Equivalence of two relations involving $\pi(x)$ and $\vartheta(x)$

Question

In Apostol's book Introduction to Analytic Number Theory, we have the following exercise

4.18. Prove that the following two relations are equivalent: \begin{align*} \text{(a)} \quad \quad & \pi(x)=\frac{x}{\log x}+O \left( \frac{x}{\log^2 x} \right).\\ \text{(b)} \quad \quad & \vartheta(x)=x+O \left( \frac{x}{\log x} \right). \end{align*}

where $\pi(x)$ is, of course, the prime counting function, and $\vartheta(x)$ is Chebychev's function $\sum_{p \le x} \log p$.

I don't quite understand what "equivalent" is supposed to mean in this context, and I thought of skipping this particular exercise, but I suppose it would be a bad habit to skip all questions that I don't understand (especially since I am self-studying). So my hope is that someone would like to help me clarify this.

Answer 1

"Equivalent" here means that you can prove that (a) implies (b), and that (b) implies (a), with significantly less effort than it takes to prove either (a) or (b).

In this particular case, I imagine that you're intended to use partial summation to derive each one from the other one.

Answer 2

Maybe this could help. Note that I use $\ln x$ for natural logarithm.

I guess someone will come up with the solution without using Chebyshev's inequalities. Perhaps you could specify in the question, whether you're allowed to used them in this exercise (i.e., whether Chebyshev's inequalities were proved in the book before this exercise).

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Lemma. $$\pi(x)\sim \frac{\vartheta(x)}{\ln x}$$

Proof. Clearly $\vartheta(x)= \sum\limits_{p\leq x} \ln p \leq \sum\limits_{p\leq x} \ln x = \pi(x) \ln x$. This implies $$\frac{\vartheta(x)}{\pi(x)\ln x} \leq 1.$$

Now let $x\geq 2$ and $0<\varepsilon<1$. Then we have $$ \vartheta(x) \geq \sum_{x^{1-\varepsilon} < p \leq x} \ln p \geq (1-\varepsilon) \ln x (\pi(x)-\pi(x^{1-\varepsilon})) \geq (1-\varepsilon) \ln x (\pi(x)-x^{1-\varepsilon}). $$ This yields (using one of Chebyshev inequalities) $$\frac{\vartheta(x)}{\pi (x) \ln x } \geq (1-\varepsilon)\left(1 - \frac{x^{1-\varepsilon}}{\pi(x)}\right) \geq (1-\varepsilon)\left(1 - \frac{x^{1-\varepsilon}\ln x}{c_2x}\right)$$ for some constant $c_2$.

If we take the limit for $x\to\infty$ we get $$\liminf_{x\to\infty} \frac{\vartheta(x)}{\pi(x) \ln x} \geq 1-\varepsilon.$$ Since $\varepsilon$ cen be chose arbitrarily small, we get $$\lim\limits_{x\to\infty} \frac{\vartheta(x)}{\pi(x) \ln x} = 1.$$