# equivalence between the three norms

Show that $$\forall x\in \mathbb R^n$$ : $$|x|_{M}\leq|x|\leq|x|_{s}\leq n|x|_{M}$$

where,

$$|x|_{M}$$ $$=$$ $$max$$ {$${|x_{1}|,...,|x_{n}|}$$} $$=$$ maximum of absolute values of the components $$=$$ sup norm of $$x$$ $$=$$ infinity norm of $$x$$

$$|x|_{s}$$ $$=$$ $$|x_{1}|$$ $$+$$ $$|x_{2}|$$ $$+$$ $$...$$ $$+$$ $$|x_{n}|$$ $$=$$ sum of absolute values of the components $$=$$ one norm of $$x$$

$$|x|$$ $$=$$ $$\sqrt{}$$ $$=$$ $$\sqrt{\sum x_{i}^2}$$ $$=$$ Euclidean norm of $$x$$ $$=$$ two norm of $$x$$

I have the following definition as a help: two norms are equivalent if exist $$K,M \in \mathbb R$$ such that $$K|x|^{*}$$ $$\leq$$ $$|x|^{+}$$ $$\leq$$ $$M|x|^{*}$$

this is what i wear so far:

$$|x|_{M}^2\leq \sum|x_{i}|^2=|x|^2$$. Thus, $$|x|_{M} \leq |x|$$ But could you give me an idea of how the others would be?

• This post math.stackexchange.com/questions/218046/… gives a generalization of the answer you are looking for. It applies to all "proper" norms - which include the 1-norm and $\infty$-norm. – firdaus Oct 25 '20 at 21:27

Let $$e_i \in \mathbb R^n$$ be the vector whose $$j$$-th entry is a $$1$$ if $$j=i$$ and $$0$$ if $$j \neq i$$. Then: \begin{align}\|x\| &= \|x_1e_1 + \cdots + x_ne_n\| \\ &\leq \|x_1e_1\| + \cdots + \|x_ne_n\| \\ &= |x_1|\|e_1\| + \cdots + |x_n|\|e_n\| = |x_1| + \cdots + |x_n|. \end{align} Also, since $$|x_i| \leq \|x\|_M$$ for all $$i$$, we have $$|x_1| + \cdots + |x_n| \leq \|x\|_M + \cdots + \|x\|_M = n\|x\|_M$$.