Let $\sigma=\sigma_{1}$ denote the classical sum-of-divisors function. Denote the deficiency of the deficient number $x>1$ by $D(x)=2x-\sigma(x)$.
Since $x>1$ is deficient, we have $D(x) \geq 1$. (Note that $D(x) \neq 0$ since deficient numbers cannot be perfect.)
Additionally, since $\sigma(x) \geq x+1$ for all $x>1$, it follows that $D(x)=2x-\sigma(x) \leq x-1$.
Thus, we have the bounds $$1 \leq D(x)=2x-\sigma(x) \leq x-1.$$
Here is my question:
Can these bounds be improved?
Added July 26 2016
There is a related MSE question here, where it is shown that the deficiency function $D(w)=2w-\sigma(w)$ satisfies $$D(xy) \leq D(x)D(y)$$ whenever $\gcd(x,y)=1$.
Added August 27 2016
I have a newer (and related) MSE question here.
No, they cannot be improved. In particular, there are infinitely many $x$ with $D(x) = 1$ and infinitely many with $D(x) = x-1$. For example, if $x = 2^{k}$, then $\sigma(x) = 2^{k+1} - 1$ and so $D(x) = 1$. On the other hand, if $x$ is prime, then $D(x) = 2x - (x+1) = x-1$.