Let have positive components and satisfy . We define the weighted power mean of order by
I claim that
Proof. Observe that
by the concavity of . Hence,
Taking the -root of both sides, we obtain that
Hence, . This establises a convenient lower bound. To prove the desired equality, we use l’hospital’s rule:
By the continuity of the exponential, we obtain that
Observe that for ,
For all , .
Proof. First, assume that and . Since the function is convex, we have by Jensen’s inequality that
We’ve already proven the cases and and and , so we’re done.
Other properties of the weighted power means include that the function is convex on . Since is continuous, it suffices to show that (see my post for why)
for all and . Observe that this expression is equal to
we conclude that
So far, we have not said anything about the limiting behavior of at . We shall redress this omission now. Define and . I claim that
Proof. Let be integers in such that and . Then for ,
and for ,
Since as and as , an application of the squeeze theorem completes the proof.