 Solution

Use logarithms to base two, so all entropies are in bits. For a discrete probability distribution $p$ and a density operator $\rho$ with eigenvalues ${\lambda }_j$, the Shannon entropy and Von Neumann entropy are respectivelyUse $0\log 0=0$. The second definition is basis-independent by the spectral theorem; for a density operator diagonal with probabilities $p_x$, it agrees with the first.

The classical relative entropy, or Kullback-Leibler divergence, of $p$ from $q$ isIt is $+\infty$ if some $p_x>0$ has $q_x=0$. The quantum relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and is $+\infty$ otherwise. Matrix logarithms are taken on the corresponding positive supports. In a common eigenbasis this reduces to classical relative entropy; both divergences are nonnegative and are generally asymmetric.

To prove Fano's inequality, let the finite alphabet size be $m\ge 2$ and let $E$ indicate an incorrect guess. Since $E$ is determined by $(X,Y)$, the chain rule for conditional entropy first givesExpanding in the opposite order givesConditioning reduces entropy, so $H(E\mid Y)\le H(E)=h_2(p_e)$, the binary entropy function. When $E=0$, knowledge of $Y$ determines $X=f(Y)$ and the conditional entropy is zero. When $E=1$, $X$ can lie in at most the $m-1$ alphabet values other than $f(Y)$. Entropy on $m-1$ values is at most ${\log }_2(m-1)$: its deficit from that number is the nonnegative relative entropy from the uniform distribution. Averaging the two cases therefore givesCombining the two expansions provesThis proof of Fano's inequality via an error indicator works for any alphabet-valued guessing rule, not just an optimal one. If $m=1$, there is no guessing uncertainty and the claim is simply $H(X\mid Y)=0$; that case is treated separately rather than writing $\log 0$.