Suppose $f(0)=0$, and $f'(x) < \frac{1}{2}$. Show $f(4) < 2$

Question

Problem: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Suppose $f(0)=0$, and $f'(x) < \frac{1}{2}$ for all $a$. Show $f(4) < 2$.

This question is intuitively pretty simple, but I'm not sure what is generally regarded as sufficiently rigorous.

Simply, if the slope of the function never reaches $\frac{1}{2}$, then combining all these small intervals will produce $f(4) <2$.

$\forall a \in \mathbb{R}, \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a} < \frac{1}{2} \implies \forall a \in \mathbb{R}, f(x)-f(a) < \frac{1}{2}(x-a)$

Letting $a=0, x=4$, we have $f(4) < 2$.

My Question:

I'm not sure if I'm skipping a ton of steps in that implication above. It looks like I just jumped to the conclusion without proving anything.

Is what I'm missing some sort of "interval argument"?

For example: If I translated using the delta-epsilon definition of limits $$\forall a \in \mathbb{R}, \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$$

to $f(x)-f(a) < \frac{1}{2}(x-a)$ for some open interval $ x \in (a- \delta, a+ \delta)$.

I would still need to show that the union of infinitely many of such small intervals satisfies $f(x)-f(a) < \frac{1}{2}(x-a)$. How do I write this rigorously?

Answer 1

Use the mean value theorem

$$\frac{f(4)-f(0)}{4-0} = f’(c) < \frac12$$

where $0<c<4$. Then $f(4)<2$ follows.

Answer 2

$f(4)-f(0)=\displaystyle{\int_{0}^{4}}f'(x)dx< $

$\displaystyle{\int_{0}^{4}}(1/2)dx =(1/2)4=2$.