MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Bohr–Mollerup theorem characterizes the Gamma function $\Gamma(x)$ as the unique function $f(x)$ on the positive reals such that $f(1)=1$, $f(x+1)=xf(x)$, and $f$ is logarithmically convex, i.e. $\log(f(x))$ is a convex function.

What meaning or insight do we draw from log convexity? There's two obvious but less than helpful answers. One is that log convexity means exactly what the definition says, no more and no less. The other is the more or less circular one that since the Gamma function is so important, any property that characterizes it is also significant.

The wikipedia article point out that "a logarithmically convex function is a convex function, but the converse is not always true" with the counterexample of $f(x)=x^2$. The only logarithmically convex examples in the article come trivially from exponentiating convex functions, and the example $\Gamma(x)$.

Let me say in advance that I'm less interested in the Gamma function than I am in the notion of log convexity, so this question is not a duplicate of

Importance of Log Convexity of the Gamma Function

A thoughtful answer by Andrey Rekalo to that question, is that functions which can be realized as finite of moments of Borel measures are log convex functions. But I'm more interested in things that are implied by (v. imply) log convexity.

My real motivation is the fact that the Riemann Hypothesis implies that the Hardy function is log convex for sufficiently large $t$. (The Hardy function $Z(t)$ is just $\zeta(1/2+it)$ with the phase taken out, so $Z(t)$ is real valued and $|Z(t)|=|\zeta(1/2+it)|$.) This is in Edward's book 'Riemann's Zeta Function' Section 8.3, in the language that RH $\Rightarrow Z^\prime/Z$ is monotonic. This says that between consecutive real zeros, $-\log|Z(t)|$ is convex.

Any insight would be welcome.

share|cite|improve this question
Maybe this paper is relevant:… – weakstar Jan 3 '11 at 20:28
log convexity leads one to conclude that the Riemann Hypothesis implies the Lindelof hypothesis. Are you perhaps thinking about a converse to Hardy's Three Circle Theorem? – Alex R. Jan 3 '11 at 20:34
@Alex - Thanks! This is probably something I should have already known, but it's the kind of answer I was looking for. – Stopple Jan 3 '11 at 20:35
Whoops not Hardy, Hadamard three circle... – Alex R. Jan 3 '11 at 20:46
up vote 4 down vote accepted

From the comments, log convexity leads one to conclude that Riemann Hypothesis implies Lindelof hypothesis. The implication of log convexity comes from Hadamard Three Circle Theorem

share|cite|improve this answer

Log convexity pops up in lots of different places (eg, the determinant of positive definite matrices is log-concave, the perron-frobenius eigenvalue of positive matrices is log convex, volumes of convex bodies are log concave, etc, etc). For a discussion having absolutely nothing to do with the gamma function, see S. Boyd and L. Vanderberghe's book "Convex optimization" (page 105). PDF available for free on Boyd's web site (google "Stephen Boyd Stanford")

share|cite|improve this answer
Igor, you are not responsible for that of course, but these log-convexity statements do not refer to the same definition of LC. In general, a log-convex function is a function $x\mapsto f(x)>0$ such that $x\mapsto\log f(x)$ is convex (as stated in the question). Instead, the log-convexity of the Perron-Frobenius eigenvalue $\lambda_{pf}$ is the convexity of $\log\lambda_{pf}(A)$ in terms of the variables $\log a_{ij}$. Something significantly different! – Denis Serre Jan 4 '11 at 6:58
I guess Denis then the lc of the Perron-Frobenius eigenvalue should be actually called "multiplicative convexity". – Suvrit Jan 4 '11 at 9:25

Many common probability densities are log-concave, which turns out to be a very useful property. For one thing, it's a way to quantify that a density is something like a normal density, perhaps sufficiently like a normal for a theorem to generalize.

Also, og-concave densities satisfy nice theorems. For example, if the PDF of a random variable is log-concave, the CDF is as well. Also, the convolution of two log-concave densities is also log-concave.

share|cite|improve this answer
Also, if you take a log-concave function of many variables and integrate some out, the result is again log-concave. This gives the convolution statement above and also Igor's one about convex bodies. Which prompts the question: is there a natural log-concave function of $n$ variables, integrable in the first $n-1$, whose integral is $\Gamma$? – Allen Knutson Jan 4 '11 at 5:31
@Allen. I had this question in mind. Could be a valuable MO question. – Denis Serre Jan 4 '11 at 6:27

In my opinion, some of the reasons that make log-convexity (l.c.) nice / useful are:

  1. If $f$ and $g$ are l.c., then $f+g$ and $fg$ are both log-convex (note log-concavity on the other hand is not closed under addition, though as Allen commented above, it is closed under convolution (Prekopá's theorem))

  2. L.c. functions are related to completely monotonic functions (i.e., functions for which $(-1)^n f^{(n)}(x) \ge 0$ for all $x > 0$, and $n \in \mathbb{N}$), in that any completely monotonic function is also l.c.; If I recall correctly, one place completely monotonic functions come up naturally is when considering Laplace transforms. Another example: the Laplace transform of any non-negative function is l.c.

  3. L.c. is closely related to multiplicatively convex functions (i.e., $f(\sqrt{xy}) \le \sqrt{f(x)f(y)}$), and it is known that the logarithmic integral is multiplicatively convex.

    Many more nice properties may be found in the book: Convex functions and their applications: A contemporary approach by C. P. Niculescu and L.-E. Persson.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.