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If $S$ is a closed subspace of a real normed space $(X,\lVert \cdot \rVert)$ and there is a vector $x \in X \setminus S$, show that exist a bounded linear funcional $l \in X^{*}\setminus \{0_{X^{*}}\}$ such that $l(s)=0$ for all $s \in S$. Here $0_{X^{*}}$ is the zero operator.

Any idea how can I prove this result?

Thanks a lot!

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    Hahn-Banach theorem. – 2017-02-12
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    Could you please give me a more detailed answer? – 2017-02-12
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    First define the functional on the linear span of $S$ and $x$. Then extend... – 2017-02-12

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HINT Apply the Hahn-Banach separation theorem to the closed set $S$ and the compact set $\{x\}$.

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    By that theorem I know that exist a bounded lineral functional $l$ and $s,t$ in $\mathbb{R}$ such that $l(x)$s\in S$. What can I conclude from this? – 2017-02-12
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    @DiegoVargas You're overloading the variable $s$. Regardless, assume (towards a contradiction) that $l(y)\neq0$ for some $y\in S$. – 2017-02-12
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    Sorry for that I didn´t realize it! Did you mean the same $l$ of the H-B theorem? Or do I need to pick up another functional? – 2017-02-12
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    @DiegoVargas The same functional. – 2017-02-12
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    Sorry I can't see any contradiction. Could you give me another hint please – 2017-02-12
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    @DiegoVargas $l(S)$ is a subspace of $\mathbb{R}$. How many subspaces of $\mathbb{R}$ are there? – 2017-02-13
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    Only $\mathbb{R}$ or $\{0\}$ if we consider it as a real vector space. As I am supposing that $l(y)\neq 0$ for some $y \in S$, the only posibility is $l(S)=\mathbb{R}$ So $l(x)\in \mathbb{R}$ implies that $x\in S$ wich is a contradiction. Is it right? I don't see how to use H-B. – 2017-02-13
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    @DiegoVargas It's almost correct. Since $l(S)=\mathbb R$ we know $l(x)$ is less than every real number, which is a contradiction. – 2017-02-13
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    Thank you very much for yor time and help! @Aweygan – 2017-02-13
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    @DiegoVargas You're welcome – 2017-02-13