Use the continuous convention . Let . Applying the allowed logarithm inequality to for givesHence . Equality forces . Since every is strictly positive, this makes the entire alphabet. Equality in every weighted logarithm inequality then forces for every . Conversely makes every summand zero. ThusThis is Gibbs inequality for relative entropy, with zero probabilities in the first argument handled explicitly.
For the uniform distribution , direct expansion givesPart (a) therefore proveswith equality exactly for for every . Thus the uniform distribution is the unique entropy maximizer.
Normalize the candidate distribution by settingThe supplied finite sum gives its meanThus it meets the constraint. Since , every other distribution with mean satisfiesNonnegativity and the equality case in part (a) proveThe maximum is unique; the entropy comparison uses the fixed mean rather than an unconstrained optimization.
The given logarithmic moment iswhere the term is zero. It is attained by the positive distributionwhose probabilities sum to one. Write , so . For any distribution with the required logarithmic moment,Consequently the unique maximizer and its entropy areThe increasing weights are consistent with the prescribed moment; they need not be a decreasing power law.
Write . For each row put ; the positive row entries form a probability distribution because the row sum is one. When , define the probability distributionPart (a), applied on this row support, gives . Its weighted sum expands asRows with contribute zero, since all in such a row vanish. Using the convention that zero-weight logarithmic terms are zero, the result isBoth stochasticity conditions have a role: row sums normalize the comparison distributions, while column sums preserve the total contribution from .
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