# Do these two pairs coincide at small time?

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 $$[0,T]$$?

$$\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$$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 $$T>0$$.