Chebyshev's theta function has linear lower and upper bounds

Put $P(y):={\prod }_{p\le y}p$ for real $y\ge 2$. We claim that $P(y)\le 4^{y-1}(y\ge 2)\mathrm{.}$ Let $q$ be the largest prime with $q\le y$. Then $P(y)=P(q)$ and $4^{q-1}\le 4^{y-1}$, so it suffices to prove the claim when $y=q$ is prime. For $q=2$ this is immediate. Let $q=2m+1$ be an odd prime, and assume inductively that $P(r)\le 4^{r-1}$ for every integer $r$ with $2\le r\le 2m$. Then $P(q)=P(m+1){\prod }_{m+1<p\le 2m+1}p\mathrm{.}$ Every prime in the second product divides $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)=\frac{(2m+1)!}{m!(m+1)!}$ by [L4], because it appears in the numerator and in neither denominator factorial. Also [L3] gives ${\sum }_{k=0}^{2m+1}\left(\genfrac{}{}{0px}{}{2m+1}{k}\right)=2^{2m+1},$ and the two equal middle terms $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)=\left(\genfrac{}{}{0px}{}{2m+1}{m+1}\right)$ therefore satisfy $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)\le 2^{2m}$. Thus $P(q)\le P(m+1)\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)\le 4^m\cdot 2^{2m}=4^{2m}=4^{q-1}\mathrm{.}$ So the claim holds for every real $y\ge 2$.

Again by [L2], the factorization of $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$ can be written as $\log \left(\genfrac{}{}{0px}{}{2n}{n}\right)\le \theta (2n)+R_n,$ where $R_n:={\sum }_{k=2}^{\lfloor {\log }_2(2n)\rfloor }\theta ((2n{)}^{1\mathrm{/}k})\mathrm{.}$ Using the trivial estimate $\theta (y)\le y\log y$ on each layer and $(2n{)}^{1\mathrm{/}k}\le \sqrt{2n}$ for $k\ge 2$, we get $0\le R_n\le \sqrt{2n}\log (2n)\lfloor {\log }_2(2n)\rfloor =o(n)\mathrm{.}$

By definition of $\theta$ and the logarithm law in [L5], $\theta (y)=\log P(y)\le \log (4^{y-1})=2(y-1)\log 2<2y\log 2$ for every real $y\ge 2$.

The lower bound in [L1] and [L5] give $\log \left(\genfrac{}{}{0px}{}{2n}{n}\right)\ge 2n\log 2-\log (2n+1)\mathrm{.}$ Combining this with step 1.2 shows $\theta (2n)\ge 2n\log 2-\log (2n+1)-R_n\mathrm{.}$ Since $\log (2n+1)+R_n=o(n)$, choose $n_0$ so large that $\log (2n+1)+R_n\le n\log 2$ for every $n\ge n_0$. Then $\theta (2n)\ge n\log 2(n\ge n_0)\mathrm{.}$

Let $x\ge 4n_0$, and put $n=\lfloor x\mathrm{/}2\rfloor$. Then $n\ge n_0$ and $2n\le x$, so by monotonicity of $\theta$ and step 2.2, $\theta (x)\ge \theta (2n)\ge n\log 2\ge \frac{\log 2}{4}x\mathrm{.}$ Step 2.1 also gives $\theta (x)\le 2x\log 2$ for every $x\ge 2$. Thus the theorem holds with $c=\frac{\log 2}{4}$, $C=2\log 2$, and $x_0=4n_0$.