Prove there exists a positive number $c$ such that $‖v‖_1≤c‖v‖_2$ for all $v∈V$.

Question

Suppose $V$ is finite-dimensional and $〈·,·〉_1$, $〈·,·〉_2$ are inner product on $V$ with corresponding norms $‖·‖_1$ and $‖·‖_2$. Prove there exists a positive number $c$ such that $‖v‖_1≤c‖v‖_2$ for all $v∈V$.

{This problem is from Axler's "Linear Algebra Done Right."}

I just wonder why all the experts' answers use an orthonormal basis to prove this, while I just thought this could be proved as follows.

Proof starts.

Consider when $v=0$.

Then, by the definition of the inner product, $\left \| v \right \|_1=\left \| v \right \|_2=0$.

Thus, for any positive number $c$, $\left \| v \right \|_1 \leq c\left \| v \right \|_2$

Now, consider $v \neq 0$. Then let $\left \| v \right \|_1=a ,\left \| v \right \|_2=b $ for an arbitrary non-zero vector $v$.

But, note that $a,b>0$ by the definition of the inner product.

Thus, for $c \geq \frac{a}{b}$, we get $a \leq c b$ $\Leftrightarrow \left \| v \right \|_1 \leq c\left \| v \right \|_2$.

Hence, for any $v\in V$, we can find a positive number $c$ s.t $\left \| v \right \|_1 \leq c\left \| v \right \|_2$. $\blacksquare$

Is there something I'm missing here?

Answer 1

Yes. You are missing the fact that what you should be proving is that there is a number $c$ such that, for every $v\in V$, $\|v\|_1\leqslant c\|v\|_2$. That $c$ that you got depends on $v$.