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How do you prove that the general norm equation $\sqrt[p]{\displaystyle\sum_{n=1}^{m} |x_n^p|}$ satisfies the inequality $\sqrt[p]{\displaystyle\sum_{n=1}^{m} |x_n^p|}\leq {\displaystyle\sum_{n=1}^{m} |x_n|}$? I think I'm pretty close, I have taken both sides to the $p$ power such that ${\displaystyle\sum_{n=1}^{m} |x_n^p|}\leq \left( {\displaystyle\sum_{n=1}^{m} |x_n|} \right) ^p$ where the RHS is the multinomial expansion $${\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{t=1}^{m}x_{t}^{k_{t}}\,,}$$ where $${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}}$$ From there, I need to somehow out a ${\displaystyle\sum_{n=1}^{m} |x_n^p|}$ from the above multinomial summation.

One side question, is this in fact the triangle inequality at all? Either way, I'm pretty sure this inequality is valid since any such expansion should result in ${\displaystyle\sum_{n=1}^{m} |x_n^p|} \leq {\displaystyle\sum_{n=1}^{m} |x_n^p|} + K$ where K is a combination of all the ways the variables $x_1, x_2, x_3 \cdots x_m$ can be multiplied together, so it must be positive since the variables each represent a nonnegative length.

Edit: I see there is a method to prove this by taking the limit of the sum after showing this for the L2 norm. However, I am still interested in knowing how to solve for this $K$.

Rithwik Sudharsan
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  • Can this K be solved for though? That would be more intuitive than taking the limit it seems to me. I don't have a restriction on the p, making one would only simplify the problem right? – Rithwik Sudharsan Jul 17 '18 at 23:04

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Assume first that $\sum_{n=1}^m|x_n|=1$. Then $|x_n|\leq1$ for all $n $. For $p\geq1$, it follows that $|x_n|^p\leq|x_n|$. You immediately get $$\sum_{n=1}^m|x_n|^p\leq\sum_{n=1}^m|x_n|=1, $$ and thus $$\tag1\left (\sum_{n=1}^m|x_n|^p\right)^{1/p}\leq1. $$

Now, for the general case, let $y_n =x_n/c$, where $c=\sum_{n=1}^m|x_n|$. Then $\sum_{n=1}^m |y_n|=1$, and by $(1) $ we have
$$\tag2\left (\sum_{n=1}^m\frac {|x_n|^p}{ c^p}\right)^{1/p}\leq1. $$ We may re-write $(2)$ as $$\tag3 \left (\sum_{n=1}^m {|x_n|^p} \right)^{1/p}\leq c, $$ which is the desired inequality.

Martin Argerami
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