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Finite log-sum inequality with support and equality cases

Statement

Let $(a_{i})_{i = 1}^{n}, (b_{i})_{i = 1}^{n}$ be finite nonnegative real vectors, with $A = \sum_{i} a_{i}$ and $B = \sum_{i} b_{i}$ . Interpret $\sum_{a_{i} > 0} a_{i} \ln (a_{i} / b_{i})$ as $+ \infty$ if some $a_{i} > 0 = b_{i}$ ; otherwise zero coordinates contribute $0$ . Interpret $A \ln (A / B)$ as $0$ if $A = 0$ and as $+ \infty$ if $A > 0 = B$ . Then

$\sum_{a_{i} > 0} a_{i} \ln \frac{a_{i}}{b_{i}} \geq A \ln \frac{A}{B} .$

When $A, B > 0$ , equality holds exactly when $a_{i} = (A / B) b_{i}$ for every $i$ . If $A > 0 = B$ , both sides are $+ \infty$ and no finite equality characterization is asserted.

Facts & Assumptions

Given: Finite nonnegative vectors $a, b$ , and $A = \sum_{i} a_{i}$ , $B = \sum_{i} b_{i}$ .

[F1]

Information units and logarithm-base convention: logarithms are natural by default, $0 \ln 0 = 0$ , and a positive numerator over zero denominator contributes $+ \infty$ .

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln$ is continuous and strictly increasing on $(0, \infty)$ and $\ln 1 = 0$ .

[F3]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $(\ln t)^{'} = 1 / t$ for $t > 0$ .

[F4]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : the product rule gives $(f g)^{'} = f^{'} g + f g^{'}$ wherever both factors are differentiable.

[F5]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: products of functions continuous at a point are continuous there.

[F6]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : if $f$ is continuous on $[u, v]$ and differentiable on $(u, v)$ , some $c \in (u, v)$ satisfies $f (v) - f (u) = f^{'} (c) (v - u)$ .

[F7]

Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval: strict convexity means $f (λ x + (1 - λ) y) < λ f (x) + (1 - λ) f (y)$ for $x \neq y$ and $0 < λ < 1$ .

[F8]

Finite Jensen inequality for a convex function and nonnegative weights summing to one: for finitely many points in an interval and nonnegative weights summing to one, a convex function at the weighted mean is at most the weighted mean of its values.

Proof

1.1F2F3F4F5F6F7algebra

Define $f (0) = 0$ and $f (t) = t \ln t$ for $t > 0$ . On $(0, \infty)$ , [F3] and [F4] give $f^{'} (t) = \ln t + 1$ , which is strictly increasing by [F2]; [F2] and [F5] also give continuity of $f$ on every positive compact interval. If $0 < u < v < w$ , [F6] on $[u, v]$ and $[v, w]$ gives secant slopes $f^{'} (c_{1})$ and $f^{'} (c_{2})$ with $c_{1} < v < c_{2}$ ; strict increase makes the first slope smaller, which rearranges to the strict convexity inequality at $v$ . For an endpoint $0 < y$ and $0 < λ < 1$ , $f (λ y) = λ f (y) + λ y \ln λ < λ f (y) + (1 - λ) f (0)$ since $\ln λ < 0$ by [F2]. Thus $f$ is strictly convex on $[0, \infty)$ .

1.2givenF1casesalgebra

If $A = 0$ , nonnegativity forces all $a_{i} = 0$ , so both sides are $0$ . If $A > 0 = B$ , all $b_{i} = 0$ and at least one $a_{i} > 0$ , so both sides are $+ \infty$ . If $A, B > 0$ but some $a_{i} > 0 = b_{i}$ , the left side is $+ \infty$ while the right side is finite. It remains to handle $A, B > 0$ with $a_{i} = 0$ whenever $b_{i} = 0$ .

2.1F1F8step 1.1algebra

For the remaining case set $J = {i : b_{i} > 0}$ , $λ_{i} = b_{i} / B$ , and $x_{i} = a_{i} / b_{i}$ for $i \in J$ . These weights are positive and sum to one, and $\sum_{i \in J} λ_{i} x_{i} = A / B$ . By [F8] and the convexity proved in step 1.1, $B f (A / B) \leq \sum_{i \in J} b_{i} f (a_{i} / b_{i})$ . Expanding $f$ turns this into the desired inequality, including coordinates with $a_{i} = 0 < b_{i}$ because $f (0) = 0$ .

2.2F7F8step 1.1choosealgebra

If all $x_{i}$ are equal, their weighted mean is $A / B$ and Jensen is an equality; together with $a_{i} = 0$ when $b_{i} = 0$ , this is exactly $a_{i} = (A / B) b_{i}$ for every coordinate. Conversely, suppose they are not all equal, so $∣ J ∣ \geq 2$ and $0 < λ_{j} < 1$ for every $j$ . For each $j \in J$ , let $μ_{i} = λ_{i} / (1 - λ_{j})$ for $i \neq j$ and $y_{j} = \sum_{i \neq j} μ_{i} x_{i}$ . Some $j$ has $x_{j} \neq y_{j}$ : otherwise $x_{j} = (A / B - λ_{j} x_{j}) / (1 - λ_{j})$ for every $j$ , forcing each $x_{j} = A / B$ . Fix such a $j$ . Then $A / B = λ_{j} x_{j} + (1 - λ_{j}) y_{j}$ ; strict convexity from step 1.1 gives $f (A / B) < λ_{j} f (x_{j}) + (1 - λ_{j}) f (y_{j})$ . By [F8] on the normalized remaining weights, $f (y_{j}) \leq \sum_{i \neq j} μ_{i} f (x_{i})$ , making the total Jensen inequality strict.

3.1step 1.2step 2.1step 2.2∎

Steps 1.2–2.2 prove the inequality in every extended-real support case and, when $A, B > 0$ , prove equality exactly at proportional vectors.

Depends on

Used by

Dependency tree · two levels

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Sources

Robert M. Gray, Entropy and Information Theory, corrected first edition (standard reference, not scraped)
Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, author prepublication (standard reference, not scraped)