# Pseudonorm Well Definedness

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.

• Use $\le$ in the definition. fixed. – GEdgar Apr 1 '18 at 14:15
• Try this to start ... $[0]$ is a vector subspace, and every $[x]$ is a coset of $[0]$. – GEdgar Apr 1 '18 at 14:18
• @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? – yoshi Apr 1 '18 at 14:47

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.