Norms on quotient spaces

Question

Let $(V, \Vert\cdot\Vert)$ be a normed space, $W$ be a closed subspace of $V$. Define $\Vert \cdot\Vert: V/W \to \mathbb{R}^+: v + W \mapsto \inf\{\Vert v + w \Vert \mid w \in W\}$. Prove that this is a norm on $V/W$.

My attempt:

I managed to show that this function is well defined, is non-degenerate and absolutely homogeneous. So I only have to show that the triangle inequality holds.

So, I must prove that $\Vert (v+ W) + (v'+W)\Vert \leq \Vert v + W \Vert + \Vert v' + W \Vert$

And I started in the following way:

$$\Vert (v + W) + (v' + W)\Vert = \Vert (v+v')+W\Vert = \inf_w \Vert v+v'+w\Vert = \inf_w \Vert (v+w) + (v' + w)\Vert $$

and in this last equality I use that $W \to W:w \mapsto 2w$ is bijective.

Continuing,

$\inf_w \Vert(v+w) + (v' + w)\Vert \leq \inf_w(\Vert v+w\Vert + \Vert v' + w\Vert) \leq \inf_w\Vert v+w\Vert + \inf_w\Vert v' + w \Vert = \Vert v+ W \Vert + \Vert v'+ W \Vert$

Is this correct? I'm asking this because I'm not really used to working with infinum.

Answer 1

$\Vert (v + W) + (v' + W)\Vert = \Vert (v+v')+W\Vert $

Is that based on (v + W) + (v' + W) = (v+v')+W? You should consider whether your instructor will consider that trivial (they may well do so, I don't know what's been established in your course). Also keep in mind that this is a bit of abuse of notation, in that you (and the original problem) are using the symbol '+' to represent both addition within V and addition within V/W.

$\inf_w \Vert v+v'+w\Vert = \inf_w \Vert (v+w) + (v' + w)\Vert $

I suppose that technically is true, but rigorously showing that it's true would take a significant amount of work. Almost as much, perhaps more, work than what you were originally asked to prove.

$\inf_w \Vert(v+w) + (v' + w)\Vert \leq \inf_w(\Vert v+w\Vert + \Vert v' + w\Vert)$

Again, true, but nontrivial. You would have to apply the triangle inequality, and show that it behaves well when combined with inf.

$\inf_w(\Vert v+w\Vert + \Vert v' + w\Vert) \leq \inf_w\Vert v+w\Vert + \inf_w\Vert v' + w \Vert $

Another nontrivial claim.

Overall, this reads more as work that takes the properties of norms and inf as given, rather than a rigorous proof.

I would start at the other end. We have $\Vert v+ W \Vert + \Vert v'+ W \Vert = \inf_w\Vert v+w\Vert + \inf_w\Vert v' + w \Vert$, but we need to recognize that the two inf's have different dummy variables. While technically, dummy variables "disappear" outside of whatever scope they're defined, and what name you choose is irrelevant outside of that scope, it's still a good idea to use different name so that you're not fooled into thinking that they're the same variable and you can combine the two infs. So we should write it as $\inf_w\Vert v+w\Vert + \inf_{w'}\Vert v' + w' \Vert$ to emphasize that these are allowed to be different w's. So now given any $\epsilon$, you can find particular $w_0$, $w'_0$ such that $\Vert v+w_0\Vert + \Vert v' + w'_0 \Vert$ is within $\epsilon$ of $\inf_w\Vert v+w\Vert + \inf_{w'}\Vert v' + w' \Vert$ (Note that this follows from the definition of inf, with a little bit of work. I'm leaving it to you to fill in the details here.) Then you can combine this with the triangle inequality to get $\Vert v+w_0\Vert + \Vert v' + w'_0 \Vert \ge\Vert v+w_0 + v' + w'_0 \Vert$ = $\Vert (v+v') + (w'_0 + w_0) \Vert$. This is, by definition, greater than or equal to $\inf_{w''}\Vert (v+v') + w'' \Vert$ (That is, the inf of an expression is, by definition, less than or equal to all possible values of the expression). Note that in this inf, I used a third dummy variable of w'' to emphasize that this w is allowed to vary independently of w and w'.

So now you just have to take $\epsilon$ to zero and show that the inequality holds.