Let $E$ be a normed space and let $L \subset E$ be a linear subspace. We put $$L^{\perp} := \{f \in E^{*} \mid \forall x \in L \; \langle f,x \rangle = 0 \}$$ Show that $${\rm dist}(f, L^{\perp}) := \sup\{ \langle f,x\rangle \mid ||x|| \leq 1, \; x \in L\}$$
Here I believe ${\rm dist}(f,L^{\perp})$ means the minimum of the set of distances from $f$ to $l \in L^{\perp}$
$\textbf{Theorem.}$ (Fenchel-Rockafeller). Let $\phi, \psi: E \to (-\infty, \infty]$ be two convex functions. Assume that there is some $x_0 \in D(\phi) \cap D(\psi)$ such that $\phi$ is continuous at $x_0$. Then, $$\begin{align*} \inf_{x \in E}\{\phi(x) + \psi(x)\} &= \sup_{x \in E^{\star}} \{-\phi^{\star}(-f) - \psi^{\star}(f)\}\\ &= \max_{f \in E^{\star}} \{-\phi^{\star}(-f) - \psi^{\star}(f)\}\\ &=-\min_{f\in E^{\star}} \{ \phi^{\star}(-f) + \psi^{\star}(f)\} \end{align*}$$
where $\phi^{\star}(f) = \sup_{x \in E, \; f \in E^{\star}} \{ \langle f, x\rangle - \phi(x)\}$ is the "conjugate function".
Note that by definition, $E^{\star}$ is equipped with the operator norm. So that, $$||x|| := \sup_{||f|| \leq 1, f \in E^{\star}} \langle f,x \rangle, \; ||f|| := \sup_{||x|| \leq 1, x \in E} \langle f,x \rangle,$$ and $\langle f, x\rangle = f(x)$ is called a dual pair and is not an inner product. We must show that $$\inf_{f \in E^{\star}} \{\phi(f) + \psi(f)\} = \sup_{x \in E} \{ - \phi^{*}(-x) - \psi^{\star}(x)\} = \sup_{||x|| \leq 1, \; x \in L}\langle f,x\rangle $$ So, $dist(f, L^{\perp}) = \inf_{f \in E^{\star}}\{||f - f_0|| \}, \; f_0 \in L^{\perp}$. Now we must define our functions $\phi$ and $\psi$.
Let $\phi(f) = ||f - f_0||$ and $\psi(f) = I_{k}(f) = \begin{cases} 0, & f\in L^{\perp} \\ \infty, & f \not\in L^{\perp} \end{cases}$
Now we write out the conjugate functions $\phi^{\star}(f), \psi^{\star}(f): E^{\star} \to (-\infty, +\infty]$.
$$\phi^{\star}(f) = \sup_{f\in E^{\star}} \{ \langle f,x \rangle - ||f - f_0||\} = \sup_{f \in E^{\star}} \{\langle f, x \rangle - ||f|| \} + \langle f, f_0 \rangle = \begin{cases} \langle f, f_0 \rangle, & ||x|| \leq 1 \\ \infty, & ||x|| > 1 \end{cases}$$ and $$\psi^{\star}(f) = \sup_{f \in E^{\star}} \{ \langle f,x \rangle - I_{k}(f)\} = \begin{cases} 0, & x \in L \\ \infty, & x \not\in L\end{cases}$$ Now I apply the theorem and substitute $\phi(x)$ and $\psi(x)$ into $\sup_{E^{\star}} \{-\phi^{\star}(-f) - \psi^{\star}(f) \}$. We are only interested in functionals in $L^{\perp}$, so we consider only $x \in L$ and $||x|| \leq 1$. Hence, $$\sup_{f\in E^{\star}} \{-\phi^{\star}(-f) - \psi^{\star}(f) \} = \sup_{f \in E^{\star}, x \in L, ||x|| \leq 1} \{-\langle f,f_0\rangle - 0 \}$$
This isn't turning out to be what the exercise is asking me to show though. Can anyone please tell me where I am going wrong with my reasoning?
The following is more straightforward that invoking Fenchel Rockafeller:
Let $\overline{B}$ be the closed unit ball.
Let $M= \sup_{x \in \overline{B} \cap L} |f(x)|$. Define $\phi(x) = f(x)$ for $x \in L$ and use the Hahn Banach theorem extend $\phi$ to all of $E$ in such a way that $\|\phi\| = M$. Note that $g = f-\phi \in L^\bot$. Then $\|f-g\| = \|\phi\| = M$, and hence $\inf_{g \in L^\bot} \|f-g\| \le \sup_{x \in \overline{B} \cap L} |f(x)|$.
For the other direction, note that if $g \in L^\bot, x \in \overline{B} \cap L$ then $\|f-g\| \ge |f(x)-g(x)| = |f(x)|$. Taking the $\sup$ of the right hand side gives $\|f-g\| \ge M$, and hence $\inf_{g \in L^\bot} \|f-g\| \ge M=\sup_{x \in \overline{B} \cap L} |f(x)|$.