Reducing the Degrees of a Graph where All Degrees are Quite Concentrated Suppose the degrees of all vertices in a graph $G$ (say, a bipartite graph) are all in a very concentrated range $[d(1 - o(1)), d(1 + o(1))]$, with $d$ an arbitrarily large constant. Here $o(1)$ is a term depending on $d$ so that the ratio of it to $d$ tends to zero as $d \to \infty$. For example, $d + \log d = d(1 + o(1))$. 
Is there a systematic way to obtain (or does there even exist) a spanning subgraph (a graph with all vertices of $G$) $G'$ of $G$ so that all degrees of all vertices in $G'$ are all in the range $[d'(1-\epsilon), d'(1+\epsilon)]$, where $d' < d$ can still be arbitrarily large and $\epsilon$ can be arbitrarily close to zero? In other words, the ratio of the maximum degree to the minimum degree remains very tightly concentrated, but the average degree has decreased.
I tried looking for a way to do so using matchings, but the best I can come up with is a construction to reduce the maximum and minimum degrees by one each. Can I do better?
 A: Thanks to Gregory J. Puleo, I have learned that the Lovasz $(g,f)$ factor theorem can be applied to conclude the following:
Theorem 1: Let $G$ be a graph whose degrees are tightly concentrated in the range $[\delta, \Delta]$. For any range $[a,b]$ with $b > a$, $G$ contains a spanning subgraph whose degrees are in the range $[a,b]$ if $\Delta / \delta < b/a$.
To prove this, we will make use of the Lovasz $(g,f)$ factor theorem. We will first need a couple of definitions. A spanning subgraph $H'$ of a graph $H = (V,E)$ is a subgraph of $H$ such that $V(H') = V(H)$. Let $g$ and $f$ be integer-valued functions on the vertices of a graph $H$, and further require that $g(v) < f(v)$ for all $v \in V(H)$. A $(g,f)$ factor is a spanning subgraph of $H$ is a graph $H'$ such that $g(v) \leq \deg(v) \leq f(v)$, for all $v \in V(H')$. The Lovasz Factor Theorem gives a necessary and sufficient condition on the existence of a $(g,f)$ factor.
Theorem 2 (Lovasz Factor Theorem): A graph $H$ has a $(g,f)$-factor if and only if
\begin{align}
\tag{1}
\sum_{t \in T} \big[\deg(t) - g(t) \big] + \sum_{s \in S} \big[f(s) \big] - e_H(S,T) \geq 0
\end{align}
for all disjoint subsets $S,T \subseteq V(H)$. Here $e_H(S,T)$ is the number of edges in $H$ going from a vertex in $S$ to a vertex in $T$.
With this, we can now prove theorem 1.
Proof of Theorem 1: Let us set $g(v) = a, f(v) = b$ for all $v \in G$. Then clearly $g(v) < f(v)$, so it remains to verify $(1)$, which in this case can be rearranged to
\begin{align}
\tag{2}
\sum_{t \in T} \big[\deg(t)\big] - a|T| + b|S| - e_H(S,T) = \sum_{t \in T} \big[\deg(t)\big] - (ar - b)|S| - e_H(S,T) \geq 0
\end{align}
where $r = |T|/|S|$. From here it's clear that if $r \leq b/a$, we would be done since the contribution of the middle term would be non-negative, and it is always the case that
$$\sum_{t \in T} \big[\deg(t)\big] - e_H(S,T) \geq 0$$
with equality only if every edge from a vertex in $T$ connects to a vertex in $S$. Let us now consider the case where $r > b/a$. Since $e_H(S,T) \leq \Delta \cdot \min(|S|, |T|)$, we have
$$\sum_{t \in T} \big[\deg(t)\big] - e_H(S,T) \geq \delta |T| - \Delta|S| = (\delta r - \Delta)|S| $$
Thus, the left-hand side of equation (2) can be lower-bounded by
$$\left(\delta r - \Delta - ar + b\right) |S| = ((\delta - a) r - (\Delta - b)) > \left(\frac{b}{a} \cdot (\delta - a) - (\Delta - b)\right)$$
But by assumption, $\Delta / \delta < b/a \implies (\Delta - b) / (\delta - a) < b/a$, so indeed the desired quantity is greater than zero. Thus by the Lovasz Factor Theorem we are done. $\square$
In particular, Theorem 1 implies that as long as $\Delta(d) /\delta(d) \to 1$, there is a spanning subgraph of $H$ with the desired property as long as $d$ is sufficiently large (as in the original question).
I have not yet worked out the algorithm to find such factors. There might also be other ways to prove theorem 1 without using such a powerful result in the Lovasz Factor Theorem. Any answers with improvements would be welcome!
