All norms are equivalent in finite dimensional vector spaces example. While researching the equivalence in finite dimensional vector spaces, I found multiple proofs of the theorem, but I still can't wrap my head around the following example:
take a finite-dimensional vector:
$v = \begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$.
If I calculate the 1-norm I get: |1| + |1| + |1| = 3.
For the 2-norm I get: $\sqrt {1^2 + 1^2 + 1^2} = \sqrt{3}$. Lastly, for the infinite norm I get 1. How are these equal? I seem to not understand what the theorem actually means.
Thanks in advance.
 A: Equivalence means that each norm is within two positive multiples of any other norm, not that they are equal. The consequence of equivalence is that they induce the same topology. In other words if a sequence converges in one norm it converges in the other.
A: Norms are equivalent if they're "close" to each other. Mathematically spoken, this means that $\exists a,b > 0\; \forall x \in E: a||x||_1 \leq ||x||_2 \leq b||x||_1$. One can show that with this definition, two norms are equivalent if and only if any sequence converging to a limit $x$ in one norm also converges to the same limit regarding the other norm. We can easily show that the $1$-norm and the $\infty$-norm are equivalent on $\mathbb{R}^n$ since obviously, $||x||_\infty = \max_{k\in\{1,...,n\}} |x_k| \leq \sum_{k=1}^n |x_k| = ||x||_1$, but also $||x||_1 = \sum_{k=1}^n |x_k| \leq \sum_{k=1}^n \max_{k\in\{1,...,n\}} |x_k| = \sum_{k=1}^n ||x||_\infty = n||x||_\infty$, so it follows that $\forall x \in \mathbb{R}^n: ||x||_\infty \leq ||x||_1 \leq n||x||_\infty$. In a similar way, we prove that the $2$-norm and the $\infty$-norm are equivalent and since equivalence is an equivalence relation (why?), it also follows that the $1$-norm and the $2$-norm are equivalent. As you said, we can prove that on finite-dimensional spaces, all norms are equivalent to each other.
