Otherwise it seems that partitioning the graph into one part V1 = V, would trivially satisfy the conclusions of the lemma.

*[Hmm, you're right. Thanks for the correction! - T.]*

Such an O(n) algorithm appears (explicitly) in the following paper of mine with Fischer and Matsliach.

http://www.cs.tau.ac.il/~asafico/regalg.pdf

That algorithm actually has the added advantage of being able to find (more or less) the smallest regular partition in the input.

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