Non-increasing continuous function with zero initial value

Question

Let $\phi : [0,\infty) \to [0,\infty)$ be continuous on $[0,\infty)$ and once differentiable in $(0,\infty)$. Assume that $\phi(0)=0$ and $\forall t \in (0,\infty), \phi'(t) \leq 0$. I want to show that $\forall t \in [0,\infty), \phi(t) = 0$.

So, I need to claim that $\forall t > 0, \phi(t) \leq \phi(0)$ but I am not sure how to rigorously show this claim. I try to use contradiction.

Assume $\exists t_{1} > 0, \phi(t_{1}) > \phi (0)$. Then, I do not know how to proceed for the details. Any hint is much appreciated!

I am sorry to ask this seemingly "trivial" question here and thank you!

Answer 1

Hint: Use the mean value theorem to derive a contradiction.

Answer 2

We have:

$(1) \quad\phi(t) \ge 0$ for all $t \ge 0.$

$(2) \quad\phi'(t) \le 0$ for all $t >0.$

$(3) \quad \phi(0)=0.$

$(2) $ shows that $ \phi$ is decreasing, hence, by $(3)$: $ \phi(t) \le \phi(0)=0$ for $t>0.$

Now $(1)$ gives the desired result.