I am trying to prove the following proposition
Let $X$ be a uniformly convex normed vector space and $A\subseteq X$ be a closed and convex subset, let $x \in X$. Show that there exists an unique $a \in A$ such that $$ \Vert x-a \Vert = \inf_{b \in A} \Vert x-b \Vert \; . $$
I have not yet thought much about existence since I guess this will follow relatively easily from the closedness of $A$.
The interesting part is the following: While thinking about uniqueness, I suddenly got the feeling that it is sufficient to proof the theorem for $x=0$. For $x = 0$ we get the following simple proof:
Assume $a,a' \in A$, $a \neq a'$ and
$$
\underbrace{\Vert x-a \Vert}_{= \Vert a \Vert} = \underbrace{\Vert x -a' \Vert}_{= \Vert a' \Vert} = \inf_{b \in A} \Vert x-b \Vert
$$
by rescaling we can set $\Vert a \Vert = \Vert a' \Vert = 1$ (I am a bit unsure about this, intuitively it seems obvious, but I don't know how to properly formalize this).
Then by the convexity of $A$: $(1-t)a+ta' \in A$ for $t \in [0,1]$, hence
$$
\frac{a}{2} + \frac{a'}{2} = \frac{a+a'}{2} \in A \; ,
$$
then we can use the uniformly convexity to get
$$
\left \Vert \frac{a+a'}{2} \right \Vert < 1
$$
and therefore
$$
\left \Vert \frac{a+a'}{2} \right \Vert < \Vert a \Vert = \Vert a' \Vert = \inf_{b \in A} \Vert b \Vert
$$
which is a contradiction to $a,a'$ being the infimum of $A$.
The problem is that I am unable to find out why $x=0$ should be sufficient, my intuition just tells my that the position of the origin is irrelevant for this theorem.
I remember a proof by Hirzebruch, showing that continuity of a linear map at zero is equivalent to continuity at an arbitrary point, his argument was of the kind "sure it this is equivalent, linearity can put the zero wherever you want".
So my question is: Is there any kind of general theorem which tells us which probiertes of normed vector spaces are of the form $$ P \text{ holds for } x=0 \Leftrightarrow P \text{ holds for any } x \; \text{?} $$
Another example of such a theorem is the compactness of the unit ball: We most often look at the closed ball with radius one around zero, but the position of the center of the closed ball should be irrelevant (I have no proof for that, so correct me if I am wrong).