Find all the continuous functions that satisfy $[f(t)]^2=F(t)-F(0)$

Question

I need to find all the continuous functions $f$ such that $$ [f(t)]^2=\int_0^t f(s) ds $$ Attempt:

Since $$ f(t)\leq [f(t)]^2+c, \, \, \forall c \geq \frac14$$ we have

$$ f(t)\leq c+ \int_0^t f(s)ds $$ Now, Gronwall's Lemma gives $$ f(t) \leq c+ \int_0^t c \,e^{t-\tau}d\tau \iff$$ $$ f(t) \leq c \, e^t, \, \,\forall c\geq \frac14$$

Given that these steps were correct, is this inequality the final answer?

Answer 1

Just observe that $f(0)=0$ and then take the derivative $$ 2ff'=f\implies f=0\lor 2f'=1,~f=\frac t2. $$ As per a hint of Ian in a prior comment, you can also combine a segment where $f=0$ and then the nullity of $f(t)(2f'(t)-1)=0$ switches to the second factor generating a solution $f(t)=\frac12\max(0, t-a)$.

Answer 2

You can use the Lemma use to prove the Fundamental Theorem of Calculus:

$$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$

Here we have $[f(t)]^2=\int_0^tf(s)ds.$

Take the $t$ derivative of both sides.

So $2f(t)f'(t)-f(t)=0=f(t)[2f'(t)-1].$

So either $f(t)=0$, or $2f'(t)-1=0\implies f(t)=t/2+c$.