Let $f(x)$ be a twice diff function such that $f''(x) \ge 0$ Let $f(x)$ be a twice diff function such that $f''(x) > 0$ and $f(x) \ge 0$ is bounded. Then show that $f(x)$ is constant.
I was going through the other answer on the website but I am not sure whether it made sense to me. So I decided to prove it on my own. Here's my attempt:
Assume that $f(x)$ is not constant then we know that we have a $x_0 \in \mathbb{R}$ such that $f(x_0) \ne f(x)$ for some $ x \in \mathbb{R}$. W l.o.g assume that $f(x_0) > f(x)$ for some $ x \in \mathbb{R}$
Then by langranges mean value theorem we have
$f(x_0)-f(x_0-h) = f'(c_1)(h)$ where $c_1 \in (x_0-h,x_0)$
Now $f'(c_1) > 0$ as $f''(x) \ge 0 $ it means that $f'$ is increasing.
Similarly in the interval $[x_0,x_0+h]$ we can find $c_2$ such that $f'(c_2) \ge f'(c_1)$. This goes on so that $f(x_0 + h)-f(x_0)=f'(c_2)(h) \ge f'(c_1)h$
Hence $f(x_0+h) \ge f'(c_1)h + f(x_0)$
Similarly we can show that $f(x_0+2h) \ge f(x_0) + 2f'(c_1)h$.
This show that $f(x)$ is unbounded which is a contradiction.
Is this proof okay?
 A: It suffices to prove that $f'=0$. Remark that the hypothesis $f\ge0$ can be relaxed to $f$ bounded.
By the absurd, take $x_0$ such that $f'\ne0$. Wlog, we can assume $f'(x_0)=c>0$, in fact if not, defining $g(x)=f(-x)$, we have $g'(x_0)=-f'(x_0)\>0$, $g''(x)=f''(x)\ge0$, $g$ bounded and $g$ constant if and only if $f$ constant.
Since $f'(x_0)=c>0$ and $f''\ge0$, $f'(x)\ge f'(x_0)=c$, $\forall x\ge x_0$. It follows
$$f(x)-f(x_0)=\int_{x_0}^x f'(t)dt\ge\int_{x_0}^x cdt=c(x-x_0),$$
that is $f(x)\ge c(x-x_0)+f(x_0)$. Hence, $f(x)\to+\infty$ as $x\to+\infty$, leading to a contradiction.
Basically what I have done it is just remarking that $f''\ge0$ means that $f$ is convex, which geometrically means that the graph of $f$ is always above the graph of its tangent (in any point). Since a straight line is not bounded from above if not constant, boundedness implies constantness.
A: I'm not sure how exactly do you conclude that $f'(c_2)\geq f'(c_1)$, but there is a bigger problem in your solution. A function is constant by definion if and only if
$$\forall x_1\neq x_2 : f(x_1)=f(x_2)$$
A negation of this sentence, in the logical meaning, would be:
$$\exists x_1, x_2 : x_1\neq x_2 \wedge f(x_1)\neq f(x_2)$$
So the first assumption, the "proof by contradiction" assumption, is incorrect. You claim that this holds for all $x$, but in fact it holds only for one specific pair of arguments.
