Given that $f:[0,1]\to [0, \infty)$ be a continuous function such that $$f(t)^2<1+2\int_{0}^{t}f(s)\:ds,\:\forall t \in [0,1] \to (1)$$ Then prove that $$f(t)<1+t ,\forall t \in [0,1]$$
I actually got the solution in this way: Differentiating $(1)$ both sides with respect to $t$ and using Leibnitz rule we have $$2f(t)f'(t)<2f(t) \to (2)$$ $\implies$ $$2f(t)\left(f'(t)-1\right)<0$$ Now since the range of $f(t)$ cannot be negative,we get $$f'(t)-1<0$$ $\implies$ $f(t)-t$ is a strictly decreasing function in $t \in [0,1]$ So we get $$f(t)-t<f(0)-0 \to (3)$$ Now in $(1)$ plugging in $t=0$ we get $$f(0)^2<1 \implies f(0)<1$$ So using $f(0)<1$ in $(3)$ we get $$f(t)-t<1 \implies f(t)<1+t, \forall t \in [0,1]$$ Now my query is in getting $(2)$. After differentiation how can we be sure that inequality is retained. Am i missing any concept there?
Let $g(t)=\int_{0}^{t}f(s)ds$ and $h(t)=\sqrt{1+2g(t)}-t-1,$ h is a continuously differentiable function.
I observe that $$h'(t)=\frac{g'(t)}{\sqrt{1+2g(t)}}-1<0.$$ So, $$h(t)\leq h(0)=0\Rightarrow 1+2g(t)\leq (1+t)^2$$ and as a consequence $$f(t)<1+t.$$