Sufficient conditions that a bipartite graph is k-regular?

Question

I have conditions on a bipartite graph $G=(V_1, V_2, E)$ that $|V_1|\ge|V_2|$ and that the degree of every vertex in $V_1$ is greater than or equal to the degree of every vertex in $V_2$. Does it immediately follow from this that $G$ is a k-regular or would further proof be required? How would you go about proving that it is?

Answer 1

Let $M = \max_{v \in V_2} d(v)$. Then since $\lvert V_1 \rvert \geq \lvert V_2 \rvert$ and for all $v \in V_1$ and $w \in V_2$, $d(v) \geq M \geq d(w)$ we have $$ e(G) = \sum_{v \in V_1} d(v) \geq \sum_{v \in V_1} M = \lvert V_1 \rvert M \geq \lvert V_2 \rvert M = \sum_{w \in V_2} M \geq \sum_{w \in V_2} d(v) = e(G)$$ So in fact we have equality thoughout. From this we deduce $\lvert V_1 \rvert = \lvert V_2 \rvert$ and for all $v \in V_1 \cup V_2$, $d(v) = M$. So the graph is regular as required.