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A positive non-log-convex solution of the Gamma functional equation

Example

Put $D : = \frac{1}{2} (\log Γ (1) + \log Γ (3 / 2)) - \log Γ (5 / 4)$ , let $A : = 1 + ∣ D ∣$ , and define $F (s) : = Γ (s) e^{A \sin (2 π s)}$ for $s > 0$ . The function $F (s) = Γ (s) e^{A \sin (2 π s)}$ is positive, satisfies $F (1) = 1$ and $F (s + 1) = s F (s)$ , differs from Gamma, and is not log-convex.

Facts & Assumptions

Given: The constants $D, A$ and the function $F$ in the Example.

[F1]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ , and $Γ (1) = 1$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

[F2]

Sine has period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[F3]

A positive function is log-convex when its logarithm is convex (Log-convex positive functions).

Verification

technique · direct

1.1

By [F2], $e^{A \sin (2 π (s + 1))} = e^{A \sin (2 π s)}$ . Together with [F1], this gives $F (s + 1) = s F (s)$ and $F (1) = 1$ ; positivity is immediate.

F1F2algebra

1.2

At $s = 1, 5 / 4, 3 / 2$ , the sine terms are respectively $0, 1, 0$ . Hence the midpoint value of $\log F$ exceeds the average of its endpoint values exactly when $A > D$ , which holds because $A = 1 + ∣ D ∣$ . By [F3], $F$ is not log-convex.

F3algebra

2.1

Since $A > 0$ , $F (5 / 4) = e^{A} Γ (5 / 4) \neq Γ (5 / 4)$ . Thus $F$ is a different positive normalized solution of the recurrence.

step 1.1step 1.2algebra∎

Depends on

Used by

FALSE: normalization and the functional equation determine the real Gamma function False statement

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 7 §4 (standard reference, not scraped)