In a Banach space X, given a norm one bounded linear functional $f$ and $c\in \mathbb{C}\backslash \{0\}$, define $H = \big\{ x\in X \,\vert\, f(x) = c\big\}$ and $\inf H$ = $\inf\_{h\in H} \|h\|$.


1. Is there a necessary condition for $\inf H = \vert\, c\,\vert$?



Infimum of norms of elements in a hyperplane
I am assuming your linear functional means "complex linear", then $\inf H = |c|$ will always be true if $\lVert f \rVert = 1$.


On the one hand, since $\lVert f \rVert = 1$, we can choose a sequence $\{x\_n\}$ of norm $1$ elements in $X$ such that $|f(x\_n)| > 0$ and $|f(x\_n)| \rightarrow \lVert f \rVert = 1$. Replacing $x\_n$ with $\dfrac{|f(x\_n)|}{f(x\_n)} \cdot x\_n$, we can assume that $f(x\_n) \in (0, + \infty)$ and $f(x\_n) \rightarrow 1$. Apparently $\dfrac{c}{f(x\_n)} \cdot x\_n \in H$ and $\left\lVert \dfrac{c}{f(x\_n)} \cdot x\_n \right\rVert = \dfrac{|c|}{|f(x\_n)|} \rightarrow |c|$, so $\inf H \le |c|$.


On the other hand, if $\inf H < |c|$, then there exists a sequence $\{ y\_n \} \subseteq H$, $f(y\_n) = c$ and $\lVert y\_n \rVert \rightarrow \inf H$, which leads to a contradiction that $|c|=|f(y\_n)| \le \lVert f \rVert \lVert y\_n \rVert \rightarrow \inf H < |c|$.


About your second question, is there really any difference between real and complex hyperplanes in a complex Banach space? I am confused about your motivation of asking this question.


This qusetion is more suitable for MSE, though.


