Find $f(t)$ : $f(t) \leq \int_0^t f(s)ds$,$0 \leq t \leq 1$

Question

Exercise :

Find all the continuous and non-negative functions $f(t)$ such that:

$$f(t) \leq \int_0^t f(s)ds$$

Attempt :

I know there are some more "brutal" solving ways to this, such as a Laplace transformation or probably Mean Value Theorem tricks, but my idea is using the Gronwall Inequality, which says :

Let $φ(t)$ be a continuous function in $[0,T]$. Suppose that $\exists k>0$ and another continuous function $f(t)$, such that :

$$ φ(t) \leq f(t) + k\int_o^tφ(u)du, \forall t \in [0,T]$$

Then : $$φ(t) \leq f(t) + k\int_0^t f(τ)e^{t-τ}dτ,\forall t \in [0,T]$$

My problem is that I cannot see how to properly use the inequality or how to fix it, so I can show what is asked.

Any help would be appreciated !

Answer 1

As an alternative to Cuteboy's solution using Taylor series, you can simply use an integrating factor. Let $F(t) = \int^t_0 f(s) ds.$ Then your inequality is $$F'(t) \le F(t) \,\,\,\,\, \text{ or } \,\,\,\,\, F'(t) - F(t) \le 0.$$ Since $e^{-t}$ is non-negative, we can multiply the inequality by $e^{-t}$: $$e^{-t}F'(t) - e^{-t} F(t) \le 0 \,\,\,\, \implies \,\,\,\, \frac{d}{dt} (e^{-t}F(t)) \le 0.$$ Now integrating from $0$ up to some $t$ gives $$e^{-t}F(t) - e^{-0}F(0) \le 0$$ but $F(0) = 0$ so this shows that $$e^{-t} F(t) \le 0 \,\,\,\,\, \implies \,\,\,\,\, F(t) \le 0.$$ However this is possible iff $f \equiv 0$.

Answer 2

I don't know if there is a way to solve this by using Gronwall, but from a gronwall you can normally only expect some rough exponential estimation. As follows is a proof of mine by using Taylor expansion: Let $F(t):=\int_0^t f(s)ds$, then $F$ is of class $C^1$, monotone increasing and $F(0)=0$. From the condition we deduce that \begin{align} F'(t)\leq F(t). \end{align} Using Taylor we obtain that for each $t$ there exists some $\tilde{t}\in[0,t]$ such that \begin{align} F(t)=F(0)+F'(\tilde{t})t=F'(\tilde{t})t\leq F(\tilde{t})t\leq F(t)t. \end{align} for all $t\in[0,1]$. Since $F$ is non negative and $t\in[0,1]$, the only possibility is $F(t)=0$. Therefore $f(t)=0$ is the only solution.

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I may find a version of Gronwall's inequality which can directly answer your question, see https://de.wikipedia.org/wiki/Gronwallsche_Ungleichung (it is in German, but the formula given here is not given in the English one).

Answer 3

Surely you realize that if you take $f=0$ and $k=1$ in

Let $\varphi$ be a continuous function on $[0,T]$. Suppose that there exist $k>0$ and a continuous function $f$ such that $$ \varphi(t) \le f(t) + k\int_0^t\varphi(u)du, \quad\forall t \in [0,T].$$ Then $$\varphi(t) \le f(t) + k\int_0^t f(\tau)e^{t-\tau}d\tau,\quad\forall t \in [0,T].$$

you get

Let $\varphi$ be a continuous function on $[0,T]$ such that $$ \varphi(t) \le \int_0^t\varphi(u)du, \quad\forall t \in [0,T].$$ Then $$\varphi(t) \le 0,\quad\forall t \in [0,T].$$

and you assume that the function is nonnegative.