# How do we show that $||x||_{\infty}\leq||x||_1\leq n||x||_{\infty}$?

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.

• 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 $||x||_{\infty}$. Commented Mar 18, 2019 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 $$\|x\|_\infty=|x_k|\leq\sum_j|x_j|=\|x\|_1.$$

• What is the justification for $$\sum_j |x_j|\leq\sum_j \| x\|_{\infty}?$$
– M B
Commented Mar 6, 2020 at 8:05
• That $\|x_j\|\leq\|x\|_\infty$ by definition. Commented Mar 6, 2020 at 14:15
• Did you use the Cauchy-Swartz inequality to get the $n$? I do not understand where this came from.
– M B
Commented Mar 6, 2020 at 23:45
• It comes from adding $n$ times the same thing. Commented Mar 7, 2020 at 0:56

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