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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.

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Generlized Mobius inversion may be helpful here. See Mike Spivey's answer here for a similar problem:… – JavaMan Sep 28 '11 at 5:48
I don't see how either of partial summation or Mobius inversion will yield an answer. Israel's hint below suggests a very straightforward solution. – anon Oct 6 '11 at 23:35
up vote 4 down vote accepted

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 $G(x)\sim cx\log(x)$.

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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 $G(x)$?

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It's possible the $O(x)$ condition was specified or implied on $[1,\infty)$, otherwise it would be redundant in light of the second condition, but it's hard to tell. – anon Oct 6 '11 at 23:37
@anon: that is the assumption I made in my answer. – robjohn Oct 7 '11 at 0:05

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