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Let $l_p=\{(x_n)\in\mathbb{R}^\mathbb{N}: \sum_{n=0}^\infty |x_n|^p<\infty\}$ and consider the following norm in $l_p$: $$\|x\|_p=\left(\sum_{n=0}^\infty|x_n|^p\right)^{1/p}$$ for $x=(x_n)_{n\in\mathbb{N}}$. I'm studying this norm, and I have to prove that the inequality $\|x+y\|_p<\|x\|_p+\|y\|_p$ holds for $x\neq \lambda y$, $\lambda>0$.

My attempt: the restriction is needed because if $y=\lambda x$ then the equality $$\|x+y\|_p=\|(1+\lambda)x\|_p=|\|x\|_p+\lambda\|x\|_p=\|x\|_p+\|y\|_p$$ holds ever.

Now I do the particular case $y=-\lambda x$, for $0<\lambda\leq 1$ (and $x\neq 0$) then: $$\|x+y\|_p=\|x-\lambda x\|_p= \|(1-\lambda)x\|_p=(1-\lambda)\|x\|_p=\|x\|_p-\lambda\|x\|_p<\|x\|_p+\|y\|_p$$ (the ''$<$'' is because $\|x\|_p>0 $)

Now if $\lambda>1$, I found that the inequality holds again with '''<'' just following a similar approach.

But how can I proceed in the general case? Thanks in advance!

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    $\begingroup$ You should seek information about Minkowski's inequality: en.wikipedia.org/wiki/Minkowski_inequality $\endgroup$
    – ajotatxe
    May 25, 2015 at 17:36
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    $\begingroup$ Notice that $||a||||b||$ looks conspicuously different from $\|a\|\|b\|$. I edited accordinglly. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    May 25, 2015 at 17:38
  • $\begingroup$ Thanks this is helpful!! I going to study the $=$ in the minkowsky's inequality! $\endgroup$
    – Valent
    May 25, 2015 at 18:01
  • $\begingroup$ It should be mentioned that the inequality doesn't hold for $0<p<1$, however a similar inequality holds in place of it: $\|x+y\|_p^p \leq \|x\|_p^p + \|y\|_p^p$. To give an example why the original inequality doesn't hold when $0<p<1$, consider $x = e_1 = (1,0,0,\dots)$ and $y=e_2 = (0,1,0,0,\dots)$. Then you have $\|x+y\|_p = (1^p+1^p)^{\frac{1}{p}} = 2^{\frac{1}{p}} > 2 = \|x\|_p + \|y\|_p$ $\endgroup$
    – JMoravitz
    May 25, 2015 at 20:03

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