I was trying to understand the following optimization problem:
$$argmin_{v \in H} {R(v) + \frac{1}{2}||v - w||^2}$$
Assume $R(v)$ is Convex, proper and semi-continuous with a unique minimizer. Notice that w is a constant vector i.e. its fixed. $H$ is just some vector space.
For ease of notation, let:
$$F(v) = R(v) + \frac{1}{2}||v - w||^2$$
Then the vector $w^*$ that minimizes the above problem must have zero as an element of the sub gradient. i.e.:
$$0 \in \partial F(w^*)$$
If we use the definition of F we get:
$$0 \in \partial(R(v) + \frac{1}{2}||v - w||^2)$$
distributing the sub-gradient we get:
$$0 \in \partial R(v) + v - w$$
This is the crucial step in my question (i.e. what my question is really about). At this point we are allowed to do "normal" algebra with the above (even though there is no equality and its in terms of sets) and produce the next step.
$$w \in \partial R(v) + v$$
I am having problems accepting that this can be done. I think the problem is that I am not sure what that step means mathematically or don't really understand intuitively what it means.
Basically, is there a rigorous and/or intuitive explanation of why we can do "normal" algebra with sets? I am just trying to justify doing algebra with sets and the $\in$ symbol.
In order to be able to do "normal" algebra in this situation, you just need to justify that $$x \in A + y$$ is equivalent to $$x - y \in A$$ where $V$ is a vector space and $x,y \in V$ and $A \subset V$.
This is easy to check. If the first statement holds, then $x = a + y$ for some $a \in A$. Now we really have an equality of elements, so we can do "normal" algebra to get $x - y = a$, which means that the second statement holds. Similarly the second implies the first.
In general, adding an element of a vector space (or even an abelian group) to a statement of the form $x \in A + y$ or even of the form $A \subset B$ results in an equivalent statement, basically because there is an inverse operation which is to subtract the element. But you can't add/subtract a subset (of more than one element) from both sides, because there is no inverse operation: The set $A - A$, if $A \subset V$, is not $\{0\}$, because $A - A = \{ a - b : a,b \in A\}$.