Duality Theorem Confusion A book that I am reading states the following theorem:
Theorem. Let $x$ be an element in a normed linear space $X$ and let $d$ denote its distance from the subspace $M (\bar{M}\neq X)$. Then
\begin{equation*}
d \triangleq \min_{m\in M} \|x-m\|_{M} = \max_{\substack{x^*\in M^\perp \\ \|x\|_{M^*} \leq 1}} <x,x^*>,
\end{equation*}
where the maximum on the right is achieved for some $x_0^*\in M^\perp$ with $\|x_0^*\|_{M^*} = 1$. If the minimum on the left is achieved for some $m_0\in M$, then $x-m_0$ is aligned with $x_0^*$. 
Proof. The dual problem is a lower bound of the primal problem since 
\begin{equation*}
\max_{\substack{x^*\in M^\perp \\ \|x\|_{M^*} \leq 1}}<x,x^*> = \max_{\substack{x^*\in M^\perp \\ \|x\|_{M^*} \leq 1}} <x-m,x^*>\ \leq \ \|x-m\|_M, \qquad \forall m\in M.
\end{equation*}
To show equality, we have to exhibit a functional $x^* \in M^\perp$ that achieves the value $d$. Let $[x+M]$ be the subspace generated by $x$ and $M$, and consider the functional $f: [x+M]\rightarrow\mathbb{R}$ defined as follows:
\begin{equation*}
f(\alpha x+m) = \alpha d, \qquad \forall \alpha\in\mathbb{R}.
\end{equation*}
Then $f$ is a bounded linear functional on $[x+M]$ with induced norm ($\|f\|_M$)
\begin{align*}
\|f\|_M &\triangleq \sup_{\alpha\in\mathbb{R},m\in M}\frac{|f(x)|}{\|\alpha x + m\|_M}\\
&= \sup_{\alpha\in\mathbb{R},m\in M}\frac{|\alpha|d}{|\alpha|\|x+\frac{1}{\alpha}m\|_M}\\
&= \sup_{m\in M}\frac{d}{\|x+m\|_M} \quad \text{Since M is a subspace}\\
&= \frac{d}{\inf_{m\in M}\|x+m\|_M} = 1.
\end{align*}
Define $x_0^*$ to be the Hahn-Banach extension of $f$. It follows immediately that $\|x_0^*\|_{M^*} = 1$, and $<x,x_0^*>=d$. The alignment follows from the fact that 
\begin{equation*}
d = \|x-m_0\|_M\|x_0^*\|_{M^*} = <x,x_0^*> = <x-m,x_0^*>
\end{equation*}
for any minimizing solution $m_0$.

Two questions:


*

*Since this theorem applies to normed linear space (and subsequently I am using it for Banach spaces), how is possible to take inner products, such as $<x,x^*>$. I thought that inner products are only defined on Hilbert spaces? It seems as though the whole theorem falls apart if an inner product is not defined.

*Further, how is there a notion of an orthogonal complement $M^\perp$ if there is no inner product?
 A: With normed spaces that are not Hilbert spaces, elements of the dual space are bounded linear functionals that take an element of the normed space and return a real value. The notation $\langle x, x^{*} \rangle$ means the evaluation of the bounded linear functional $x^{*}$ (an element of the dual space) at $x$ in the original normed space. 
If it happens that the space we start with is a Hilbert space, then all of the bounded linear functionals $f(x)$ are of the form
$f(x)=\langle x, f \rangle=\mbox{inner product}(x,f)$
for some element $f$ in the Hilbert space. Thus in the Hilbert space case, the notation $\langle x, f \rangle$ corresponds exactly to the inner product of $x$ and $f$.
This choice of notation can be very confusing to students if they've previously seen $\langle x, y \rangle$ for the inner product of $x$ and $y$.  Some books reserve $(x,y)$ for the inner product in Hilbert spaces to avoid that confusion. 
The definition of orthogonality in a Hilbert space is that two elements of the space, $x$, $y$, are orthogonal if $(x,y)=0$.  This extends to orthogonality of $x$ in a Banach space $S$ and $y$ in the dual space $S^{*}$, if $\langle x,y \rangle=0$.  If $M$ is a subset of a Banach space $S$, then the $M^{\perp}$ is the set of $y$ in $S^{*}$ such that $\langle x, y \rangle=0$ for every $x$ in $M$.  Notice that $M^{\perp}$ lives in the dual space $S^{*}$ rather than in $S$.   
See for example Chapter 5 of David G. Luenberger's Optimization by Vector Space Methods.
