On page 93 in Introduction to Analysis by Maxwell Rosenlicht, Problem 22 tasks one with the following:
Let $V, V'$ be normed vector spaces and $f: V \to V'$ a linear transformation. Prove the following statements.
(a) If $f$ is continuous at one point it is continuous everywhere, and in fact is uniformly continuous.
(b) $f$ is continuous iff the set $\{ \| f(x)\|/\|x\| : x \in V, x\neq 0 \}$ is bounded.
(c) $f$ is continuous if $V$ is finite dimensional. (Hint: Use a basis of $V$.)
(d) The set of all infinite sequences of real numbers with only a finite number of nonzero terms is a normed vector space if we define $$ (x_1, x_2, x_3, \ldots) + (y_1, y_2, y_3, \ldots) = (x_1+y_1, x_2+y_2, x_3 + y_3, \ldots) \\ c(x_1,x_2,x_3,\ldots) = (cx_1, cx_2, cx_3, \ldots) \\ \| (x_1, x_2, x_3, \ldots)\| = \max \{ |x_1|, |x_2|, |x_3|, \ldots \} $$
and the map sending $(x_1, x_2, x_3, \ldots)$ into $(x_1, 2x_2, 3x_3, \ldots)$ is a one-one linear transformation of this normed vector space onto itself that is not continuous.
I am focusing here on (c) in Problem 22. In short my solution went along the lines of:
Let $\|\cdot \|$ denote the abstract norm on $V$. Let $\{v_1, \ldots, v_n\}$ be a basis of $V$.
The map $x \mapsto \| x \|$, where $x\in V$, is continuous with respect to the topology induced by the norm $\| \cdot \|_1$
Via compactness, this enables us to conclude there is a real $M > 0$ such that $M \| x \|_1 \leq \| x \| $ for each $x \in V$.
Therefore for any $x \in V$: if $\|x\| < \delta$, then $ \| x \|_1 < \frac {\delta}M $.
Hence we have $\|f(x)\| \leq \sup\{ \| f(v_i) \| \}_{i=1}^n \cdot \| x\|_1 < \sup\{ \| f(v_i) \| \}_{i=1}^n \cdot \frac {\delta}M $, which can be made smaller than any beforehand given $\epsilon > 0$. $\square$
But in the subsequent Problem 23, we are tasked with following:
Use Problem 22 to prove that if $V$ is a finite dimensional vector space over $\mathbb{R}$ and $\| \|_1$, $\| \|_2$ are two norm functions on $V$ (i.e., real-valued functions such that $(V,\| \|_1)$ and $(V, \| \|_2)$ are normed vector spaces), then there exist positive real numbers $m,M$ such that $m \leq \frac { \| x\|_1 }{ \| x\|_2 } \leq M$ for all nonzero $x \in V$.
We are asked to use Problem 22 to solve Problem 23. Is this not strange, or am I misunderstanding something?
In order to solve Problem 22 (c), I had to use the property in Problem 23.
Does this mean Rosenlicht had in mind a different solution of Problem 22 (c)? One not invoking the property in Problem 23?
I have been looking online and have not find a proof that does not rely on the property in Problem 23, namely the equivalence of norms in finite dimensional vector spaces.
Does anyone know if Problem 22 (c) can be solved without invoking equivalence of norms? Thank you.
Best Answer
Let $\text{dim}(X) = n$ and there is a basis $\{e_1, \dots, e_n\}$ for $X$. We can write $x = \sum \zeta_i e_i$ and derive $$ \begin{align*} \|f(x)\|_Y =\left\|\sum\zeta_if(e_i)\right\|_Y &\leq \sum |\zeta_i|\|f(e_i)\|_Y \\ &\leq \max_k \|f(e_k)\|_Y \sum|\zeta_i| \end{align*} $$ Try to prove the following lemma:
Lemma
Using the lemma, we have $$ \sum |\zeta_i| \leq \frac1c \left\|\sum\zeta_ie_i\right\|_X = \frac1c \|x\|_X $$ for some $c>0$. Now you have $$ \|f(x)\|_Y \leq \alpha\|x\|_X $$ where $\alpha = \frac1c \max_k \|f(e_k)\|_Y$.
From this, you can try to prove that $f$ is continuous if it is bounded.