Let $\alpha, \beta:\mathbb R_+\to [0,1]$ be continuous and decreasing functions s.t. $\alpha(0)=1=\beta(0)$ and $\alpha, \beta$ are continuously differentiable on $(0,\infty)$ satisfying for some $c>0$ $$\frac{c}{\sqrt{t}}\le\alpha'(t), \beta'(t)\le 0,\quad \forall t>0.$$ Let $f,g :\mathbb R_+\to [0,1]$ be continuous s.t. for all $t\ge 0$ \begin{eqnarray} \\alpha\beta\_t &\le& C\sqrt{t} \big(\\alpha\beta\_t+\fg\_t\big) \\ \fg\_t &\le& C\sqrt{t}\left(\\alpha\beta\_t+\fg\_t + \int_0^t\alpha'(s)\beta'(s)ds\right), \end{eqnarray} where $C>0$ is some constant and $\\cdot\_t$ denotes the uniform norm of functions on $[0,t]$. Can we prove the existence of $T>0$ s.t. $(\alpha,f)=(\beta,g)$ over the interval $[0,T]$?
1 Answer
$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$The answer is no.
E.g., suppose that for all small enough $t>0$ we have \begin{equation*} f(t)=t,\quad g(t)=0,\quad \al'(t)=\frac1{\sqrt t}\,\Big(1+\sin\frac1t\Big),\quad \be'(t)=\frac1{\sqrt t}, \end{equation*} with $\al(0)=1=\be(0)$.
Then all your conditions hold. In particular, for $t\downarrow0$,
\begin{equation*}
\al(t)\be(t)=\Big\int_{1/t}^\infty\frac{\sin z}{z^{3/2}}\,dz\Big=O(t^{3/2})
\end{equation*}
and
\begin{equation*}
\int_0^t\al'(s)\be'(s)\,ds=\int_{1/t}^\infty\frac{\sin z}{z^{3/2}}\,dz\asymp t^{1/2},
\end{equation*}
so that
\begin{equation*}
\begin{aligned}
\\al\be\_t &\le C\sqrt{t} \big(\\al\be\_t+\fg\_t\big) \\
\fg\_t &\le C\sqrt{t}\left(\\al\be\_t+\fg\_t + \int_0^t\al'(s)\be'(s)ds\right)
\end{aligned}
\tag{1}\label{1}
\end{equation*}
for some real $C>0$ and all $t$ in some right neighborhood of $0$. So, inequalities \eqref{1} hold for some real $C>0$ and all real $t\ge0$.
Yet, $f\ne g$ in any right neighborhood of $0$. So, $(\al,f)\ne(\be,g)$ on $[0,T]$, for any real $T>0$.

$\begingroup$ Thank you very much Iosif for your example. Indeed, my family is contaminated by covid and I have to stay at home (with my laptop left in office). So sorry for not replying in time $\endgroup$– GJC20Jul 12, 2022 at 12:09

$\begingroup$ @GJC20 : Thank you for your response. Best wishes to you and your family! $\endgroup$ Jul 12, 2022 at 13:48