Take for example the measure $\mu(n)=n^2$ on $\{1, \ldots, N\}$ and a random variable $X$ distributed according to the probability obtained by normalizing $\mu$. 


Does there exists a constant $K>0$ such that $H(X \mid X \in C) \leq K.H(X)$ for every subset $C \subset \{1, \ldots, N\}$ and every $N$ ?


I know how to prove that $H(X \mid X >n) \leq H(X)$ as well as $H(X \mid X <n) \leq H(X)$ for every $n$, and I am under the impression (intuitively) that the equality $H(X \mid X \in C) \leq H(X)$ should be more or less true for every $C$, and should be asymptotically true for large $N$.


But $H(X \mid X \in C) \leq K.H(X)$ would be enough for my goal, and I really don't know how to get such a $K$. I think this should be more generally true for $\mu(n)=f(n)$ for a class of convex functions $f$, but I would already be happy to know how to deal with the $n^2$ example.



Estimating entropy conditional to an event
For $\mu(n)=n^2$, a quick calculation (\*) gives $H(X)\ge (1-\mathcal o(1))\log N$. On the other hand, a probability carried by a finite set $C$ has entropy at most $\log |C|$ (entropy of the equiprobable measure), so that $H(X|X\in C)\leq\log N$. This should be enough for your inequality, with $K\to 1$ for large $N$.


(\*) Let $a(N)=\sum\_1^N n^2=(2N^3+3N^2+N)/6$. By definition, $H(X)=\sum\_1^N \frac{n^2}{a(N)}\log\frac{a(N)}{n^2}=\log a(N)-\frac2{a(N)}\sum\_1^N n^2\log n$.


Now, majorize $\sum\_1^N n^2\log n$ : as $x^2\log x$ is increasing on $[1,\infty)$, you have $\sum\_1^N n^2\log n\le\int\_1^{N+1}x^2\log x\ dx$.


But $x^2\log x$ is the derivative of $\frac{x^3}{3}\log x-\frac{x^3}{9}$, so that $\int\_1^{N+1}x^2\log x\ dx=\frac{(N+1)^3}{3}\log(N+1)-\frac{(N+1)^3}{9}+\frac19$.


Finally $H(X)\ge\log a(N)-\frac{2(N+1)^3}{3a(N)}\log(N+1)+\frac{2(N^3+3N^2+3N)}{9a(N)}$.


