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

metric-spacesnormed-spacesreal-analysis

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.

Best Answer

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 $$ \|x\|_\infty=|x_k|\leq\sum_j|x_j|=\|x\|_1. $$