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+\|f-g\|\_t\big) \\
\|f-g\|\_t &\le& C\sqrt{t}\left(\|\alpha-\beta\|\_t+\|f-g\|\_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 $[0,T]$?



Do these two pairs coincide at small time?
$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$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+\|f-g\|\_t\big) \\
\|f-g\|\_t &\le C\sqrt{t}\left(\|\al-\be\|\_t+\|f-g\|\_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 $T>0$.


