Can we find a real number $a>0$ such that $f(t)>a>0$ and $a$ does not depend of the value of $t$. Let $f$ be a strictly positive real function defined on the interval $(0,1)$. Let $t∈(0,1)$ be fixed. 

Can we find a real number $a&gt;0$ such that $f(t)&gt;a&gt;0$ and $a$ does not depend of the value of $t$.
If $t$ is fixed as you say, the existence of a function $f$ defined on $(0,1)$ is irrelevant, and the requirement "does not depend on the value of $t$" is vacuous. All we have is that the real number

$$f(t)$$

is positive and obviously there is an $a$ such that

$$f(t)&gt;a&gt;0.$$



Anyway, $a$ must depend on the value of $f(t)$, for example as $a=\dfrac{f(t)}2$. You can't pick a suitable value for $a$ without knowing $f(t)$, because $\mathbb R_+^*$ has no minimum.
There is no such $a$. Fix $t \in (0,1)$ and imagine that there is some constant $a$ that doesn't depend on $t$ such that 
$$f(t)&gt; a &gt; 0.$$

Since $a$ doesn't depend on $t$ we must have $f(t') &gt; a &gt; 0$ for any other $t' \in (0,1).$ 

Now consider any function which is strictly positive such that $f(t) \rightarrow 0$ as $t \rightarrow 0$ (e.g. $f(x) = x$ on $(0,1)$) then taking the limit as $t \rightarrow 0$ in 
$$ f(t) &gt; a &gt; 0$$
gives us 
$ 0 \geq a \geq 0.$ Therefore $a = 0$ which contradicts the fact that $a &gt; 0.$