How to prove the use of boundedness may be the key to this problem Let $f$ be a bounded convex function on $(0,+\infty)$,prove that $f$  is decreasing and
$$\displaystyle\lim_{x\to+\infty}xf'_+(x)=0.$$
Geometrically speaking, the significance of the first problem is obvious, but I don't know how to analyze and prove it.I try to use the following inequality$$\dfrac{f(x_2)-f(x_1)}{x_2-x_1}\leqslant\dfrac{f(x_3)-f(x_1)}{x_3-x_1}\leqslant\dfrac{f(x_3)-f(x_2)}{x_3-x_2}.$$I don't know how to take advantage of boundedness...
 A: The use of boundedness
Suppose $f$ is not decreasing. Then there are numbers $0\le a<b$ such that $f(a)<f(b)$. Let $c=b-a$.
By convexity, $f(a+2c)+f(a)\ge 2f(a+c)$ and so $$f(a+2c)-f(a+c)\ge f(a+c)-f(c).$$
Similarly, $$f(a+3c)-f(a+2c)\ge f(a+2c)-f(a+c)$$ and so on. Therefore, summing these inequalities, $$f(a+nc)-f(a)\ge n(f(a+c)-f(a))$$ and so $f(a+nc)$ is unbounded as $n$ tends to infinity.
This contradiction proves that $f$ is decreasing.
A: For the first part we need that $f$ is bounded above, e.g. by $M$: For $0 < x < y < z$ we have
$$
 f(y) \le \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(z) \le \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x} M \, .
$$
Now take the limit for $z \to \infty$ to conclude that $f(y) \le f(x)$, i.e. $f$ is decreasing.
For the second part we need that $f$ is also bounded below: $f$ is  decreasing which implies that $L = \lim_{x \to \infty} f(x)$ exists. Now write the convexity condition for $0 < x/2 < x < y$ as
$$
 \frac{f(x)-f(x/2)}{x-x/2} \le  \frac{f(y)-f(x)}{y-x} \le 0 \, .
$$
Taking the limit $y \to x^+$ gives
$$
 \frac{f(x)-f(x/2)}{x-x/2} \le f'_+(x) \le 0 \\
\implies 2 \bigl(f(x) - f(x/2) \bigr) \le x f'_+(x) \le 0 \, .
$$
For $x \to \infty$ the left-hand side converges to $2 (L-L) = 0$, and it follows that $\lim_{x \to \infty} x f'_+(x) = 0$.
