Let $\left(X,||\cdot||\right)$ be an $\mathbb{R}$ normed space and $Y$ be a subspace of $\left(X,+,\cdot\right)$ such that , if $f \in X^*= \mathbb{B} (X, \mathbb{R})$ with $f|_{Y}=\mathbb{O}$ then $f=\mathbb{O}$. Prove that $Y$ is dense on $\left(X,||\cdot||\right)$.
Solution
Suppose that $Y$ is not dense on $\left(X,||\cdot||\right)$ . Then, the set $\overline{Y}\neq X$ is a closed subspace of $\left(X,||\cdot\right)$ and according to Hahn-Banach - theorem, there exists $f\in X^*$ such that $f|_{\overline{Y}}=\mathbb{O}$ and $||f||=1$ . But now,
$$y\in Y\Rightarrow y\in \overline{Y}\Rightarrow f(y)=0\Rightarrow f|_{Y}=\mathbb{O}$$
and according to the hypothesis, $f=\mathbb{O}$, a contradiction, since $||f||=1$.
The exercise can also be found in mathimatikoi.org
Solution
Suppose that $Y$ is not dense on $\left(X,||\cdot||\right)$ . Then, the set $\overline{Y}\neq X$ is a closed subspace of $\left(X,||\cdot\right)$ and according to Hahn-Banach - theorem, there exists $f\in X^*$ such that $f|_{\overline{Y}}=\mathbb{O}$ and $||f||=1$ . But now,
$$y\in Y\Rightarrow y\in \overline{Y}\Rightarrow f(y)=0\Rightarrow f|_{Y}=\mathbb{O}$$
and according to the hypothesis, $f=\mathbb{O}$, a contradiction, since $||f||=1$.
The exercise can also be found in mathimatikoi.org
No comments:
Post a Comment