Real-valued harmonic function as NOT the difference of two nonnegative harmonic functions The following is a question that I found on an old qualifying exam:
Find a real-valued harmonic function on the unit disk $\mathbb{D}$ that cannot be written as the difference of two nonnegative harmonic functions.
 A: *

*For any harmonic function, the average   over the circle $|z|=r$ stays the same for all $r<1$, namely it is the value at $0$. 

*If $u=v-w$, then $|u|\le |v|+|w|$. 

*Combine 1 and 2 to conclude: if $u=v-w$ where $v$ and $w$ are nonnegative harmonic functions, then the average   of $|u|$ on the circle $|z|=r$ is bounded by a constant independent of $r$. 

*It remains to find a harmonic function $u$ for which the circular averages of $|u|$ are unbounded. One example is $\operatorname{Re}((1-z)^{-2})$. Indeed, when $|\arg (1-z)|<\pi/8$ it admits the estimate 
$$\operatorname{Re}(1-z)^{-2} > \frac{1}{\sqrt{2}} |1-z|^{-2}$$ 
Since the intersection of the circle $|z|=r$ with the sector  $|\arg (1-z)|<\pi/8$ contains an arc of length of order $r$ on which $|1-z|$ is of order $r$, the claim  follows.

A: This is not a complete solution - 
Every non-zero harmonic function $u$ can be written as
$ u = \dfrac{|u|+u}{2} - \dfrac{|u|-u}{2}$. At points where $u>0$, by the identity : $u = u - 0$ and when $u<0$ then $u = 0 - (-u)$. In both the cases, the functions are non-negative. An alternative argument is $|u|\geq u$ and $|u|\geq -u$ for all real valued $u$ and hence the difference functions are always non-negative.   
So every real valued harmonic function, $u$ can be written as the difference of two non-negative harmonic functions when $|u|$ is also harmonic. This certainly allows $u$  to be from the set of linear polynomials in $x$ and $y$ that is $u = ax+by$ where $a, b \in \mathbb{R}$ as $|u|$ is not harmonic and hence the above identity does not apply.
