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Let $(E,\|\cdot\|)$ a normed space and consider $M \subseteq E$ a closed vectorial subspace. Consider in $E$ the equivalence relation $x \equiv y \iff x-y \in M$, and let $E/M$ the quotient set. The equivalence class of $x$ is the set $x+M = \{ x+m | m \in M \}$. Show that $E/M$ is a vectorial space with the opperations $(x+M) +' (y+M) = (x+y) + M$ and $\lambda \cdot'(x+M) = \lambda x + M$. Show that the expression $\|x+M\| = \inf \{ \|x+m\| \mid m \in M\}$ defines a norm in $E/M$.

Please verify my progress in proving $E/M$ is a vectorial space:

  1. Well-definiteness of $+'$ and $\cdot'$:

Suppose $x+M = x'+M$ and $y+M = y'+M$. Then, by definition of set $x+M$, $x=x'$ and $y=y'$, then $x+y = x'+y'$.

Suppose $x+M = x'+M$. Then $x=x'$. Hence $\lambda (x+M) = \lambda x + M = \lambda x' + M = \lambda (x'+M)$.

  1. Axioms of $+'$:

$$(x+M) +' (y+M) = (x+y) +' M = (y+x) +' M = (y+M) +' (x+M)$$

$$((x+M) +' (y+M)) +' (z+M) = ((x+y)+M) +' (z+M) = (x+y+z)+M$$ $$(x+M)+'((y+M)+'(z+M)) = (x+M) +' +((y+z)+M) = (x+y+z)+M$$

$$M +' (x+M) = x+M = (x+M) +' M$$

$$(-x+M) +' (x+M) = M = (x+M) +' (-x+M)$$

  1. Axioms of $\cdot'$:

$$0 \cdot' (x+M) = M$$

$$1 \cdot' (x+M) = x+M$$

$$(\lambda \mu)\cdot'(x+M) = \lambda \mu x + M = \lambda\cdot'(\mu x + M) = \lambda\cdot' ( \mu \cdot'(x+M))$$

  1. Axioms of Distributivity:

$$\lambda \cdot'((x+M) +' (y+M)) = \lambda\cdot'((x+y)+M) = \lambda(x+y) + M = \lambda x + \lambda y + M = (\lambda x + M) +' (\lambda y + M) = \lambda \cdot' (x+M) +' \lambda \cdot' (y+M)$$

$$ (\lambda + \mu) \cdot' (x+M) = (\lambda + \mu)x + M = (\lambda x + \mu x) + M = (\lambda x + M) +' (\mu x + M) = \lambda \cdot'(x+M) +' \mu \cdot'(x+M)$$

Then I need to prove that the expression $\|x+M\| = \inf \{ \|x+m\|\mid m \in M\}$ defines a norm in $E/M$. I'm not sure how to conclude positive definiteness neither if what I did on homogeneity and triangle inequality is right.

Positive definiteness: $\|x+M\| \geq 0$, since $\|x+m\| \geq 0$ and and so the infimum has to be $0$ or bigger. If $\inf \{ \|x+m\| \mid m \in M \} = 0$, then $\forall \epsilon>0 \exists y \in \{ \|x+m\| \mid m \in M \} $ such that $ 0 \leq y < \epsilon$. I need to conclude that $x=0$, but I'm not sure how.

Homogeneity: $||\lambda\cdot'(x+M)|| = ||\lambda x + M || = \inf \{ || \lambda x + m|| | m \in M \} = \inf \{ || \lambda x + \lambda m|| | m \in M \}$ (since $\lambda \in M)$ $ = \inf \{ || \lambda (x + m)|| | m \in M \} = |\lambda| \inf \{ || x + m|| | m \in M \} = |\lambda|||x+M||$

Triangle Inequality:

$ || (x+M) +' (y+M)|| = \inf \{ || (x+y) + m|| | m \in M \} \leq \inf \{ || (x+y) + 2m|| | m \in M \} \leq \inf \{ ||x+m|| + ||y + m|| | m \in M \} = \inf \{ ||x+m|| | m \in M \} + \inf \{||y + m|| | m \in M \} $

Thanks.

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For positive definiteness you should use that $M$ is assumed closed. So if $\|x+m_k\|\rightarrow 0$, $m_k\in M$ then show (1) that the sequence $(m_k)_k$ is Cauchy, (2) that it therefore converges to some $m\in M$ and (3) finally $x=-m\in M$.

In homogeneity you should distinguish the case $\lambda=0$ and $\lambda\neq 0$.

In the triangle inequality perhaps better to say for the last part: $$\inf\{\|x+y+m\| : m\in M\}=\inf\{\|x+m+y+n\| : m,n\in M\} \leq \inf\{\|x+m\|+\|y+n\| : m,n\in M\} = \inf\{\|x+m\| : m\in M\}+\inf\{\|y+n\| : n\in M\}$$

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