Entropy and total variation distance Let $X$, $Y$ be discrete random variables taking values within the same set of $N$ elements. Let the total variation distance $|P-Q|$ (which is half the $L_1$ distance between the distributions of $P$ and $Q$) be at most $\epsilon$. What is a standard, easily provable bound on the difference between the entropies of $X$ and $Y$?
(It is easy to see that such a bound must depend not only on $\epsilon$ but also on $N$.)
 A: $\newcommand{\de}{\delta}
\newcommand{\ep}{\epsilon}$
Note that $p\ln p-p$ is decreasing in $p\in[0,1]$, so that $p\ln p-p\le q\ln q-q$ and hence $p\ln p-q\ln q\le p-q=|p-q|$ if $0\le q\le p\le1$. 
Next, take any real $c\ge1$. Note that $g(p):=p\ln p+cp$ (with $g(0):=0$) is convex in $p\in[0,1]$. So, assuming $0\le p\le q\le1$ and $q\ge e^{-c}$, we have 
$g(p)\le g(0)\vee g(q)=0\vee g(q)=g(q)$ and hence 
$p\ln p-q\ln q\le c(q-p)=c|p-q|$. 
Thus, 
\begin{equation}
 p\ln p-q\ln q\le c|p-q|
\end{equation}
whenever $0\le p,q\le1$ and $q\ge e^{-c}$. 
Also, $-q\ln q$ is increasing in $q\in[0,e^{-1}]$ and hence in $q\in[0,e^{-c}]$, so that $-q\ln q\le ce^{-c}$ for $q\in[0,e^{-c}]$. Also, $p\ln p\le0$ if $0\le p\le1$. 
Therefore, the difference between the entropies of $Q=(q_i)_{i=1}^N$ and $P=(p_i)_{i=1}^N$ is 
\begin{equation}
 \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i=S_1+S_2+S_3,
\end{equation}
where 
\begin{align*} 
S_1 &:=\sum_{i\colon q_i\ge e^{-c}} (p_i\ln p_i-q_i\ln q_i)\le \sum_{i\colon q_i\ge e^{-c}}c|p_i-q_i|
\le c\de \quad\text{if}\ \de\ge\|P-Q\|_1, \\
S_2 &:= \sum_{i\colon q_i< e^{-c}}p_i\ln p_i\le0, \\ 
S_3 &:= \sum_{i\colon q_i< e^{-c}}(-q_i\ln q_i)\le\sum_{i\colon q_i< e^{-c}}ce^{-c}\le Nce^{-c}.  
\end{align*}
So, taking now $c=\ln\frac N\de$ and assuming $N\ge e\de$, we see that 
\begin{equation}
 \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le 
 c\de+Nce^{-c}
 =2\de\ln\frac N\de.
\end{equation}
Taking here $\de=2\ep$ and noting that $N\ge1$, we conclude that 
\begin{equation}
 \sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le 
4\ep\ln\frac{N}{2\ep}
\end{equation}
if $\ep\le1/(2e)$. 
A: $\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\R}{\mathbb{R}}$
Here is yet another answer providing the exact bound. This answer is perhaps a bit more elementary than the excellent answer given by user Algernon. Another advantage of this approach is that it produces the exact bound for any convex function $f$ in place of the function $p\mapsto p\ln p$.
To preserve the history of the question, I have also retained my previous answer, which used different (if somewhat similar) ideas and provided a suboptimal bound. 
Take indeed any convex function $f\colon[0,1]\to\R$ and consider the difference 
\begin{equation}
 \De:=\sum_1^N f(p_i)-\sum_1^N f(q_i)
\end{equation}
between the "generalized" entropies of $Q=(q_i)_{i=1}^N$ and $P=(p_i)_{i=1}^N$. 
We want to find the exact upper bound on $\De$ subject to the given conditions on $(P,Q)$. 
In what follows, $(P,Q)$ is a point satisfying these conditions. 
Without loss of generality (wlog), for some $k\in\{1,\dots,N\}$ we have $p_i\ge q_i$ for $i\le k$, $p_i\le q_i$ for $i\ge k+1$, and $q_1\ge\cdots\ge q_k$, so that 
\begin{equation}
 \ep=\sum_1^k(p_i-q_i)=\sum_{k+1}^N(q_i-p_i)>0. 
\end{equation}
Let $p_i^*:=q_i$ for $i=2,\dots,k$ and $p_1^*:=q_1+\ep[=\sum_1^k p_i-\sum_2^k q_i\le1]$. Then the vector $(p_1^*,\dots,p_k^*)$ majorizes (in the Schur sense) the vector $(p_1,\dots,p_k)$ and still satisfies the condition $p_i^*\ge q_i$ for $i\le k$. Also, since $f$ is convex, $\sum_1^k f(p_i)$ is Schur convex in $(p_1,\dots,p_k)$. So, wlog $(p_1,\dots,p_k)=(p_1^*,\dots,p_k^*)$. 
In particular, $p_1>q_1$ and $q_m\ge p_m$ for all $m=2,\dots,N$. 
Moreover, wlog $p_m=0$ for any $m=2,\dots,N$. Indeed, take any $m=2,\dots,N$ with $p_m>0$ and replace $p_1,q_1,p_m,q_m$ respectively by $p_1+t,q_1+t,p_m-t,q_m-t$, where $t:=p_m\in(0,1-p_1]$; then all the conditions on $P,Q$ will still hold. After this replacement, $\De$ will change by the sum of the nonnegative expressions $[f(p_1+t)-f(q_1+t)]-[f(p_1)-f(q_1)]$ and $[f(q_m)-f(p_m)]-[f(q_m-t)-f(p_m-t)]$; this nonnegativity follows by the convexity of $f$. Making such replacements for each $m=2,\dots,N$ with $p_m>0$, we will change $\De$ by a nonnegative amount, and will also get $p_m=0$ for all $m=2,\dots,N$ indeed. 
Thus, wlog 
\begin{gather}
 p_1=1=q_1+\ep,\quad p_i=0\ \forall i=2,\dots,N,\quad 
 \sum_2^N q_i=\ep. 
\end{gather}
Since $\sum_2^N f(q_i)$ is Schur convex in the $q_i$'s, wlog
\begin{align}
 \De&=f(1)-f(q_1)+\sum_2^N f(0)-\sum_2^N f(q_i)  \\ 
 &\le f(1)-f(1-\ep)+(N-1)f(0)-(N-1)f(\tfrac\ep{N-1})=:\De_{f;\ep,N}.   
\end{align}
The bound $\De_{f;\ep,N}$ on $\De$ is obviously exact, since it is attained when $p_1=1=q_1+\ep$ and $q_2=\cdots=q_N=\tfrac\ep{N-1}$. 
In the particular case when $f(p)=p\ln p$ (with $f(0)=0$), the exact bound $\De_{f;\ep,N}$ becomes $H(\ep)+\ep\ln(N-1)$, where $H(\ep):=\ep\ln\frac1\ep+(1-\ep)\ln\frac1{1-\ep}$. 
A: Claim. If $\|P-Q\|\leq\varepsilon\leq\frac{1}{2}$, then $|H(P)-H(Q)| \leq H(\varepsilon) + \varepsilon\log N$.
Proof.
Let $\varepsilon':=\|P-Q\|$.
Let $(X,Y)$ be an optimal coupling of $P$ and $Q$, so that
\begin{align}
   \mathbb{P}(X\neq Y) = \|P-Q\| \;.
\end{align}
Using a standard construction, we can assume that $X$ and $Y$ have the particular form
\begin{align}
   X &:=
      \begin{cases}
         Z & \text{if $B=0$,} \\
         \tilde{X} & \text{if $B=1$,}
      \end{cases} &
   Y &:=
      \begin{cases}
         Z & \text{if $B=0$,} \\
         \tilde{Y} & \text{if $B=1$,}
      \end{cases}
\end{align}
where $B$, $Z$ and $(\tilde{X},\tilde{Y})$ are independent and $B\sim\text{Bern}(\varepsilon')$.
Note that
\begin{align}
   H(X|B) \leq H(X) \leq H(B) + H(X|B) \;.
\end{align}
For $H(X|B)$ we can write
\begin{align}
   H(X|B) &= \varepsilon' H(X|B=1) + (1-\varepsilon') H(X|B=0) \\
   &= \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) \;.
\end{align}
Thus,
\begin{align}
   \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) &\leq H(X)
   \leq H(B) + \varepsilon' H(\tilde{X}) + (1-\varepsilon') H(Z) \;,
   \tag{$\clubsuit$}
\end{align}
and similarly,
\begin{align}
   \varepsilon' H(\tilde{Y}) + (1-\varepsilon') H(Z) &\leq H(Y)
   \leq H(B) + \varepsilon' H(\tilde{Y}) + (1-\varepsilon') H(Z) \;.
   \tag{$\spadesuit$}
\end{align}
Combining ($\clubsuit$) and ($\spadesuit$) we get
\begin{align}
   |H(X)-H(Y)| &\leq
      H(B) + \varepsilon' |H(\tilde{X}) - H(\tilde{Y})| \\
   &\leq H(\varepsilon') + \varepsilon' \log N \\
   &\leq H(\varepsilon) + \varepsilon \log N \;,
\end{align}
as claimed.
QED

Edit [2018--09--17] (following Iosif Pinelis's comment).
Refining the above reasoning a little bit, we can get the better bound
\begin{align}
   |H(P)-H(Q)|\leq H(\varepsilon) + \varepsilon\log(N-1) \;.
\end{align}
Indeed, let $\Sigma$ denote the $N$-element set that is the support of $P$ and $Q$.  As before, let $\varepsilon':=\|P-Q\|$, and let us discard the trivial cases $\varepsilon'\in\{0,1\}$, so that $0<\varepsilon'<1$.
Recalling from the construction of an optimal coupling, define for $a\in\Sigma$,
\begin{align}
   R_0(a) &:= P(a)\land Q(a) &
      P_0(a) &:= P(a)-R_0(a) \\
   & &
      Q_0(a) &:= Q(a)-R_0(a) \;.
\end{align}
Observe that $R_0$, $P_0$ and $Q_0$ are non-negative functions and
\begin{align}
   \sum_{a\in\Sigma}R_0(a)=1-\varepsilon' \qquad\text{and}\qquad
      \sum_{a\in\Sigma}P_0(a)=\sum_{a\in\Sigma}Q_0(a)=\varepsilon' \;.
\end{align}
Thus, $\tilde{R}:=R_0/(1-\varepsilon')$, $\tilde{P}:=P_0/\varepsilon'$ and $\tilde{Q}:=Q_0/\varepsilon'$ are probability distributions on $\Sigma$ satisfying
\begin{align}
   P(a) &= (1-\varepsilon')\tilde{R}(a) + \varepsilon'\tilde{P} \\
   Q(a) &= (1-\varepsilon')\tilde{R}(a) + \varepsilon'\tilde{Q} \;.
\end{align}
If we now choose $Z\sim\tilde{R}$, $\tilde{X}\sim\tilde{P}$, $\tilde{Y}\sim\tilde{Q}$ and $B\sim\text{Bern}(\varepsilon')$ independently, we have a coupling as promised above.
Back to the inequality, observe that since both $P$ and $Q$ are non-negative and normalized, there necessarily exist $a,b\in\Sigma$ such that $P_0(a)=0$ and $Q_0(b)=0$.  This means that each of $\tilde{P}$ and $\tilde{Q}$ is in fact supported on a strict subset of $\Sigma$.  Hence,
\begin{align}
   |H(\tilde{X})-H(\tilde{Y})| &\leq \max\{H(\tilde{P}),H(\tilde{Q})\}
      \leq \log(N-1)
\end{align}
and the (updated) claim follows as before.
Note.
As the example H A Helfgott gave in the comments ($\require{enclose}\enclose{horizontalstrike}{N=0}$, $\require{enclose}\enclose{horizontalstrike}{X\sim\text{Bern}(\varepsilon)}$ and $\require{enclose}\enclose{horizontalstrike}{Y\sim\text{Bern}(0)}$) shows, this refined bound is sharp at least when $\require{enclose}\enclose{horizontalstrike}{N=2}$.
Iosif Pinelis gave the following example in his comment below which shows that the refined bound is sharp for every $N$ and $\varepsilon$:
Let $\Sigma:=\{1,2,\ldots,N\}$ and
\begin{align}
   P(a) &:=
      \begin{cases}
         1 & \text{if $a=1$,}\\
         0 & \text{otherwise,}
      \end{cases} &
   Q(a) &:=
      \begin{cases}
         1-\varepsilon & \text{if $a=1$,}\\
         \varepsilon/(N-1) & \text{otherwise.}
      \end{cases}
\end{align}
Then, $\|P-Q\|=\varepsilon$ and $|H(P)-H(Q)|=H(Q)=H(\varepsilon)+\varepsilon\log(N-1)$.
