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I must be missing something simple...

For a bilinear map $B: X \times Y \rightarrow Z$, the norm of the map is defined as $$\|B\| :=\sup_{\|x\| \leq 1, \hspace{1mm} \|y\| \leq 1} \|B(x, y)\|_Z.$$ This definition implies the following inequality: $$\|B(x, y)\|_Z \leq \|B\| \cdot\|x\|_X \cdot \|y\|_Y.$$ This definition is not what I would have guessed. I would have guessed the following definition: $$\|B\| :=\sup_{\|x\| = 1, \hspace{1mm} \|y\| = 1} \|B(x, y)\|_Z$$ which I thought would imply the same inequality: $$\|B(x, y)\|_Z \leq \|B\| \cdot\|x\|_X \cdot \|y\|_Y.$$ Can someone tell me why my definition of choice would fail to imply the inequality of interest that inspired the definition.

BenB
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1 Answers1

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They are actually equivalent.

Clearly $$\sup_{\|x\| = \|y\| = 1} \left\| B(x,y)\right\| \le \sup_{\|x\|, \|y\| \le 1} \left\| B(x,y)\right\|$$

Conversely, let $x,y \in X$ such that $\|x\|, \|y\| \le 1$.

We have:

$$\left\|B(x,y)\right\| = \underbrace{\|x\|\|y\|}_{\le 1}\left\|B\left(\frac{x}{\|x\|},\frac{y}{\|y\|}\right)\right\| \le \left\|B\left(\frac{x}{\|x\|},\frac{y}{\|y\|}\right)\right\| \le \sup_{\|u\| = \|v\| = 1} \left\| B(u,v)\right\|$$

Taking the supremum over all $x,y \in X$ such that $\|x\|, \|y\| \le 1$ we obtain

$$\sup_{\|x\|, \|y\| \le 1} \left\| B(x,y)\right\| \le \sup_{\|x\| = \|y\|= 1} \left\| B(x,y)\right\|$$

mechanodroid
  • 46,490
  • This looks solid. I had a similar justification. I was worried that I was missing some technical point, but it appears that the definitions are equivalent. I am curious why the definition with the inequalities is more common, as for linear maps of a single variable the definition of the norm is typically given as the supremum over the unit sphere. – BenB Mar 21 '18 at 21:13
  • @BenB For a linear map $A$ the four of these are quite common:

    $$|A| = \sup_{|x| \le 1}|Ax| = \sup_{|x| = 1}|Ax| = \sup_{x \ne 0} \frac{|Ax|}{|x|} = \inf{M > 0 : |Ax| \le M|x|, \forall x \in X}$$

    Also if $X$ is an inner product space then

    $$|A| = \sup_{|x|, |y| \le 1}\left|\langle Ax, y\rangle\right| = \sup_{|x|= |y| = 1}\left|\langle Ax, y\rangle\right|$$ which is very similar to the definition of norm $|B|$.

     – mechanodroid Mar 21 '18 at 21:24