in normed space hyperplane is closed iff functional associated with it is continuous 

E is a normed linear space .
i have two questions 
Q1 why the complement of H is nonempty 
Q2 How then the functional is continuous??
Thanks
 A: $Q1$: There is an implicit assumption in the theorem, and that is that $f$ is not identically zero. If $f=0$ then $\{x|f(x)=\alpha\}$ is either empty or the whole Banach space $X$, neither of which is a hyperplane. 
$H^c =\{ x|f(x)\neq \alpha\}$. Suppose $H^c$ is empty. Then $f(x)\neq \alpha$ is impossible, so $f(x) = \alpha$ for all $x$. This contradicts linearity of $f$. (Easy to check.)
$Q2$: For linear functionals $f$ over a Banach space $X$, continuity at $x\in X$ is equivalent to contiuity at $0\in X$, which is equivalent to local boundedness at $0\in X$, i.e., that
$$ \|f\| = \sup_{z\in B(0,1)} \frac{|f(z)|}{\|z\|} < +\infty. $$
Given that $ f(x_0 + r z) < \alpha$ for all $z \in B(0,1)$ we have also that $f(x_0 - r z)<\alpha$, since $-z\in B(0,1)$ if and only if $z\in B(0,1)$. Thus, $-f(z) = f(-z) < \tfrac{1}{r}(\alpha - f(x_0))$, as well as $f(z) < \tfrac{1}{r}(\alpha - f(x_0))$, and
$$ |f(z)| < \frac{1}{r}(\alpha - f(x_0)) $$
for all $z\in B(0,1)$. It follows that
$$ \|f\| \leq \frac{1}{r}(\alpha - f(x_0)). $$
