# Any two norms on a finite dimensional $\Bbb K$-linear space are equivalent.

Any two norms on a finite dimensional $\Bbb K$-linear space are equivalent where $\Bbb K = \Bbb R$ or $\Bbb C$.

I have assumed WLOG $\Bbb K = \Bbb R$. Let $(X,\| \cdot \|)$ be a finite dimensional $\Bbb R$-linear space with $\dim (X)=n$. Then $X \simeq \Bbb R^n$. Let $\| \cdot \|_1$ and $\| \cdot \|_2$ be two norms on $X$. Consider two norms ${\| \cdot \|^{*}}_1$ and ${\| \cdot \|^{*}}_2$ on $\Bbb R^n$defined by ${\| Tx \|^{*}}_1 = \|x\|_1$ and ${\|Tx\|^{*}}_2 = \|x\|_2$. Then ${\| \cdot \|^{*}}_1 \equiv {\| \cdot \|^{*}}_2$ on $\Bbb R^n$. Now how can I proceed? Please help me.

Thank you very much.

• What you have written so far makes very little sense. What is $T$? – Eric Wofsey Aug 18 '18 at 17:17
• $T$ is an isomorphism between $X$ and $\Bbb R^n$. – Dbchatto67 Aug 18 '18 at 17:18
• Is this a typo that you have written $\|\cdot\|^*_1$ twice, or do you only have a single norm there? Also: why is $\|\cdot\|^*_1\equiv\|\cdot\|^*_2$? Do you already assume the equivalence of norms of $\Bbb R^n$ (I assume $\equiv$ means equivalence)? – M. Winter Aug 18 '18 at 17:22
• This is a textbook theorem. – Lord Shark the Unknown Aug 18 '18 at 17:26
• Yeah @M.Winter you have rightly guessed. I have made it correct. – Dbchatto67 Aug 18 '18 at 17:43

Two given norm over $\mathbb{R}^n$, let's say $N_1$ and $N_2$ are said to be equivalent if there exist two constants $C_1, C_2$ such that $$C_1 N_2(x) \leq N_1(x) \leq C_2 N_2(x) \quad \forall x \in \mathbb{R}^n$$
Considering $N_1$ to be the max norm, and $N_2 = N$ a generic norm due to this being an equivalence realtion it's enough to show that $|| \cdot ||_\infty \sim N$.
To this aim we first prove that $N$ is uniformly continuos. $$| N(x) - N(y) | \leq N(x - y) = N \left( \sum_{i = 1}^n (x_i - y_i)\mathbb{e}_i \right) \leq \sum_{i = 1}^n |x_i - y_i| N(\mathbb{e}_i) \leq M ||x - y||_\infty$$
Defining $M = \sum_{i = 1}^n N(\mathbb{e}_i)$. We have that $N$ is uniformly continuos and in the unitary sphere (with respect to the $|| \cdot ||$ norm - that is a compact in $\mathbb{R}^n$) has maximum and minimum value $C_1, C_2$ hence $$C_1 \leq N\left( \frac{x}{ ||x||_\infty} \right) \leq C_2 \quad \Rightarrow \quad ||x||_\infty C_1 \leq N(x) \leq ||x||_\infty C_2$$ That is the thesis