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click to see the picture of one related page from the paper

this is the link of the whole paper

What I cannot really clearly understands is the content bellow: the shortcut of the inequality

If you don’t want to see the pictures,here is the main description: Let $$N_{a,b}^{\rho}(f)=\left(\sum\limits_{k \in \mathbb{Z}}\rho^{bk}\mu(f \ge \rho^k)^{b/a}\right)^{1/b},\rho > 1,a,b >0,\\f \ \text{is measurable}, \mu \ \text{is the Lebesgue measure}.$$ Then there is the inequality $$ N_{r,u}^{\rho}(f) \leq \left( N_{s,v}^{\rho}(f) \right)^{\nu}\left(N_{t,w}^{\rho}(f) \right)^{1-\nu},\\ \text{with}\ \frac{1}{r}=\frac{\nu}{s}+\frac{1-\nu}{t}, \frac{1}{u}=\frac{\nu}{v}+\frac{1-\nu}{w},\nu \in (0,1). $$

This can be used to show that $$ ||f||_{a,b^{‘}} \leq 2^{\frac{2}{b}}||f||_{a,b},\ for \ \ 0<b\leq b^{‘}\leq \infty $$ Mark:make sure that you don’t mix the symbol v with $\nu$.

I’ll be sincerely appreciated if anyone can solve the problem.

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  • $\begingroup$ I think that it might be better to link directly to the paper - math.univ-toulouse.fr/~ledoux/BCLS.pdf - rather than including pictures. I have edited the post to include the part from your second screenshot. If needed, you can edit it further to make clear which part is problematic; to include additional context, etc. $\endgroup$ Jan 29, 2018 at 7:30
  • $\begingroup$ It’s my first time to ask question there,so may not show the problem friendly.Thanks for your advise,I’ve make the problem clear just now. $\endgroup$ Jan 29, 2018 at 8:06
  • $\begingroup$ I fixed the numerators in the formulae for $\frac1r,\frac1u$, else the inequality was non-homogeneous and obviously wrong, now it is true. $\endgroup$ Jan 29, 2018 at 11:00
  • $\begingroup$ I did't get the idea?What do you mean for that "I fixed the numerators in the formulae for $1/r,1/u$, else the inequality was non-homogeneous and obviously wrong"? $\endgroup$ Jan 29, 2018 at 11:55
  • $\begingroup$ it was written $1/r=1/s+(1-\nu)/t$ instead of $1/r=\nu/s+(1-nu)/t$, the same for second formula. $\endgroup$ Jan 29, 2018 at 12:28

2 Answers 2

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Denote $c_k=\mu(f \ge \rho^k)$, actually we do not need anything about these non-negative numbers. Use the Hölder inequality $$\left(\sum a_k\right)^\alpha\left(\sum b_k\right)^\beta\geqslant \left(a_k^{\alpha/(\alpha+\beta)}b_k^{\beta/(\alpha+\beta)}\right)^{\alpha+\beta}$$ for non-negative numbers. In our situation $\alpha=\nu/v$, $\beta=(1-\nu)/u$, $a_k=\rho^{vk}c_k^{v/s}$, $b_k=\rho^{wk}c_k^{w/t}$.

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  • $\begingroup$ I've solved the problem under the instruction of my tutor,it was really a pleasent time and I planned to go back to answer this question by myself.And it's amazing that you've solved the problem in such a short time and even with the almost same method! I'm really thankful for your answer. $\endgroup$ Jan 29, 2018 at 12:26
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Denote $h_k(r)=\rho^k \mu(f \geq p^k)^{\frac{1}{r}}$,then $h_k(r)=h_k(s)^{\nu}h_k(t)^{(1-\nu)}$, use Sobolev inequality $\sum\limits_{k}a_kb_k \leq (\sum\limits_{k}a_k^p)^{\frac{1}{p}}(\sum\limits_{k}a_k^p)^{\frac{1}{q}},with \frac{1}{p} + \frac{1}{q}=1,p>1.$

In our situation,we let $p=\frac{v}{u\nu},q=\frac{w}{u(1-\nu)},a_k=h_k(s)^{u\nu},b_k=h_k(t)^{u(1-\nu)}.Then$ \begin{aligned} N_{r,u}^{\rho}(f)=\left(\sum\limits_{k \in \mathbb{Z}}h_k(r)^u\right)^{\frac{1}{u}} &= \left(\sum\limits_{k \in \mathbb{Z}}h_k(s)^{u\nu}\cdot h_k(t)^{u(1-\nu)}\right)^{\frac{1}{u}}\\ &\leq \left[\left(\sum h_k(s)^v \right)^{\frac{u\nu}{v}}\left(\sum h_k(t)^w \right) ^{\frac{u(1-\nu)}{w}}\right]^{\frac{1}{u}}\\ &=\left(N_{s,v}^{\rho}(f) \right)^{\nu}\left(N_{t,w}^{\rho}(f) \right)^{1-\nu} \#. \end{aligned} $\int_Ef^{r\cdot \alpha/(\alpha - 1)}d\mu \le (CW_n(f^{\alpha/(\alpha - 1)})^{\vartheta}||f^{\alpha/(\alpha - 1)}||_s \le C^{r\vartheta}||f||_1 $

\begin{aligned} \int_E \sum_{j=1}^{N}\frac{(b|f|^{\alpha/(1-\alpha)})^j}{j!} &= \sum_{j=1}^{N} \frac{b^j}{j!}\int_E|f|^{j\cdot \frac{\alpha}{\alpha - 1}}d\mu \le \sum_{j=1}^{N}\frac{b^j}{j!}C^{j\vartheta(j)}||f||_1 \end{aligned} &\le

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