If we consider the complete bipartite graph $K_{n,m},$ where $n,m$ are natural numbers with $n < m,$ then the maximum degree of a vertex in this graph is $=m$ and there are at least $n$ many vertices having degree $m$, and so the number of "kings" (as per the definition above) in this graph is $ = n,$ and the total number of vertices $=n+m.$

Now, since $n,m$ can be anything (with $n \le m$), so one can take $n,m$ to be such that $n/(n+m) > \frac{1}{3},$ and thus in this case we see that the number of kings in this graph is $> \frac{1}{3} \cdot$ the number of vertices in the graph.

For such graphs you can just set $n=m-1$ to see that you cannot get a bound $\le 1/2$ for instance; you may be interested to see that the bound $1/2$ is attained asymptotically thus implying that the "asymptotic" bound of $1/2$ cannot be reduced any further.

If we consider the complete bipartite graph $K_{n,m},$ where $n,m$ are natural numbers with $n < m,$ then the maximum degree of a vertex in this graph is $=m$ and there are at least $n$ many vertices having degree $m$, and so the number of "kings" (as per the definition above) in this graph is $ = n,$ and the total number of vertices $=n+m.$

Now, since $n,m$ can be anything (with $n \le m$), so one can take $n,m$ to be such that $n/(n+m) > \frac{1}{3},$ and thus in this case we see that the number of kings in this graph is $> \frac{1}{3} \cdot$ the number of vertices in the graph.

For such graphs you can just set $n=m-1$ to see that you cannot get an "asymptotic" bound of $< 1/2.$

If we consider the complete bipartite graph $K_{n,m},$ where $n,m$ are natural numbers with $n \le m,$ then the maximum degree of a vertex in this graph is $=m$ and there are at least $n$ many vertices having degree $m$ (the number of such vertices will be more if $n=m$), and so the number of "kings" (as per the definition above) in this graph is $ \ge n,$ and the total number of vertices $=n+m.$

Now, since $n,m$ can be anything (with $n \le m$), so one can take $n,m$ to be such that $n/(n+m) > \frac{1}{3},$ and thus in this case we see that the number of kings in this graph is $> \frac{1}{3} \cdot$ the number of vertices in the graph. (For instance taking $n=m,$ or $n=m-1$ one can see what happens.)

Setting $n=m$ also shows that one can have every vertex to be a king, thus implying the tightness of the bound $1$ for $\frac{\text{number of kings}}{\text{number of vertices}},$ this graph will be regular of course if the upper bound is attained; for non-regular graphs you can just set $n=m-1$ to see that you cannot get a bound $\le 1/2$ for instance; you may be interested to see that the bound $1/2$ is attained asymptotically thus implying that even for non-regular graphs, the "asymptotic" bound of $1/2$ cannot be reduced any further.

If we consider the complete bipartite graph $K_{n,m},$ where $n,m$ are natural numbers with $n < m,$ then the maximum degree of a vertex in this graph is $=m$ and there are at least $n$ many vertices having degree $m$, and so the number of "kings" (as per the definition above) in this graph is $ = n,$ and the total number of vertices $=n+m.$

Now, since $n,m$ can be anything (with $n \le m$), so one can take $n,m$ to be such that $n/(n+m) > \frac{1}{3},$ and thus in this case we see that the number of kings in this graph is $> \frac{1}{3} \cdot$ the number of vertices in the graph.

For such graphs you can just set $n=m-1$ to see that you cannot get a bound $\le 1/2$ for instance; you may be interested to see that the bound $1/2$ is attained asymptotically thus implying that the "asymptotic" bound of $1/2$ cannot be reduced any further.