1
$\begingroup$

The question is as follows (Its not very long - most of it is just definitions!):

A nonnegative real-valued function $\| \cdot \|$ defined on a vector space $X$ is called a pseudonorm if $\|x + y\| \le \|x\| + \|y\|$ and $\|\alpha x\| = |\alpha|\; \|x\|$. Define $x \cong y$, provided $\|x - y\| = 0$. Define $X/\cong$ to be the set of equivalence classes of X under $\cong$ and for $x \in X$ define $[x]$ to be the equivalence class of $x$. Show that $X/\cong$ is a normed vector space if we define $\alpha[x] +\beta[y]$ to be the equivalence class of $\alpha x + \beta y$ and define $\|[x]\|=\|x\|$.

I think there's a bunch of stuff to do here:

  • Show linearity of classes is well defined

  • Show classes, so defined, form a vector space

  • Show the norm is well defined

  • Show the norm, so defined, satisfies the usual properties

My questions are:

  1. Are these the things to do in the problem?

  2. If so, I can do the last 3, but I'm getting mixed up on the first -- arguing that linearity of classes is well-defined. Usually I start with 2 elements that are in the same class and show the operation defined on one is equal to operation defined on the other. But here we have a binary operation - so I'm confused about how this argument should proceed. It occurred to me that can show $\alpha[x]+\beta[y] = [\alpha x + \beta y]$ as sets but I'm not sure this is sufficient.

$\endgroup$
  • 1
    $\begingroup$ Use $\le$ in the definition. fixed. $\endgroup$ – GEdgar Apr 1 '18 at 14:15
  • 1
    $\begingroup$ Try this to start ... $[0]$ is a vector subspace, and every $[x]$ is a coset of $[0]$. $\endgroup$ – GEdgar Apr 1 '18 at 14:18
  • $\begingroup$ @GEdgar can you elaborate? Part of my problem is I'm having trouble with the logic of the argument and what I need to show. I can see that your hint will show equality as sets -- but is that enough? $\endgroup$ – yoshi Apr 1 '18 at 14:47
1
$\begingroup$

Let $x,y,x',y' \in X$ such that $[x] = [x']$ and $[y] = [y']$. To verify that addition and scalar multiplication is well defined, we wish to show that $[\alpha x + \beta y] = [\alpha x' + \beta y']$ for any scalars $\alpha, \beta$.

We have

$$\|\alpha x + \beta y - (\alpha x' + \beta y')\| \le |\alpha|\|x-x'\| + |\beta|\|y-y'\| = 0$$

because $\|x-x'\| = \|y-y'\| = 0$ by assumption. Therefore $[\alpha x + \beta y] = [\alpha x' + \beta y']$ so we can define

$$[x] + [y] = [x+y]$$ $$\alpha[x] = [\alpha x]$$

It is similar for the norm. Assume that $[x] = [x']$ and we wish to show that $\|x\| = \|x'\|$.

We have

$$\|x\| \le \|x-x'\| + \|x'\| = \|x'\|$$ $$\|x'\| \le \|x'-x\| + \|x\| = \|x\|$$

so $\|x\| = \|x'\|$. Therefore, we can define $\|[x]\| = \|x\|$.

Now it should be straightforward to check that the operations and the norm satisfy the required axioms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.