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Consider Holder conjugate exponents $p$ and $q$, i.e., $\frac{1}{p} + \frac{1}{q} = 1$. Multiplying through by $pq$ gives $p + q = pq$. So conjugate exponenets are just those real numbers whose product and sum are the same. I have two questions.

  1. Is this phenomenon important enough for some notion in algebra that generalizes it? For instance, given a ring $(R,+,*)$, the pair $(x,y)$ is call blah if $x*y = x+y$.

  2. Is there some insight into why these pairs are precisely the ones for which Holder's inequality is true?

H_R
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    I've come to the conclusion that it's better to take the norm to be $(\int |u|^{1/p})^p$, which makes the exponent algebra so much simpler and less contorted: you just get $p+q=1$ in Hölder's inequality. Sadly it'll never catch on. – Chappers Jul 09 '18 at 19:29

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I can give a relatively good insight for your second question. We know that Holder's inequality relies on Young's inequality: $$\forall a,b \geq 0: ab \leq \frac{a^p} {p}+\frac{b^q}{q}$$

This is related to the concavity of the logarithm function. You can prove Young's inequality by considering $f(x)=\log(x)$ and using Jensen's inequality. After applying Jensen's inequality, you will see that for it to work you must have $\frac{1}{p}+\frac{1}{q}=1$. This answers your second question about why this pair is special, I think.

A possible generalization of Holder's inequality should probably first deal with Young's inequality. Quoting from Wikipedia:

If $f$ is a convex function and its Legendre transform (convex conjugate) is denoted by $g$, then $$ab \leq f(a) + g(b)$$ This follows immediately from the definition of the Legendre transform.

stressed out
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