H
Hamdiken
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Hamdiken Asks: Showing an inequality in vector space.
Let $E$ be a normed vector space. We define a linear form $T:E\rightarrow \mathbb R$. Denote $H=Ker(T)$.
$H$ is the continuous inverse of a closed set ($\{0\}$), hence closed.
My guess is that since the set $\{a\}$ is bounded, and since $H$ is closed in a vector space, which means it's also bounded, gives the result that $a+H$ is closed. As for the $0$, we have for every $y\in a+H$, $T
=T(a+x)=T(a)+T(x)=1$. However $T(0)=0\neq1$, so $0\notin a+H$.
Since $0\notin a+H$, there exists a neighbourhood $V_0$ of $0$ such that $V_0\cap(a+H)=\emptyset$, hence the existence of an open ball $B(0,r)$ such that $B(0,r)\cap(a+H)=\emptyset$.
This is where I stopped. I thought of the orthogonal projection but didn't understand how to proceed.
Any help and correction of the answers above will be highly appreciated.
Let $E$ be a normed vector space. We define a linear form $T:E\rightarrow \mathbb R$. Denote $H=Ker(T)$.
- Show that if $T$ is continuous, then $H$ is a closed set.
$H$ is the continuous inverse of a closed set ($\{0\}$), hence closed.
- Suppose that $H$ is closed. Let $a\in E$ such that $T(a)=1$. Show that the set $a+H$ is closed and does not contain $0$.
My guess is that since the set $\{a\}$ is bounded, and since $H$ is closed in a vector space, which means it's also bounded, gives the result that $a+H$ is closed. As for the $0$, we have for every $y\in a+H$, $T
=T(a+x)=T(a)+T(x)=1$. However $T(0)=0\neq1$, so $0\notin a+H$.- Deduce the existence of $r>0$ such that $B(0,r)\cap(a+H)=\emptyset$.
Since $0\notin a+H$, there exists a neighbourhood $V_0$ of $0$ such that $V_0\cap(a+H)=\emptyset$, hence the existence of an open ball $B(0,r)$ such that $B(0,r)\cap(a+H)=\emptyset$.
- Show that for every $x\in B(0,r)$, we have $$|Tx|\leq 1$$
This is where I stopped. I thought of the orthogonal projection but didn't understand how to proceed.
Any help and correction of the answers above will be highly appreciated.
