# let $f:[0,\infty) \to [0,\infty)$ be a cont. diff s.t $f(0)=0$ and $f'$ … then $\forall s,t \geq 0 ; 0 \leq f(s+t) \leq f(s)+f(t)$

let $f:[0,\infty) \to [0,\infty)$ be a contionus function , differential such thatt $f(0)=0$ and $f'$ is non-negative and monotone decreasing

then
$$\forall s,t \geq 0 ; 0 \leq f(s+t) \leq f(s)+f(t)$$

definition

f' is monotone decreasing so $$x_1 \leq x_2 \Rightarrow f'(x_1)\geq f'(x_2)$$

Attempt 1

The hint is to use the fundemental theorem of calc. along with an derivative non negative. not suprise if its a duplicate cannot find it.

But Tried to use mean value theorem

\begin{aligned} f'(c)t &=\frac{f(s+t)-f(t)}{t} \\f'(c)t+f(s)&=f(s+c) \end{aligned} $f'(c) >0$... go into cases stuck not sure where to go or if took a wrong turn somewhere.

hand sketch of what is going on

Since $f'$ is nonnegative and monotone decreasing, we know that $f'>0$ and as such $f$ is increasing. Because $f(0)=0$, this shows the first inequality, that is, $f(s+t)\geq 0$ whenever $s,t \geq 0$.

For the other inequality, I would like you to consider the integrals:

\begin{aligned} &\int_{0}^{s+t}f'(x)\,dx=f(s+t)-f(0)=f(s+t)\\ &\int_{0}^sf'(x)\,dx=f(s)\\ &\int_{0}^tf'(t)\,dx=f(t)\\ \end{aligned}

(because $f(0)=0$)

The second inequality may thus be rewritten as:

$$\int_{0}^{s+t}f'(x)\,dx\leq \int_{0}^sf'(x)\,dx + \int_{0}^tf'(t)\,dx$$

or, rearranging and assuming without loss of generality that $s \leq t$:

$$\int_{t}^{s+t}f'(x)\,dx\leq \int_{0}^sf'(x)\,dx$$

Now this inequality should be fairly obvious: we're integrating both functions over intervals with the same length ($s$), but the interval in the LHS is strictly to the right of the interval on the RHS. Since $f'$ is monotone decreasing, the integrand on the LHS is smaller at each point than the integrand on the RHS.