let $B(,)$ and $B'(,)$ be inner products over $\mathbb R$. show there is $c \in \mathbb R$ s.t $B(u,u) \leq cB'(u,u)$

Question

As the title says, let $B(,)$ and $B'(,)$ be inner products over $\mathbb R^n$. show there is $c \in \mathbb R^n$ s.t $B(u,u) \leq cB'(u,u)$ for all $u \in \mathbb R^n$.

I have been thinking about this for an hour now but still haven't found any leads. I'm thinking maybe because $B,B'$ are symmetric positive definite bilinear form, then there is a basis $C$ of $\mathbb R^n$ such that $[B]_C$ and $[B']_C$ are diagonal, but it hasn't gotten me far.

I'd appreciate some help with this.

Answer 1

The key point is that $\mathbb{R}^n$ has a basis, as you noticed. Let me correct the details of your attempt (and introduce a simpler norm to work with):

Let $(x_1, \dots, x_n)$ be a basis of $\mathbb{R}^n$, let $x = a_1 x_1 + \dots + a_n x_n$, then we define the $\infty$-norm as: $$N(x) = \|x\|_\infty = \max(|a_1|, \dots, |a_n|)$$

Then for any norm $\| \cdot \|$:

$$\begin{align}\|a_1 x_1 + \dots + a_n x_n\| & \leq \|a_1 x_1\| + \dots + \|a_n x_n\| \\ & = |a_1| \cdot \|x_1\| + \dots + |a_n| \cdot \|x_n\| \\ & \leq N(x) \cdot \|x_1\| + \dots + N(x) \cdot \|x_n\|\\ & = \alpha N(x) \end{align}$$

where $\alpha = \|x_1\| + \dots + \|x_n\|$.

On the other hand, assume that there is no $\beta > 0$ such that for any $x$, $N(x) \leq \beta \|x\|$. Then for any $n \in \mathbb{N}$, there exists $x(n)$ such that $N(x(n)) > n \|x(n)\|$. Consider $y(n) = \frac{1}{N(x(n))} x(n)$ such that: $$N(y(n)) = 1$$ $$\|y(n)\| < \frac 1 n$$

By the Bolzano-Weierstrass theorem, since $y(n)$ is bounded (first equation), it has a convergent subsequence towards a limit $l$. By the second equation, $\|l\| = 0$ and therefore $l = 0$. Therefore $N(0) = 1$ by the first equation: this is absurd. Therefore there exists $\beta > 0$ such that $N(x) \leq \beta \|x\|$ for every $x$.

Now given any two norms $\|\cdot\|$ and $\|\cdot\|'$, and $\alpha, \beta, \alpha', \beta'$ such as before, we have:

$$\|x\| \leq \alpha N(x) \leq \alpha \beta' \|x\|'$$

Answer 2

I'm not 100% sure on this answer, I'd like someone to review it.

let $(\phi_1,\phi_2,...\phi_n)$ be a basis of $\mathbb R^n$, so for every $x \in R^n$ we get that $$x=\sum_{i=1}^n \alpha_i \phi_i$$

Let $N(x)$ be the norm: $$N(x)=\sqrt{\sum_{i=1}^n |\alpha_i|^2}$$

I will show that $N(x)$ is an upper bound of all norms. Meaning $||x|| \leq \beta N(x)$ where $||.||$ is some norm and $\beta \in \mathbb R$. It follows that all norms are bound by one another, meaning either $||x|| \leq ||x||^{'} \leq N(x)$ or $||x||^{'} \leq ||x|| \leq N(x)$

let $||.||$ be some norm.

$$||x||=||\sum_{i=1}^n \alpha_i \phi_i|| \leq \sum_{i=1}^n |\alpha_i|* ||\phi_i|| \leq \sqrt{\sum_{i=1}^n |\alpha_i|^2}*\sqrt{\sum_{i=1}^n ||\phi_i||^2}=\sqrt{\sum_{i=1}^n ||\phi_i||^2}*N(x)$$

So overall: $$||x|| \leq \beta N(x), \beta=\sqrt{\sum_{i=1}^n ||\phi_i||^2}$$

Q.E.D