Does $f(t) \leq - \int_0^t f(s) ds \implies f \leq 0$? I don't know if $f(t) \leq - \int_0^t f(s) ds \implies  f \leq 0$ is true or not (f is a continuous function with $f(0) = 0$). The range of $t$ is $[0,\infty[$
I can show that it implies that $\int_0^t f(s) ds \leq 0$ (by multiplying by $e^{t}$ and integrating) but there are function f that are not negative everywhere such that $\int_0^t f(s) ds \leq 0$ (really negative next to $0$ and then positive but small).
For the context, I am trying to get information about $g \geq 0$ when I know that $g(t) + \int_0^t g(t) \leq t$. I would like to obtain a kind of gronwall bound, with a "small" contant (not exponential) as:   $g(t)\leq C t$ or better something with not depend on $t$.
 A: Concerning your main question, one way to construct counterexamples is to take $f$ to be negative on the time interval $[0,5]$ in such a way that the inequality is satisfied, for instance $f(t)=-e^{t}+1$; then bring the function back to zero in another interval $[5,6]$ (in such a way that $f(t)\leq 0$ on $[5,6]$ and $f(6)=0$). At that point, you would have
$$ f(6)=0,\qquad \int_0^6f(s)ds\leq -e^{5}+6. $$
Now you can choose $f$ to be very small, say less that $1$ on the interval $[6,7]$, and then constantly equal to zero on $[6,+\infty[$. The integral of $f$ is still very negative, so that $-\int_0^tf(s)ds\geq e^5-7>2$ for $t\in[6,\infty[$, but $f$ is less than $1$ on the same interval.
Putting everything together, you end up with a counterexample to the main inequality. Explicityly, you can define $f$ to be
$$ f(t)=\left\{\begin{aligned}-e^{t}+1,\quad&t\in[0,5]\\-(1-e^5)(t-6),\quad &t\in[5,6]\\\sin(\pi(t-6)),\quad&t\in[6,7]\\0,\quad&t>7\end{aligned}\right. $$
Edit. Note in particular that you can never have a bound of the form $f(t)\leq h(t)$ for some function $h$. In fact, you can scale my counterexample in such a way that $f$ is insanely negative on the interval $[0,t/2]$, so that $-\int_0^tf(s)ds$ grows to be insanely positive on the same interval; then bring $f$ back to zero on an interval $[t/2,t-\varepsilon]$, and then make $f$ grow in such a way that $f(t)=h(t)+1$. If $\varepsilon$ is small enough, you can do this in such a way that $-\int_0^tf(s)ds$ still remains large.
While I was writing this, I figured that the above inequality is linear, so that if you find a function $f$ that satisfies it, it is also satisfied for $2f$, $3f$,...
