Projections onto closed and convex sets

Question

I have to prove that if $A$ is convex and closed set, then $z=P_A(x)$ for all $z\in A$ if and only if $\langle x-z, z-y\rangle \geq 0$ for all $y\in A$

I have following proof which is not much complicated, but I don't understand few things.

If $g(\theta)=||x-((1-\theta)z+\theta y)||^2, \theta \in R, z=P_A(x), y\in A$ is quadratic function of the variable $\theta$ and it has minimum at $\theta =-\frac{\langle x-z,z-y\rangle}{||z-y||^2}$

Now there is a part that I don't understand:

For $z=P_A(x)$, from convexity of a set A, we get $g(0)\leq g(\theta)$ for all $\theta \in [0,1]$, so $\theta_{min} \leq 0$.

I know why $g(0)\leq g(\theta)$ (I can see it by simply putting $0$ in function), but I don't know how convexity of $A$ caused that and why did we take $\theta$ from $[0,1]$.

The rest of the proof is ok.

Would anybody try to make this clear to me?

Answer 1

If $z$ is the projection of $x$ to $A$, then it's the closest point to $x$ in $A$. If $y$ is any other point in $A$ and $0\leq\theta\leq 1$, then $(1-\theta)z+\theta y$ is in $A$ because $A$ is convex. (Notice that this would not follow if $\theta$ were outside $[0,1]$.) So the distance from $x$ to this point $(1-\theta)z+\theta y$ in $A$, must be at least the distance from $x$ to $z$. Square both sides of that inequality (because squared distances are algebraically nicer than distances in Euclidean space), and you get $g(\theta)\geq g(0)$. So $g(\theta)$, with $\theta$ restricted to $[0,1]$, takes its minimum value at $\theta=0$.