Considering $$\infty\geq p\geq q\geq1$$
How can I show that the $p$-norm is smaller or equal to the $q$-norm?
I can only show the case for $p=1, q=2$, but have no idea how to show others. Thank you!
For $p=1$, $q=2$, Take square on 1-norm and 2-norm, then $(x_1+x_2+n_3+...+x_n)^2 \geq (x_1^2+x_1^2+...+x_n^2)$ So I can have the conclusion for this particular case.
Let $x=(x_1,\ldots,x_n)$ be a non nul vector. Let $\alpha = \frac{1}{\|x\|_q}$, then $0\le \alpha |x_i|\le 1$ for all $i$. Thus $(\alpha |x_i|)^p \le (\alpha |x_i|)^q$, hence, using norm homogeneity: $$\sum_i (\alpha |x_i|)^p \le \sum_i(\alpha |x_i|)^q \le\alpha^q \sum_i( |x_i|)^q \le 1\,,$$ so: $$(\sum_i (\alpha |x_i|)^p)^{1/p} \le \alpha \|x\|_p \le 1\,,$$ and balancing the last inequality, $$ \|x\|_p \le \frac{1}{\alpha} \le \|x\|_q\,.$$
If $x$ is the nul vector, the inequality is trivial.
