Baby rudin theorem 8.19

Question

Theorem:

If $f$ is a positive function on $(0,\infty)$ such that

(a)$ f(x+1) = xf(x),$

(b) $f(1) = 1$,

(c) $\log f$ is convex,

then $f(x) = \Gamma(x)$.

Proof:

Since $\Gamma$ satisfies (a), (b), and (c), it's enough to prove that $f(x) $ is uniquely determined by (a),(b),(c), for all $x>0.$ By (a), it's enough to do this for $x \in (0,1).$

Put $\varphi = \log f$. Then

$\varphi(x+1)$ = $\varphi(x)$ + logx (0<x<$\infty$),

$\varphi(1)$ = 0, and $\varphi$ is convex. suppose 0 < x < 1, and n is a positive integer.

By ($\varphi(x+1) = \varphi(x) + \log x $ ($0<x<\infty$)), $\varphi(n+1) = \log (n!)$.

Consider the difference quotients of $\varphi$ on the intervals $[n,n+1], [n+1,n+1+x],[n+1,n+2]$. since $\varphi$ is convex

$$\log n \leq \frac{\varphi(n+1+x)-\varphi(n+1)}{x}\leq \log(n+1).$$

I don't understand how the last equation comes from the convexity of $\varphi$.

Any help would be appreciated.

Answer 1

If $\phi$ is convex function and $s<y<z<v$ then $$\frac{\phi (y) -\phi (s) }{y-s}\leq \frac{\phi (z) -\phi (y) }{z-y}\leq \frac{\phi (v) -\phi (z) }{v-z}$$ put in this inequality $s=n, y=n+1 , z= n+1+x, v=n+2$ and yóu obtain it.

Answer 2

You actually want to use the inequality:

$$\frac{\phi(y)-\phi(s)}{y-s}\leq \frac{\phi(z)-\phi(y)}{z-y} \leq \frac{\phi(v)-\phi(y)}{v-y}$$ which holds for convex $\phi$ and $s<v<y<z$.

Plug in $s=n,y=n+1,z=n+1+x,v=n+2$.

You can understand why the inequality is true by drawing a picture. The expressions above are the slopes of the secant lines for intervals where the upper end of the interval always increases and since the second derivative is positive the first derivative is increasing so the slopes of the secant lines are increasing.