1
$\begingroup$

Let $x>0$; I have to solve the following inequality $$ \Gamma(x)\le\frac1x $$ Now \begin{align*} \Gamma(x)\le\frac1x \Longleftrightarrow x\Gamma(x)-1\le0 \end{align*} But calling $F(x):=x\Gamma(x)$ the previous inequality holds iff $$ F(x)-F(1)\le0 $$

and here maybe I should exploit some convexity property; I know that the Gamma function (on $\Bbb R_{>0}$) is positive log-convex, thus it's convex. The identity is also convex (non strictly); and even though the product of two convex functions in not necessarely so, I think it's true when the two functions are non negative; thus $F$ would be convex, but from here I can't continue!

EDIT: Noticing that $x\Gamma(x)=\Gamma(1+x)$, we shoud prove that $\Gamma(1+x)\le1$, and looking at the graphof the Gamma, this is true iff $1\le1+x\le2$, i.e. $0\le x\le1$ (then we should consider $x\neq0$, since we're dealing with the Gamma function); but how can I prove this analitically?

$\endgroup$
  • $\begingroup$ Exercise 3.32.(a) on p. 119 in the book quoted by math.stackexchange.com/questions/27571/… provides the following sufficient condition: let $f,g: \mathbf{R} \rightarrow \mathbf{R}$ be convex, both nondecreasing (or non increasing) and positive on an interval; then $f \cdot g$ is convex. $\endgroup$ – mlc Mar 27 '17 at 19:13
  • $\begingroup$ I didn't notice your edit at first: maybe you have written it while I was writing my answer... I do think that the variations of $\Gamma$ function imply that $\Gamma^{-1}(0,1)$ is $(1,2)$ when a rigorous proof. $\endgroup$ – Jean Marie Mar 27 '17 at 20:03
1
$\begingroup$

You already have all the key ingredients:

  1. $\log\Gamma$ is convex and $\Gamma$ is positive, hence $\Gamma$ is convex on $\mathbb{R}^+$;
  2. Actually $\Gamma''(x)\geq c > 0$ for any $x\in\mathbb{R}^+$;
  3. If $f(x)$ is a strictly convex and $C^2$ function on $\mathbb{R}^+$ the equation $f(x)=k$ cannot have more than two solutions; $\Gamma(x+1)=1$ has the solutions $x=0$ and $x=1$;
  4. It follows that $\forall x\in(0,1)$ we have $\Gamma(x+1)<1$, as well as $\Gamma(x+1)>1$ for any $x>1$.
$\endgroup$
1
$\begingroup$

As $x>0$, your inequality is equivalent to $x \Gamma(x) \leq 1$, itself equivalent, using the functional equation of $\Gamma$ function, to:

$$\Gamma(x+1) \leq 1$$

But $\Gamma$ function is decreasing on $(0,x_0)$, then increasing on $(x_0,+\infty)$ with $\Gamma(1)=\Gamma(2)=1$ (see for example (https://math.stackexchange.com/q/3247), where $x_0$ is explicited as being $\approx 1.46$).

Thus (1) is true for $x+1$ outside interval $[1,2]$, Equivalently $x$ has to be outside $[0,1]$, i.e $x>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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