- Two norms $\def\norm#1{\lVert#1\rVert}\norm\cdot_1$ and $\norm\cdot_2$ are equivalent iff $\;\exists\;c_1,c_2>0$ such that $c_1\norm x_1\le \norm x_2\le c_2\norm x_1$
Show that $\norm x_1=\sum_{i=1}^n \lvert x_i \rvert$ and $\norm x_2=\left ( \sum_{i=1}^n x_i^2 \right )^{1/2}$ are equivalent.
It looks like $c_2$ is $1$, and that this can be proven with induction. But what could $c_1$ be?
edit
I mean what I don't understand is the following: If I square both terms, and expand $(\sum_{i=1}^n \lvert x_i \rvert)^2 = \left ( \sum_{i=1}^n x_i^2 \right ) + \sum_{i=1}^n x_i(x_k+\dotsb+x_n)_{x_k\neq x_i}$. However, the second term grows with $n$, so how can $\frac{\norm x_1}{\norm x_2} \leq C$ at all?