Does $\varphi(x)\le\int_0^x\varphi(t)dt$ for all $x\in[0,\infty)$ imply $\varphi\equiv0$. Let $\varphi$ be a nonnegative and continuous function on $[0,\infty)$ and such that $$\varphi(x)\le\int_0^x\varphi(t)dt$$ for all $x\in[0,\infty)$. Can we infer from here that $\varphi\equiv0$.
By taking limit on both sides, I got $\varphi(0)\le 0$. As $\varphi$ is nonnegative, I can say $\varphi(0)=0$. But what then. I could have progressed if I could show that $\varphi(x)\le \varphi'(x)$, but it is not quite evident from the given condition.
 A: Consider $\varphi$ on the interval $[0,1]$. Then by the Extreme Value Theorem $\varphi$ has a maximum value on it and assume it attains it for $M \in [0,1]$. Then by the condition we have:
$$\varphi(M) \le \int_0^M \varphi(x) dx  \le \int_0^M \varphi(M) dx = \varphi(M) \cdot M$$
If $M=1$ then from the obtained equality in we get that $\varphi(x) = \varphi(M)$ for all $x \in [0,1]$, but this gives us that $\varphi(x) = \varphi(0) = 0$ for all $x \in [0,1]$. Otherwise if $M \in [0,1)$ we have that $\varphi(M) = 0$ and as $0 = \varphi(M) \ge \varphi(x) \ge 0$ for $x \in [0,1]$ we have that $\varphi(x) = 0$ on $[0,1]$
Now similarly consider $\varphi$ on $[1,2]$. It has a maximum value $N$. Then by the condition:
$$\varphi(N) \le \int_0^N \varphi(x) dx  = \int_1^N \varphi(M) dx = \varphi(N) \cdot (N-1) \implies \varphi(N) = 0$$
Here we used the fact that $\int_0^1 \varphi(x) dx = 0$ and that $N-1 \le 1$ with similar arguement as above.
Therefore going by a interval of a length $1$ at a time we can prove that $\varphi(x) = 0 $ on $[0, \infty)$
A: Using the MVT we can rewrite the inequality as
$$\varphi(x)\le \varphi(\xi)x,\quad\forall x\in[0,\infty)\tag1$$
for some $\xi\in[0,x]$. Then:


*

*If $x< 1$ and $\varphi(x)\neq 0$, then by the continuity and non-negativity of $\varphi$ we have that
$$0<\varphi(x)/x\le\varphi(\xi_1)\implies \varphi(\xi_1)/\xi_1\le\varphi(\xi_2)\implies\cdots\implies\varphi(\xi_k)/\xi_k\le\varphi(\xi_{k+1})$$
Thus observe that the sequence defined by $(\xi_k)$ is decreasing and bounded below by zero so it converges to some point in $[0,1]$. Hence applying limits to both sides of $\varphi(\xi_k)/\xi_k\le\varphi(\xi_{k+1})$ we find that
$$\begin{matrix}\infty\le\varphi(0),\quad\text{whenever }\lim\xi_k= 0\\\text{or}\quad\varphi(c)/c\le\varphi(c),\quad\text{whenever }\lim \xi_k=c>0\end{matrix}$$
In any case we reach a contradiction so we conclude that $\varphi(x)=0$ when $x\in[0,1)$.

*Assume, from the previous part, that $\varphi(x)=0$ for $x\in[0,1)$. Then, by continuity, $\varphi(1)=0$ also. Suppose now that $\varphi(x)>0$ for some $x\in(1,2)$. Then the original inequality implies
$$\varphi(x)\le\int_1^x\varphi(t)\,\mathrm dt\implies\varphi(x)=(x-1)\varphi(\xi),\quad\forall x\in[1,2)\tag2$$
for some $\xi\in[1,x]$. Thus repeating the same recursion than in the previous part we have that
$$\begin{matrix}\infty\le\varphi(1),\quad\text{whenever }\lim\xi_k= 1\\\text{or}\quad\varphi(c)/(c-1)\le\varphi(c),\quad\text{whenever }\lim \xi_k=c>1\end{matrix}$$
In any case we find that necessarily $\varphi=0$ also in $[1,2)$. 

*Finally, by the previous cases, using strong induction on the intervals $[k,k+1)$ it can be shown that $\varphi=0$ for all $x\in[0,\infty)$, as desired.

For sake of completion I will add a simple version of the Gronwall's lemma that I know from Analysis II of Amann and Escher: 

Consider a non-empty interval $J$, $t_0\in J$, $\alpha,\beta\in[0,\infty)$ and a continuous function $y:J\to[0,\infty)$ that satisfies
$$y(t)\le\alpha+\beta\left|\int_{t_0}^ty(s)\,\mathrm ds\right|,\quad\forall t\in J$$
then $$y(t)\le\alpha e^{\beta|t-t_0|},\quad\forall t\in J$$

Then applying this theorem to the exercise we can see that $\alpha=0$, consequently $\varphi=0$.
A: The usual technique is to consider $$g(x) =e^{-x}\int_{0}^{x}\varphi(t)\,dt$$ and then $$g'(x) =e^{-x} \left(\varphi(x) - \int_{0}^{x}\varphi(t)\,dt\right)\leq 0$$ so that $g$ is decreasing and hence $g(x) \leq g(0)=0$. But $g$ is non-negative and hence $g(x) =0$ for all $x\geq 0$ and therefore $\int_{0}^{x}\varphi(t)\,dt=e^{x} g(x) $ and it's derivative $\varphi(x)$ vanish for all $x\geq 0$.
A: By Grönwall's inequality (integral form), 
$$
\varphi(x) \leq \int_0^x \varphi(t) dt \implies \varphi(x) \leq 0,
$$
thus $\varphi \equiv 0$.
