Is $\| f \|_{\infty} = \| f \|_D$? Let $K$ be a metric space. Denote $C(K) = \{ f:K \rightarrow \mathbb{R} : f \text{ is continuous} \}$ and $D(K) = \{ f:K \rightarrow \mathbb{R}:f=u-v, \text{ where }u,v \geq 0 \text{ are bounded and lower semicontinuous functions.} \}$
Let $\| f \|_D =\inf \{ \big| \sum_{n=1}^{\infty} |f_n| \big|_{\infty}: (f_n) \subseteq C(K) \text{ and } f = \sum_{n=1}^{\infty}f_n \text{ pointwise.} \}$
Note that the norm $\| f \|_{\infty}$ is the sup-norm. 
Question: For every $f \in D(K),$ is $\| f \|_{\infty} = \| f \|_D$ true? 
My attempt: For every $\varepsilon>0,$ there exists a sequence $(f_n) \subseteq C(K)$ such that $f = \sum_{n=1}^{\infty}f_n$ pointwise  and  $\| \sum_{n=1}^{\infty}|f_n| \|_{\infty} \leq \| f \|_D + \varepsilon.$
Clearly $\| f \|_{\infty} = \| \sum_{n=1}^{\infty}f_n \|_{\infty} \leq \| \sum_{n=1}^{\infty}|f_n| \|_{\infty} \leq \| f \|_D + \varepsilon.$
Since $\varepsilon$ is arbitrary, we have $\| f \|_{\infty} \leq \| f \|_D.$ 
For the reverse inequality $\| f \|_D \leq \| f \|_{\infty},$ I have no idea what to do.  
 A: You should consider putting this question on math overflow.  On line 2 of page 4 of this paper farmaki it says that the two norms are not equivalent.
There are a few counterexamples on page 52 of this paper rosenthal. I did not read all the details though. It seems you could take $[0,1]$ and a closed nowhere dense set there (for example the Cantor set $D$) and define $f=1$ in $[0,1]\setminus D$ and $f=-1$ on $D$. Then $\Vert f\Vert_D=3$. Good luck reading the paper! It does not look easy.
A: This is a partial result: $\|f\|_D\le \sup(u+v)$ where $u,v$ are as in the definition of $D(K)$. The remaining part of the puzzle is whether $u,v$ can be chosen so that $\sup(u+v) = \sup|f|$ (this would be true if one can take $uv=0$, but that is far from clear.) 
Every nonnegative lower semicontinuous function $u$ is the pointwise limit of an increasing sequence of nonnegative continuous functions. Indeed, let 
$$u_n(x) = \inf_{y\in K} (u(y)+nd(x,y))$$
Then $0\le u_n(x)\le u(x)$, and $u_n$ is Lipschitz  continuous with constant $n$, as the infimum of a family of $n$-Lipschitz functions $x\mapsto u(y)+nd(x,y)$. And since the points with $d(x,y)>n^{-1}u(x)$ don't contribute to the infimum, we have
$$
u_n(x) \ge \inf_{d(x,y)\le n^{-1}u(x)} u(y) \to u(x) \quad \text{as }n\to\infty 
$$
by virtue of the lower semicontinuity of $u$. 
Thus, we can write $u = \sum_{n=1}^\infty \tilde u_n$ where $\tilde u_1=u_1$ and $\tilde u_n = u_{n}-u_{n-1}$ for $n>1$. This is a series of nonnegative continuous functions. Note that $\sup \sum_n |\tilde u_n| = \sup u$.
Then represent $v=\sum_{n=1}^\infty \tilde v_n$ in the same fashion, and note that $f=sum_n \tilde u_n-\tilde v_n$ with $\sup \sum_n |\tilde u_n-\tilde v_n| \le  \sup (u+v)$.
