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

Since

,

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.

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