# Prove that $\Vert\cdot\Vert_2$ and $\Vert\cdot\Vert_3$ are equivalent on $\mathbb R^n$ without using that all norms on $\mathbb R^n$ are equivalent

Prove that $$\Vert\cdot\Vert_2$$ and $$\Vert\cdot\Vert_3$$ are equivalent on $$\mathbb R^n$$ without using the fact that all norms on $$\mathbb R^n$$ are equivalent.

In other words I need to show that there exist $$m,M>0$$ such that $$m\Vert x\Vert_3\le\Vert x\Vert_2\le M\Vert x\Vert_3$$ for all $$x \in \mathbb R^n$$. Where $$\Vert(x_1,\dots,x_n)\Vert_k= \sqrt[k]{\sum_{i=1}^n|x_i|^k}$$

Attempt: I tried using induction to prove that $$\Vert x\Vert_2\ge\Vert x\Vert_3$$ for all $$x\in\mathbb R^n$$ but it got really messy, so I am wondering if there is a better way to do this? Also i have no idea how to do the other inequality.

If $$\|y\|_2 \leq 1$$ then $$\|y\|_3 \leq 1$$ too because $$|y_i| \leq 1$$ and so $$|y_i|^{3} \leq |y_i|^{2}$$. Now take any non-zero $$x$$ and put $$y=\frac x {\|x\|}$$ to conclude that $$\|x\|_3 \leq \|x\|_2$$. Thus we can take $$m=1$$.

Next note that $$\|y\|_3 \leq 1$$ then $$\|y\|_2 \leq 1=\sqrt n$$ because $$|y_i| \leq 1$$ for all $$i$$. Hence we can take $$M=\sqrt n$$.

• I am sorry why does $|y_i|^3\le|y_i|^2$ prove that $\Vert y\Vert_3\le 1$? – Nasal Nov 16 '19 at 13:20
• @Nasal I am saying that $\sum |y_i|^{2} \leq 1$ implies $\sum |y_i|^{3} \leq 1$. – Kavi Rama Murthy Nov 16 '19 at 13:23