Use the continuous convention . Let . Applying the allowed logarithm inequality to for gives
Hence . 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. Thus
This is Gibbs inequality for relative entropy, with zero probabilities in the first argument handled explicitly.
For the uniform distribution , direct expansion gives
Part (a) therefore proves
with equality exactly for for every . Thus the uniform distribution is the unique entropy maximizer.
Normalize the candidate distribution by setting
The supplied finite sum gives its mean
Thus it meets the constraint. Since , every other distribution with mean satisfies
Nonnegativity and the equality case in part (a) prove
The maximum is unique; the entropy comparison uses the fixed mean rather than an unconstrained optimization.
The given logarithmic moment is
where the term is zero. It is attained by the positive distribution
whose probabilities sum to one. Write , so . For any distribution with the required logarithmic moment,
Consequently the unique maximizer and its entropy are
The 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 distribution
Part (a), applied on this row support, gives . Its weighted sum expands as
Rows with contribute zero, since all in such a row vanish. Using the convention that zero-weight logarithmic terms are zero, the result is
Both stochasticity conditions have a role: row sums normalize the comparison distributions, while column sums preserve the total contribution from .

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