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Let \(\displaystyle{\left(X,||\cdot||\right)}\) be an \(\displaystyle{\mathbb{R}}\) - normed space and \(\displaystyle{Y}\) a

subspace of \(\displaystyle{\left(X,+,\cdot\right)}\) such that, if \(\displaystyle{f\in X^{\star}=\mathbb{B}(X,\mathbb{R})}\) with \(\displaystyle{f|_{Y}=\mathbb{O}}\) , then

\(\displaystyle{f=\mathbb{O}}\) . Prove that \(\displaystyle{Y}\) is dense on \(\displaystyle{\left(X,||\cdot||\right)}\) .

Suppose that \(\displaystyle{Y}\) is not dense on \(\displaystyle{\left(X,||\cdot||\right)}\) . Then,

the set \(\displaystyle{\overline{Y}\neq X}\) is a closed subspace of \(\displaystyle{\left(X,||\cdot\right)}\) and

according to \(\displaystyle{\rm{Hahn-Banach}}\) - theorem, there exists \(\displaystyle{f\in X^{\star}}\) such that

\(\displaystyle{f|_{\overline{Y}}=\mathbb{O}}\) and \(\displaystyle{||f||=1}\) . But now,

\(\displaystyle{y\in Y\implies y\in \overline{Y}\implies f(y)=0\implies f_{Y}=\mathbb{O}}\) and according to the

hypothesis, \(\displaystyle{f=\mathbb{O}}\), a contradiction, since \(\displaystyle{||f||=1}\) .