Hanner's inequalities in the theory of $L^p$ spaces (see http://en.wikipedia.org/wiki/Hanner's_inequalities) look hard to come-up with at the first glance. Their proof (say, the one in Lieb & Loss "Analysis", Theorem 2.5.) gives no intuition (at least for me) how they come about. How does one see that these inequalities turn up naturally? Do you know a proof which at the same time hints to how one starts considering Hanner inequalities. I hope this is not too vague of a question. Both Wikipedia and Lieb & Loss mention that Hanner's inequalities have to do with uniform convexity of $L^p$ spaces, but from that alone I cannot see how they arise "naturally".
1 Answer
First, a probably unsatisfactory answer to your first question. I believe that Hanner proved his inequalities because he was investigating the modulus of convexity of $L^p$. So possibly he formed a guess on whatever basis, saw that guess would be correct if what have come to be known as Hanner's inequalities were true, then found that he could prove the inequalities. (I haven't actually looked at Hanner's paper so I might be wrong about the history, and in any case this is the kind of thought process that is seldom recorded in papers.)
A better answer is that Hanner's inequalities generalize the parallelogram identity in a very natural way.
As for your second question, I don't know that this really fits what you're asking for either, but this paper contains the most natural-looking proof of Hanner's inequalities that I've seen.
Added: On further reflection, the best answer to the first question combines my first two paragraphs above. The parallelogram identity very precisely expresses the uniform convexity of $L^2$. So in investigating the modulus of convexity of $L^p$, it is natural to look for some inequality that generalizes the parallelogram identity to $L^p$.
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2$\begingroup$ Evidently I missed that reference when I wrote my Hanner-less proof of uniform convexity of $L^p$ (arxiv.org/abs/math.FA/0502021 and dx.doi.org/10.1090/S0002-9939-06-08366-3). Thanks for pointing it out. $\endgroup$ Commented Jun 10, 2010 at 18:18