Unicity of solution of $\inf_{d\in X}\{\langle z,d\rangle:\||d\|| \leq 1\}$

Question

Let (X, ⟨·, ·⟩) Hilbert space y ∥ · ∥ the induced norm by the inner product ⟨·, ·⟩ in X. For a element z $\in$ $X/{0}$. Consider the problem

$\inf_{d\in X}\{\langle z,d\rangle:||d|| \leq 1\}$

Show that for all $z\neq 0$, the problem has a unique solution.

I have already proved the existence of the solution by the Weiertrass theorem, but I still have to prove the uniqueness of the solution. I have tried to show it by the common method: By contradiction, let us assume that there exists $d_{1},d_{2}\in B(0,1)$ such that

$\inf\{\langle z,d_{1}\rangle\} = \inf\{\langle z,d_{2}\rangle\} = \alpha$

I have tried to play with the properties of the norm and the definition of the problem, but I have not yet come to a contradiction that is clear to me. ¿Can you give me a hint, please?

Answer 1

Let $d$ be any solution. Let us show that $d=-\frac z {\|z\|}$. First note that $ \langle z, d \rangle \leq \langle z, -\frac z {\|z\|} \rangle$. This gives $\langle z, d \rangle \leq -\|z\|$. Now $\|z+\|z\|d\|^{2}=\|z\|^{2}+\|z\|^{2}\|d\|^{2}+2 \Re \langle z, \|z\|d \rangle\leq 0$. Hence $d=-\frac z {\|z\|}$.

Incidentally, it is easy to check that $-\frac z {\|z\|}$ is indeed a solution, so existence is also proved.

Answer 2

Any $d \in X$ can be written uniquely as $d = \alpha z+ y$, where $y \bot z$. Then $\|d\|^2 = \alpha^2 \|z\|^2 + \|y\|^2$.

Then $\langle z, d \rangle = \alpha \|z\|^2$. If $\|d\| \le 1$ then the corresponding $\alpha$ can have values in the $[-{1 \over \|z\|}, {1 \over \|z\|}]$, and hence the $\min$ value of $\langle z, d \rangle$ is $-\|z\|$. If $\langle z, d \rangle = -\|z\|$, then we must have $y=0$.

In particular, the minimising $d$ is unique.