Theorem 1   All norms on finite-dimensional vector spaces over ${\mathbb{R}}$ or ${\mathbb{C}}$ are equivalent.

Proof. Any such finite-dimensional space is really just the same as ${\mathbb{R}}^n$ so we can talk about just those spaces. That is, any finite-dimensional vector space over ${\mathbb{R}}$ or ${\mathbb{C}}$ is isomorphic to ${\mathbb{R}}^n$ for some $n$ (note that ${\mathbb{C}}$ is just isomorphic to ${\mathbb{R}}^2$ as a vector space over ${\mathbb{R}}$ ). To see this, just write any element of the space in terms of the basis and then define the isomorphism to take that basis to the standard basis in ${\mathbb{R}}^n$ and then extend linearly.

First let's show that if two norms are equivalent on the unit sphere (all $\vec{x}$ such that $\|\vec{x}\|=1$ with respect to some fixed norm, for example the standard Euclidean norm) then they are equivalent everywhere. We can write any $\vec{x} \in {\mathbb{R}}^n$ as a multiple of some scalar $\gamma \geq 0$ and a vector on the unit sphere, say $\vec{x_0}$ , that is $\vec{x} = \gamma \vec{x_0}$ . Then when suppose we have two equivalent norms, say $\|\cdot\|_a$ and $\|\cdot\|_b$ , on the unit sphere

So the norms are equivalent everywhere.

Suppose we are working with the 2-norm. Now we want to show that any other norm is a continuous function with respect to the 2-norm. Take an arbitrary finite-dimensional space $X$ and an arbitrary norm $\|\cdot\|$ . Also suppose that $\{ \vec{b_i} \}_1^n$ is a basis of $X$ and so an element $\vec{x} \in X$ may be written as $\vec{x} = \sum_1^n x_i \vec{b_i}$ . Now given an $\epsilon > 0$ , choose $\delta > 0$ such that $\| \vec{x} - \vec{y} \|_2 < \delta$ (the Euclidean distance is less then $\delta$ ) implies that \begin{equation*} \max \{ | x_i - y_i | \} < \frac{\epsilon}{\sum_{i=1}^{n} \|\vec{b_i}\|} \end{equation*}In fact we can just choose $\delta$ to be the right side of the above inequality. Now we note that the triangle inequality immediately also yields the inequality $|\,\|\vec{x}\| - \|\vec{y}\|\,| \leq \|\vec{x}-\vec{y}\|$ . So \begin{equation*} \begin{split} \big|\,\|\vec{x}\| - \|\vec{y}\|\,\big| & \leq \|\vec{x}-\vec{y}\| \\ & = \left\| \sum_{i=1}^n x_i \vec{b_i} - \sum_{i=1}^n y_i \vec{b_i} \right\| \\ & = \left\| \sum_{i=1}^n (x_i-y_i) \vec{b_i} \right\| \\ & \leq \sum_{i=1}^n |x_i-y_i|\, \| \vec{b_i} \| \\ & \leq \left( \max_i |x_i-y_i| \right) \sum_{i=1}^n \| \vec{b_i} \| \\ & < \frac{\epsilon}{\sum_{i=1}^{n} \|\vec{b_i}\|} \sum_{i=1}^n \| \vec{b_i} \| \\ & = \epsilon . \end{split} \end{equation*}And so $\|\cdot\|$ is a continuous function.

Suppose we are given two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ , we know that they are both continuous functions with respect to the 2-norm. And so the function defined as \begin{equation*} f(\vec{x}) := \frac{\|\vec{x}\|_a}{\|\vec{x}\|_b} \end{equation*}is a continuous function on the unit sphere (with respect to the 2-norm). This function is continuous except perhaps at 0, but we don't care about the value at zero. On the unit sphere however $f(\vec{x})$ is continuous and thus achieves a maximum and a minimum since the unit sphere is compact. Let's call the minimum and maximum, $\alpha$ and $\beta$ respectively. Then for any $\vec{x}$ on the unit sphere we have

And so the norms are equivalent on the unit sphere and thus as we shown above, everywhere.