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For a non-negative sequence of real numbers $t_1, \dotsc, t_n$ with $\sum_{i=1}^n t_i = 1$ and $r\neq 0$, define $M_r(a) = (\sum_{i=1}^n t_i a_i^r)^{1/r}$ ($a = (a_1, \dotsc, a_n)$ non-negative). Show that if $r < 0 < s < t$ and $1/r + 1/s = 1/t$ then $$M_t(ab) \geq M_r(a)M_s(b)\ ,$$ for nonnegative sequences $a,b$, where $ab = (a_1b_1, \dotsc, a_nb_n)$. Deduce that if $0 < r < 1$ then $$ M_r(a+b) \geq M_r(a)+M_r(b)\ .$$


The above is exercise 15, chapter 1, of Bollobás's Linear Analysis book. The first statement is simple enough to show. The second statement is that proven in, for example, this answer. However, whenever Bollobás says "deduce", there has thus far always been a very immediate way to obtain the next statement from what has been established thus far (if not always easy to see), and here, I very much cannot see it. Any suggestions for how to do this deduction (rather than prove it from scratch, which is not an issue)?

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  • $\begingroup$ how is $M_r$ defined for negative $r ?$ You specified in the beginning that $r>0.$ $\endgroup$
    – dezdichado
    Commented Jan 3 at 22:40
  • $\begingroup$ also what is $ab$ for two vectors ? $\endgroup$
    – dezdichado
    Commented Jan 3 at 22:45
  • $\begingroup$ Defined the same as for $r>0$, and the latter is an elementwise product. Will edit when I'm by a computer. $\endgroup$
    – Oxonon
    Commented Jan 3 at 22:54
  • $\begingroup$ ok. Generally, Holder implies Minkowski for $p>1$ and is easily derived. So I imagine one could mimic the idea to reproduce a similar proof for $p<1.$ $\endgroup$
    – dezdichado
    Commented Jan 3 at 22:58
  • $\begingroup$ Yes that was indeed what I attempted. Maybe I just made a mess of it. $\endgroup$
    – Oxonon
    Commented Jan 3 at 22:58

1 Answer 1

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$$M_p^p(a+b) = \sum_{i=1}^n(a_i+b_i)^p = \sum_{i=1}^na_i(a_i+b_i)^{p-1}+\sum_{i=1}^nb_i(a_i+b_i)^{p-1}=M_t^t(a^{1/t},(a+b)^{(p-1)/t}) + M_t^t(b^{1/t},(a+b)^{(p-1)/t}).$$ But $$M_t(a^{1/t},(a+b)^{(p-1)/t})\geq M_r(a^{1/t})M_s((a+b)^{(p-1)/t})$$ and so: $$M_p^p(a+b)\geq\left(M_r^t(a^{1/t})+M_r^t(b^{1/t})\right)M_s^t((a+b)^{(p-1)/t})$$ with $r<0<s<t$ and $\frac 1t = \frac 1s+\frac 1r.$ Now, just take $t=1$ and $r=p$, which gives $s = \dfrac{p}{p-1}<0$ and it simplifies to: $$M_p^p(a+b)\geq(M_p(a)+M_p(b))M_p^{p-1}(a+b)$$ and we are done. Note that I did not bother to include the weight variables since they don't really affect anything, or just think of a finite, counting measure. Also, I used $p$ since $r,p,s$ are reserved for the inequality we are going to use.

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    $\begingroup$ Thanks, I didn't think to take $t=1$ and got very tangled. $\endgroup$
    – Oxonon
    Commented Jan 3 at 23:57
  • $\begingroup$ yeah interpolation inequalities in general is very annoying like that. Also,it looks like I switched $r$ and $s$. $\endgroup$
    – dezdichado
    Commented Jan 4 at 0:26

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