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Use a fixed logarithm base $b>1$, and write $c={\log }_{b}e=1/\ln b$. All relative entropies below use this same base. Adopt $0\log (0/q)=0$ and $p\log (p/0)=+\infty$ for $p>0$.

For Gibbs inequality, if $q(x)=0<p(x)$ at some point, the relative entropy is infinite and the required conclusion is immediate. Otherwise let $S=\{x:p(x)>0\}$. The elementary inequality $\ln u\le u-1$, with equality only at $u=1$, gives, for every $x\in S$,ConsequentlyThis argument applies to a countable discrete probability distribution too: the difference between the two sides of the pointwise inequality is nonnegative, and the negative part of the entropy summand is bounded by $cq(x)$, hence summable. Thus no subtraction of divergent series is involved. Equality forces both ${\sum }_{S}q=1$ and equality in every pointwise inequality; hence $q(x)=p(x)$ on $S$ and $q=0=p$ off $S$. Conversely $p=q$ plainly gives zero. Therefore $\boxed{D(p\Vert q)\ge 0,D(p\Vert q)=0⟺p=q}$.

For the log-sum inequality, set $F={\sum }_{A}f$, $G={\sum }_{A}g$, and normalize to the probability distributions $p_A=f/F$, $q_A=g/G$. When $A$ is nonempty, positivity ensures $F,G>0$, and direct expansion givesThus normalization reduces the log-sum inequality to Gibbs inequality. Equality holds exactly when $f/F=g/G$. The empty-set case is the zero identity under the usual zero-mass convention.

For the final bound, partition the alphabet using $A=\{x:p(x)>q(x)\}$ and put $a=p(A)$, $d=q(A)$. Since the total signed difference is zero,Apply the log-sum inequality separately to $A$ and its complement. The same normalization argument works for countable sets of finite total mass, using the countable Gibbs inequality just proved. It yields the binary partition bound for relative entropyZero block masses are interpreted by the stated conventions; a positive mass divided by zero makes the bound infinite. The given binary estimate now impliesThis is Pinsker's inequality. Equivalently, for the total variation distance $\Vert p-q{\Vert }_{\mathrm{TV}}=\frac{1}{2}{\sum }_x|p(x)-q(x)|$, the bound is $D(p\Vert q)\ge 2{\log }_{b}e\Vert p-q{\Vert }_{\mathrm{TV}}^2$. The factor ${\log }_{b}e$ keeps the statement valid for either bits or natural-logarithm units.

Choose an independent set uniformly and let $X_i$ indicate whether it contains vertex $i$. Then $H(X)={\log }_{2}i(G)$. Put $P_i=N(i)\cap [i-1]$ and $b_i=|P_i|$. Form a multiset containing every nonempty $P_i$ and $b_i$ copies of singleton $\{i\}$. Each coordinate $j$ occurs exactly $d_j$ times: in $d_j-b_j$ later-neighbor sets and in $b_j$ singleton copies.

For a set $A$ in this multiset, weight it by $1/{\min }_{j\in A}{d}_j$. These weights form a fractional cover, since each of the $d_j$ occurrences of $j$ has weight at least $1/d_j$. Apply the upper Madiman-Tetali inequality. Since every $j\in P_i$ precedes $i$, one has $d_j\ge d_i$; enlarging that set's weight to $1/d_i$ and removing conditioning only enlarges its nonnegative contribution. For singleton terms retain the useful conditioning on $P_i$. ThusFor $b_i>0$, let $q_i=\mathbb{P}(X_{P_i}=0)$. Every nonzero neighbor assignment forces $X_i=0$, while the all-zero assignment allows at most two values. The maximum entropy on a finite alphabet givesThe last inequality is the binary entropy maximization formula, or the log-sum inequality, with group sizes $2^{b_i}$ and $2^{b_i}-1$. If $b_i=0$ the whole term is zero, also agreeing with the formula. Exponentiating proves the degree-ordered independent-set entropy bound

For nonnegative $a_i,b_i$, with $a={\sum }_i{a}_i$ and $b={\sum }_i{b}_i$, the log-sum inequality iswith the usual extended-value conventions. Equality holds when $a_i/b_i$ is constant wherever $a_i>0$.

Let $0\le \lambda \le 1$ and put $P=\lambda P_1+(1-\lambda )P_2$ and $Q=\lambda Q_1+(1-\lambda )Q_2$. For each alphabet symbol $x$, apply the log-sum inequality to $a_1=\lambda P_1(x)$, $a_2=(1-\lambda )P_2(x)$ and the corresponding $b$ values. Summing over $x$ giveswhich is joint convexity in $(P,Q)$.

For nonnegative numbers $a_i,b_i$, put $a={\sum }_i{a}_i$ and $b={\sum }_i{b}_i$. The log-sum inequality iswith $0\log (0/b)=0$ and the usual extended-value convention when a denominator vanishes. Equality holds precisely when $a_i/b_i$ is constant over the indices with $a_i>0$.

Let $K(y\mid x)$ be a Markov kernel, and let $P_Y=PK$, $Q_Y=QK$ be the output probability distributions. Applying the log-sum inequality for each $y$ to $a_x=P(x)K(y\mid x)$ and $b_x=Q(x)K(y\mid x)$ givesSumming over $y$ and using ${\sum }_{y}K(y\mid x)=1$ yields the data processing inequality for relative entropy

With Total variation distance defined by $d_{\mathrm{TV}}(P,Q)={\sup }_A|P(A)-Q(A)|$ and relative entropy computed using natural logarithms, Pinsker's inequality statesIf $P$ is not absolutely continuous with respect to $Q$, the relative entropy is infinite and the bound is vacuous. A small relative entropy therefore forces small discrepancies in probabilities of all events. For example, $P=(0.6,0.4)$ and $Q=(0.5,0.5)$ have Total variation distance $0.1$ and relative entropy $0.6\log 1.2+0.4\log 0.8\simeq 0.02014$, giving an upper bound about $0.1003$.

A short proof reduces to a binary event. Grouping the probability mass inside and outside $A$, the log-sum inequality gives $D(P\Vert Q)\ge d(p\Vert q)$, with $p=P(A)$ and $q=Q(A)$. For $0<q<1$, the binary relative entropy as a function of $p$ has minimum zero at $p=q$ and second derivative $1/[p(1-p)]\ge 4$. Integrating this derivative bound twice gives $d(p\Vert q)\ge 2(p-q{)}^2$. Endpoint cases follow by limits. Taking the supremum over $A$ proves the inequality. Using base-two logarithms changes the constant by a factor of $\log 2$.