Consider a normed vector space $(\mathbb{R}^n,\mathbb{R},||\cdot||_p)$. For any $x\in\mathbb R^n$, we want to show the following:

$$||x||_{\infty}\leq||x||_1\leq n||x||_{\infty}.$$

I remember from metric spaces that we can define $u_i=|x_i-y_i|$ for all $i$ where $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$. Hence, we have $||x||_{\infty}=\max\{u_1,\ldots,u_n\}$ and $||x||_1=\sum_{i=1}^{n}u_i$.

By the fact $|x_i-y_i|\leq\max|x_i-y_i|$ for all $i$, we have that $$u_1+\cdots+u_n\leq\max u_1+\cdots+\max u_n$$ which proves the second inequality.

I am not sure how to show the first inequality rigorously. Intuitively it makes sense that the sum of all positive differences of $x$ and $y$ have to be greater than the maximum of one difference.

I'd appreciate any hint or help. Thank you.

  • 1
    $\begingroup$ Your understanding of the definitions of norms is not correct, check en.wikipedia.org/wiki/Norm_(mathematics)#p-norm, but your reasoning is (almost) correct for the second inequality. Correct it using the right definitions of norms. The first inequality follows because you sum over all $|x_i|$ in $||x||_1$, but only take the largest $|x_i|$ in $||x||_{\infty}$. $\endgroup$ – B.Swan Mar 18 at 2:50

You are putting a $y$ in the mix that you don't need.

As you say, you have $$ \|x\|_1=\sum_j|x_j|\leq\sum_j\|x\|_\infty=n\|x\|_\infty. $$ Also, if $k$ is such that $\|x\|_\infty=|x_k|$, then $$ \|x\|_\infty=|x_k|\leq\sum_j|x_j|=\|x\|_1. $$


By definition, $$||x||_{\infty}=\max\{|u_1|,\ldots,|u_n|\} = |u_p|,$$ for some $1 \leq p \leq n$. Clearly, $$|u_p| \leq \sum_1^n |u_i| = ||x||_{1}.$$


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