Proof a corollary on Hahn-Banach theorem A corollary says
Let $F\subset E,$ is a closed linear subspace of $E$, where $E$ is a normed vector space. Then it exists a linear functional $f\in E^*,f\neq 0$ such that
$$ \langle x,f \rangle = 0, \forall x\in F .$$
That is, every closed linear subspace $F$ is contained by the kernel of some non-trivial continuous linear functional $f$. 
I tried to prove this as follows.
Hahn-Bananch theorem implies for each $x_0\in E \setminus F $,
$$\exists f\in E^* \Longrightarrow  \langle x,f \rangle \le \alpha \leq \langle x_0, f \rangle, \forall x\in F .$$
But how to preceed ? 
Any one helps ?
Thank you in advance
 A: Recall the quotient space $E/F$, whose elements are of the form
$$[x]:=x+F=\{x+y\in E\,:\,y\in F\}$$
where $x\in E$. This is always a vector space, but given a normed structure on $E$ it is not necessarily obvious how to introduce a related norm on $E/F$. However, if $F$ is closed then we can define a norm as
$$\|[x]\|_{E/F}:=\inf_{y\in F}\|x+y\|$$
It is not immediately obvious that this is even a well-defined map, so we shall demonstrate this. Suppose $x,y\in E$ are such that $[x]=[y]$. Then there exists $z\in F$ such that $y=x+z$. Observe that
$$\{\|x+w\|\,:\,w\in F\}=\{\|x+z+w\|\,:\,w\in F\}$$
and hence $\|[x]\|_{E/F}=\|[y]\|_{E/F}$, so the map is well-defined. It should be reasonably obvious that $\|\alpha[x]\|_{E/F}=|\alpha|\|[x]\|_{E/F}$, and the triangle inequality is also not too difficult: given $x,y\in E$ and $u,v\in F$ we have
$$\|[x+y]\|_{E/F}\le\|x+y+u+v\|\le\|x+u\|+\|y+v\|,$$
so taking infima over all $u,v\in F$ we get $\|[x+y]\|_{E/F}\le\|[x]\|_{E/F}+\|[y]\|_{E/F}$. To show $\|\cdot\|_{E/F}$ is indeed a norm it suffices to show that if $\|[x]\|_{E/F}=0$ then $[x]=[0]$. Suppose $\|[x]\|_{E/F}=0$ and let $(y_n)$ be a sequence in $F$ such that $\|x+y_n\|\to0$. Then $y_n\to-x$ as $n\to\infty$, so since $F$ is closed we have $x\in F$. In particular, this implies $[x]=[0]$. Hence $\|\cdot\|_{E/F}$ is indeed a norm.
Now use the most common corollary to Hahn-Banach: given a normed vector space $X$ and $x\in X$, $x\ne0$ there exists $x^*\in X^*$ such that $\|x^*\|=1$ and $\langle x^*,x\rangle=\|x\|$. Let $[x_0]\in E/F$, $[x_0]\ne[0]$, so by the stated result there is $f\in(E/F)^*$ such that $\|f\|=1$ and $\langle f,[x_0]\rangle=\|[x_0]\|_{E/F}$.
Define $\phi:E\rightarrow\mathbb{K}$ by $\phi(x)=\langle f,[x]\rangle$. Clearly $\phi$ is linear, and $\|\phi(x)\|\le\|f\|\|[x]\|_{E/F}\le\|f\|\|x\|$ so $\phi\in E^*$. If $x\in F$, then $\phi(x)=\langle f,[x]\rangle=\langle f,[0]\rangle=0$. However $\phi\ne0$ since $\phi(x_0)=\|[x_0]\|_{E/F}>0$. This completes our proof.
