About $l_2$ and $l_\infty$ Norms

Question

There is a theorem about norms that says $$||\textbf{x}||_\infty\le||\textbf{x}||_2\le\sqrt{n}||\textbf{x}||_\infty$$

In a text I found an illustration of this.

How does it correspond to the theorem?

Answer 1

The illustration shows that in $\mathbb{R}^2$ the Euclidean unit ball, i.e., the set $\{x\in\mathbb{R}^2:\|x\|_2\leq 1\}$ is contained in the unit square, i.e. the set $\{x\in\mathbb{R}^2:\|x\|_{\infty}\leq 1\}$. So if $\|x\|_2\leq 1 $ then $\|x\|_{\infty}\leq 1$, which implies of course that $\|x\|_{\infty}\leq \|x\|_2$ for every $x$, because given any $x\in\mathbb{R}^2$, $x\neq 0$, the point $x/\|x\|_2$ belongs to the Euclidean unit ball, hence also to the unit square, hence $\|x/\|x\|_2\|_{\infty}\leq 1$, hence $\|x\|_{\infty}\leq \|x\|_2$.

It also shows that if you dilate the unit square by a factor of $\sqrt{2}$, then the inclusion is reversed, that is, the set $\{x:\|x\|\leq\frac{1}{\sqrt{2}}\}$ is contained in the Euclidean unit ball, which of course implies that $\|x\|_2\leq\sqrt{2}\|x\|_{\infty}$ for every $x$, the proof being very similar to the one above.

The next step is to generalize this to higher dimensions. And so it happens, that for every $x\in\mathbb{R}^n$, the inequalities you wrote $$\|x\|_{\infty}\leq \|x\|_2\leq\sqrt{n}\|x\|_{\infty}$$ are valid. This is because the Euclidean ball in $\mathbb{R}^n$, i.e., the set $\{x\in\mathbb{R}^n:\|x\|_2\leq 1\}$ is contained in the unit cube, i.e, the set $\{x\in\mathbb{R}^n:\|x\|_{\infty}\leq 1\}$. (prove it!) and if you dilate the unit cube by a factor of $\sqrt{n}$, you get the reverse inclusion. To see this, note simply that $$x_1^2+\cdots + x_n^2\leq n\max_i x_i^2$$ hence $\|x\|_2\leq\sqrt{n}\|x\|_{\infty}$.

Answer 2

Recall that every norm satisfies $\|\alpha x\|=|\alpha| \|x\|$ (absolute homogeneous property). Hence in order to show that $\|x\|_1 \ge \|x\|_2$ (with $\| \cdot \|_1$ and ,$\| \cdot \|_2$ two norms) , it is enough to show that the boundary of the unit ball for the first norm is inside the unit ball of the second one.

In other words : let $B_1,B_2$ the unit balls for $\| \cdot \|_1,\| \cdot \|_2$ respectively. If $\partial B_1 \subset B_2$, then $\|x\|_1 \ge \|x\|_2$.

Indeed, if $\partial B_1 \subset B_2$ :

$$\forall x, \|x\|_1=1 \implies \|x\|_2 \le 1 \implies \|x\|_2 \le\|x\|_1.$$

Now, being given any $x$. We have $\frac x {\|x\|_1} \in \partial B_1$ which implies $\left|\left|{\frac x {\|x\|_1}} \right| \right|_2 \le \left|\left|{\frac x {\|x\|_1}} \right| \right|_1$.

Thus, by absolute homogeneous :

$$\|x\|_2 \le\|x\|_1.$$

Remark : if $\| \cdot \|$ is a norm and $K>0$, then $K \| \cdot \|$ is also a norm.

Answer 3

Every point $x$ on the circle falls inside the outer square, which means that $\|x\|_\infty \leq 1 \implies \|x\|_\infty \leq \|x\|_2$.

Also, every point $x$ on the circle falls outside the inner square, which means that $\|x\|_\infty \geq 1/\sqrt{n} \implies \sqrt{n} \|x \|_\infty \geq \| x \|_2$.

If these inequalities are true for vectors $x$ that satisfy $\|x\|_2 = 1$, then it follows that the inequalities are true for all vectors.