In our lecture, we were ask to simplify the expression $x+y+|x-y|$.
This is how I simplified it:
If $x\ge y$, $|x-y|=x-y$ and $x+y+|x-y|=x+y+(x-y)=2x$.
If $x<y$, $|x-y|=y-x$ and $x+y+|x-y|=x+y+(y-x)=2y$.
Now
$x+y+|x-y|=\begin{cases}2x&\text{if}~x\geq y\\ 2y&\text{if}~x<y\end{cases}$ .
The solution was correct but our lecturer also continued this by telling that we can now also conclude from the solution that
$x+y+|x-y|=2max(x,y)$.
I'm not sure, however, how we came to this conclusion?
You came to the same conclusion, because if $x \ge y$, $2x = 2\max(x,y)$ and if $x < y$, $2y = \max(x,y)$ as well.
This is simpler as we use one "formula" instead of two depending on domain. (though $\max$ implicitly also is of this form, of course).
The point of this formula is how we can express the max in this lattice as a formula in terms of the linear operation and the norm: $\max(x,y) = \frac{1}{2}((x+y) + |x-y|)$. For the minimum a related formula exists.
If $x \geq y$, then what is $\max(x, y)$? In this case, what are $2\max(x, y)$ and $x + y + |x - y|$ equal to?
You have proved that
$$x+y+|x-y|=\begin{cases}2x&\text{if}~x\geq y\\ 2y&\text{if}~x<y\end{cases}$$
Note that if $x\ge y$ then $x$ is the maximum, so $2x$ is twice the maximum.
Otherwise $ y$ is the maximum and $2y$ is twice the maximum.
