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The Chebyshev theta function is one of the two Chebyshev functions and is the natural logarithm of the primorial:where both the sum and the product range over prime numbers.

For a probability mass function $Q$ of full support on a finite alphabet and a real function $g$,where $D_e$ uses natural logarithms. The unique maximizing probability mass function is the exponential tilt $P^{\ast }(x)=Q(x)e^{g(x)}/{\mathbb{E}}_Q{e}^g$.

For $x>0$, concavity of the natural logarithm givesEquivalently, ${\log }_{b}x\le ({\log }_{b}e)(x-1)$ for $b>1$.

For an event of probability $p$, the log odds are the natural logarithm of its odds:

The log odds ratio is the natural logarithm of an odds ratio. For two independent binomial groups with positive event and non-event counts $(a,b)$ and $(c,d)$, its estimate is $\log (ad/(bc))$, with first-order variance $1/a+1/b+1/c+1/d$ by the delta method. Zero cells need a method that handles boundary data; the displayed normal approximation is not valid there.

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.

Since the natural logarithm is increasing, comparison with an integral givesThe lower bound is $n\log n-n+1$. Dividing all three expressions by $n\log n$ and applying the squeeze theorem gives

Taking the natural logarithm of the finite Euler product and using the Taylor seriesgivesThe double series converges absolutely, and the Mertens second theorem therefore makes the right sideforExponentiating provesThis is the Mertens third theorem.

First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so

Regularity at infinity excludes the natural logarithm and every positive radial power. ThusThe constant $a_0$ and the decaying coefficients $c_n,d_n$ remain arbitrary.

Regularity at the origin excludes the natural logarithm and every negative radial power. ThusThe coefficients $a_0,a_n,b_n$ remain arbitrary.

Put $p_i=\mathbb{P}(X_i=1)$ and $\lambda ={\sum }_i{p}_i$. A relative-entropy form of the Poisson approximation bound for dependent Bernoulli variables isThe last two terms form the total correlation; they vanish when the Bernoulli variables are independent.

Here $D_e$ and $H_e$ use natural logarithms.

To prove the bound, let $Q_i$ be the Poisson distribution with mean $p_i$ and let $Q={⨂}_i{Q}_i$. Expanding the Kullback-Leibler divergence against this product law givesThe supplied one-dimensional estimate bounds the first sum by ${\sum }_i{p}_i^2$. Under the addition map, $P_{X_1^n}$ becomes $P_{S_n}$, while the Sum of independent Poisson random variables under $Q$ has the Poisson distribution with mean $\lambda$. The data processing inequality for relative entropy proves the displayed result.

If a bound directly in the paper's unhalved total-variation norm is desired, Pinsker's inequality also gives

Write the Chebyshev theta function asTaking natural logarithms in part a givesand henceBecause $\sqrt{n}\log n/n\to 0$, the right-hand side is at least $n/2$ for every sufficiently large integer $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the floor function. Then $2n\le x$ and $n\ge x/3$, so the monotonicity of $\vartheta$ gives

Suppose that the positive integer $n$ is a decisive number. For every prime number $p<\sqrt{n}$, the composite number $p^2$ satisfies $1<p^2<n$. It therefore cannot be coprime to $n$, so $p\mid n$.

The product of all prime numbers below $\sqrt{n}$ consequently divides $n$. Taking natural logarithms givesChanging the strict endpoint can remove at most one prime term. By part b, for every sufficiently large $n$ the right-hand side is at least, for example, $\sqrt{n}/13$. This is impossible for sufficiently large $n$, sinceThus every decisive number lies below one fixed bound, and only finitely many integers can do so.

Writing $P,Q$ for the laws of $X,Y$, respectively, and using Jensen inequality for the concave natural logarithm,This is the lower-bound half of the Gibbs variational principle for relative entropy.