Consider a n dimensional space, it is known (Wikipedia) that for p>r>0, we have
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I have two questions about the above inequality.
(\bf 1). The first is how to show \|x\|_p\leq\|x\|_r when p,r\leq1. When p>r\geq1, we can define f(s)=\|x\|_s,\,\,s\geq1 and find out that f'(s)=\|x\|_s\left\{-\frac{1}{s^2}\log(\sum_i|x_i|^s)+\frac{1}{s}\frac{\sum_i|x_i|^s\log(|x_i|)}{\sum_i|x_i|^s}\right\}.
Then by the concavity of the \log function, we can see that \frac{\sum_i|x_i|^s\log(|x_i|)}{\sum_i|x_i|^s}\leq \log\left(\sum_i\frac{|x_i|^s}{\sum_j|x_j|^s}\cdot|x_i|\right).
Let y_i=\frac{|x_i|^s}{\sum_j|x_j|^s}, it is easy to see \|y\|_{s^*}\leq1, where s^*\geq1 and 1/s+1/s^*=1. Then, the Hölder's inequality leads to
\frac{\sum_i|x_i|^s\log(|x_i|)}{\sum_i|x_i|^s}\leq \log\left(\sum_i\frac{|x_i|^s}{\sum_j|x_j|^s}\cdot|x_i|\right)= \log\left(\sum_iy_i\cdot|x_i|\right)\leq\log(\|x\|_s\|y\|_{s^*})\leq\log\|x\|_s.
Therefore, we can conclude f'(s)\leq0 and \|x\|_p\leq\|x\|_r is satisfied. However, when p,r<1, we do not have s^*\geq1 and \|y\|_{s^*}\leq1. The last step does not work any more.
({\bf 2}). My second question is how to show \|x\|_r\leq n^{(1/r-1/p)}\|x\|_p. In fact, I was trying to show this by solving the following optimization problem:
\max_{\|x\|_p\leq1} \|x\|_r.
But seems it is difficult to derive a closed form solution. The objective function is non-smooth. Is there any elegant way to solve the above optimization problem?
Can anyone give me a hint? Thanks a lot.
Answer
This answers your first question
As for the second question. Consider Holder inequality
\sum\limits_{i=1}^n |a_ib_i|\leq \left(\sum\limits_{i=1}^n |a_i|^{s/(s-1)}\right)^{1-1/s}\left(\sum\limits_{i=1}^n |b_i|^s\right)^{1/s}
with a_i=1, b_i=|x_i|^r, s=p/r.
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