Let $H$ be a Hilbert space, and $x_1, x_2 \in H$ and $r_1, r_2 > 0$ satisfying $\|x_1 - x_2\| = r_1 + r_2$. Define $B_j := \{x \in H : \|x - x_j\| \leq r_j\}$ for $j = 1, 2$.
If $H$ were just $\mathbb R^2$ (or $\mathbb R^n$), then it is clear that $B_1 \cap B_2$ contains only one element. Now, I would like to know whether this holds true for general Hilbert space $H$.
I believe this is false, but failed to give a counterexample. I could easily come up with an example for $L^1([0,1])$: let $x_1(t) = 1$ and $x_2(t) = 0$ for all $t \in [0,1]$ and $r_1 = r_2 = 1/2$. Then $(1/2+\epsilon)\chi_{[0,1/2]} + (1/2-\epsilon)\chi_{[1/2,1]} \in B_1 \cap B_2$ for any $\epsilon \in [0, 1/2)$.
Now, I tried to give a similar counterexample for $L^2([0,1])$, but I am unable to give an explicit example.
I would appreciate if you could help me with this problem. If the statement is indeed false, please give a counterexample. If the statement is true, then please give some outline of the proof. I would also appreiciate any reference.