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

Question

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

Answer 1

$Q1$: 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 $$ \|f\| \leq \frac{1}{r}(\alpha - f(x_0)). $$