Proving a function is less or equal to a polynomial

Question

Lets say we got a differentiable function F: ℝ → ℝ and f(0) = a, f'(x) ≤ b for any x > 0.

Now, how to prove f(x) ≤ bx + a for any x ≥ 0 ?

It seems to be kinda simple, but I'm stuck.

Answer 1

Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) \ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,\infty)$ and since $h(0) = 0$, we must have $h(x) \ge 0$ for all $x\ge 0$. In other words, $g(x) \ge f(x)$ for all $x\ge 0$.

Answer 2

Integrals are monotone, hence you can just integrate $f'(x) \leq b$.

Answer 3

If $h(x)\le g(x)$ then $\int_a^b h(x)dx \le \int_a^b g(x)dx$

Answer 4

Given $x\gt 0$, we can find $\xi\in (0,x)$ such that $$f'(\xi)=\frac {f (x)-f (0)}{x-0}.$$ Since $\xi\gt 0$, $f'(\xi)\leq b$,$\implies \dfrac {f (x)-f (0)}{x-0}\leq b $. Substituting $f (0)=a $, we get $f (x)\leq bx+a $.

$x=0$ case is trivial.