believe that this would satisfy Robert Israel's concern.
Let $\epsilon>0$. By the second condition, there is a $B_\epsilon$, so that for $x>B_\epsilon$,
$$
(c-\epsilon)x<T(x)<(c+\epsilon)x
$$
Break up the sum into two parts
$$
\sum_{x\ge n>x/B_\epsilon}\left|T\left(\frac{x}{n}\right)\right|\le\sum_{x\ge n>x/B_\epsilon}c_1\frac{x}{n}\tag{1}
$$
$$
\sum_{n\le x/B_\epsilon}(c-\epsilon)\frac{x}{n}\le\sum_{n\le x/B_\epsilon}T\left(\frac{x}{n}\right)\le\sum_{n\le x/B_\epsilon}(c+\epsilon)\frac{x}{n}\tag{2}
$$
Since $\displaystyle\sum_{n\le x}\frac{1}{n}=\log(x)+\gamma+O\left(\frac{1}{x}\right)$, we get that
$$
\begin{align}
\left|\sum_{n\le x}T\left(\frac{x}{n}\right)-cx\log(x)\right|
&\le\epsilon x(\log(x)-\log(B_\epsilon))+\gamma x+c_1x\log(B_\epsilon)+O\left(1\right)\\
&\le x\log(x)\left(\epsilon+\frac{\gamma+c_1\log(B_\epsilon)+O\left(\frac{1}{x}\right)}{\log(x)}\right)
\end{align}
$$
which says that $G(x)\sim cx\log(x)$.
