How would I show that $\|\cdot\|_3$ and $\|\cdot\|_\infty$ are equivalent norms on $\mathbb R^2$?

I understand that to say two norms are equivalent, then there exist two real constants, $m,M$ such that,

$$m\|\cdot\|_3\le\|\cdot\|_\infty\le M\|\cdot\|_3$$

And that if we were to sketch the norms $\|\cdot\|_\infty=\|\cdot\|_3=1$ then we could stretch or shrink them to fit into each other, again, by the constants $m,M$.

However, I am not to sure what $\|\cdot\|_3$ exactly looks like, and haven't had much luck with a graphing calculator, and so I am not entirely sure how to go about rigorously finding the constants, having not worked with $\|\cdot\|_3$ at all before and not being able to sketch it.

Can anybody help me to discern its shape as well as with finding $m,M$?


It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$.

It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$.

In terms of the shape of $||.||_3$, you really just want to sketch the curve $||(x,y)||_3=1$. This curve is $(x^3+y^3)^{1/3}=1$, and should remind you of the circle equation. Think about how it might differ from a circle.


Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $||x||_3^3=\sum_{i=1}^n|x_i|^3$.

Using the fact that $\forall 1\leq i\leq n$, $|x_i|\leq||x||_{\infty}$, we can chose $m=n^{-\frac{1}{3}}$.

On the other hand:

$$||x||_{\infty}^3=\left(\max_{1\leq i\leq n}|x_i|\right)^3=\max_{1\leq i\leq n}|x_i|^3\leq\sum_{i=1}^n|x_i|^3=||x||_3^3$$

So you can chose $M=1$.


We have the following:

Suppose $z:=(x,y)\in R^2$.

First prove that $| x |^3 + | y |^3\leq(| x | + | y |)^3$.

Therefore, $2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3$.

On the other hand, $|x|^3+|y|^3\geq \Vert z \Vert_\infty^3$.

Thus, $\Vert z \Vert_3 \geq \Vert z \Vert_\infty$.

Summing up:

$2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3 \geq \Vert z\Vert_\infty$.

In general, for dimension $n$ and $1\leq p\leq \infty$:

$n \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_p \geq \Vert z\Vert_\infty$.

Hence, you can take $m:= \frac{1}{n}$ and $M:=1$.


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