I've seen variations of this problem around, but none that quite match my situation. I'm asking because the given solution to this problem is much more complicated. Given a function $f:[0, \infty) \longrightarrow [0,\infty)$ with $f''(x) < 0 \; \forall \; x \in [0,\infty)$, I want to show $f$ is non decreasing $(f'(x) \geq 0)$ and $f$ is uniformly continuous.
For a contradiction, suppose $f'(x_0) < 0$. Since $f'(x)$ is always decreasing, for any $y > x_0$ we have $f'(y) < f'(x_0)$. In an integral, this means $$\int_{x_0}^yf'(x)dx \leq f'(x_0)(y-x_0) < 0$$
I have a feeling that's where my mistake is. Using familiar properties of integrals, we apply this to get
$$\int_0^y f'(x)dx = \int_0^{x_0}f'(x)dx + \int_{x_0}^yf'(x)dx \leq \int_0^{x_0}f'(x)dx + f'(x_0)(y-x_0)$$
The FTOC says $$f(y) - f(0) \leq f(x_0) - f(0) + f'(x_0)(y-x_0)$$ $$f(y) - f(x_0) \leq f'(x_0)(y-x_0)$$
This object on the right goes to $-\infty$ as $y \longrightarrow \infty$, which says there is some $y$ for which $f(y) - f(x_0) < -f(x_0)$. But then $f(y) < 0$. This contradicts the fact that $f$ is always positive. Is this right? Is there a way to do this with MVT and not integrals?
For uniform continuity, choose $\varepsilon > 0 $. Since $f'(x)$ is always decreasing, we have $f'(0) > f'(x) > 0 $ for all $x$. Let $\delta = \varepsilon/f'(0)$. For any pair $x<y$, the MVT gives us a $c \in (x,y)$ so that $$f(y) - f(x) = f'(c)(y-x)$$
But then $|y-x| < \delta$ implies $$|f(y)-f(x)| = f'(c)(y-x) < f'(0)\delta = \varepsilon$$
and we're done.
If $g'<0$ then by the MVT, $g$ is strictly deceasing. Applying this with $g=f',$ we see that $f'$ is strictly decreasing. Now if $f'(x_0)<0$ then for any $x>x_0,$ the MVT says there is $y\in (x_0,x)$ with $$f'(x_0)>f'(y)=\frac {f(x)-f(x_0)}{x-x_0},$$ implying $$f(x)<(x-x_0)f'(x_0)+f(x_0).$$ But the RHS above goes to $-\infty$ as $x\to\infty,$ contrary to the requirement that every $f(x)\ge 0$.
For uniform continuity of $f$: We have $0\le f'(y)\le f'(0)$ for all $y\ge 0.$ Given $e>0,$ take $d>0$ such that $d\cdot f'(0)<e.$ Now if $0\le x<x',$ there is $y\in (x,x')$ with $$\left|\frac {f(x)-f(x')}{x-x'}\right|=|f'(y)|=f'(y)\le f'(0). $$ Therefore $$|x-x'|<d\implies |f(x)-f(x')|\le |x-x'|\cdot f'(0)\le d\cdot f'(0)<e.$$