Subgraph of G whose maximum degree and minimum degree are of the same order Consider a graph $G$ with max degree $\Delta_G$, min degree $\delta_G$ and average degree $d_G$. 

Is it possible to obtain a subgraph of $G$, say $G'$, such that  $\Delta_{G'} = c_1d_{G}$, $\delta_{G'} = c_2d_{G}$ , where $c_1,c_2$ are constants

EDIT 1: Also the size of the resultant subgraph $G'$ must be constant times the size of original graph $G$.

My attempt : 

Let $X$ be the random variable which denotes the degree of any randomly chosen vertex. Therefore, $d_G = E[X]$. Using markovs's inequality, we have

$$P(X &gt; 3E[X]) &lt; \frac{1}{3} $$
 and 
$$P(X &lt; \frac{E[X]}{3}) = P(\frac{1}{X} &gt; \frac{3}{E[X]}) &lt; \frac{E[\frac{1}{X}]*E[X]}{3} &lt; \frac{1}{3}$$

Using union bound, we have

$$P(\frac{E[X]}{3} &lt; X &lt; 3E[X]) &gt; 1/3$$.

Using the above equation, we can show that there exist $|V|/3$ vertices in $G$ whose degree is between $d_G/3$ and $3d_G$.

Does the graph obtained by the above $|V|/3$ vertices solve the problem? If not, can we modify the technique to obtain a solution?

Thanks in advance!
If you want to find an induced subgraph of "sufficient size" i.e not of constant size, then this is not possible. Consider the star $K_{1, n}$, which has average degree $2-1/n$.  Take any subset of the vertices. 


if the subset contains the central node, then the max degree equals the size of the subset. If this subset is more than a constant size, then it violates your condition.
if the subset does not contain the central node then the max degree is zero, which again violates your condition.