Infimum is uniquely achieved problem in Hilbert space Let $M \subset H$ be a closed linear subspace that is not reduced to $\{0\}$. Let $f \in H$, $f \notin M^\perp$. Prove that
$$m=\inf_{u\in M\\ \|u\|=1}(f,u)$$
is uniquely achieved.
I was thinking about varying $ u \in H $ in the expression $(f,u)$ and via Riesz's representation theorem to find a functional $\varphi$ such that $\varphi(u) = (f,u)$ and then use this corollary

the problem is that the set $\{ u \in M: \|u\|=1\}$ is not convex,
and also this result does not guarantee uniqueness. Any better ideas?
 A: Hint: A standard result tells us that we can write
$$
f=f_M+f_{M^\perp}\ ,
$$
where $\ f_M\in M\ $ and $\ f_{M^\perp}\in M^\perp\ $.  For any $\ u\in M\ $ with $\ \|u\|=1\ $, we have
$$
\left|(f,u)\right|= \left|\left(f_M,u\right)\right| \le\left\|f_M\right\|\ ,
$$
and hence
$$
(f,u)\ge-\left\|f_M\right\|
$$
if $\ (f,u)\ $ is real. But if
$\ v=-\frac{f_M}{\left\|f_M\right\|}\ $, then $\ v\in M\ $, and
$$
(f,v)=-\left\|f_M\right\|\  
$$
Also, if $\ w\ $ is any other unit member of $\ M\ $, with $\ (f,w)= -\left\|f_M\right\|\ $, what can you then conclude about the value of $\ \|w-v\|\ $?
A: I will assume that you are dealing with a real Hilbert space.
Let $c=\inf \{(f,u): u \in M , \|u\|\leq 1\}$. The corollary can be applied to this and we get a unique $u_0 \in M$ such that $\|u_0\| \leq 1$ and $c=f(u_0) \leq f(u)$ whenever $u \in M$ and $\|u\| \leq 1$. Claim: $\|u_0\|=1$. For this first observe that $c$ cannot be $0$. If $c=0$ then $0 \leq (f,u)$ whenever $u \in M$ and $\|u\| \leq 1$ and changing $u$ to $-u$ we see that $(f,u)\leq 0$ whenever $u \in M$ and $\|u\| \leq 1$. But then $f \in M^{\perp}$ a contradiction.
Rest is simple:  If $\|u_0\|<1$ and $f(u_0) >0$ take $u =(1+\epsilon) u_0$ with a small enough to get a  contradiction (to the fact that $f(u_0$ minimizes $(f,u)$). If $f(u_0) <0$ take $u =(1-\epsilon) u_0$ with a small enough to get a  contradiction.
We have proved our claim that $\|u_0\|=1$ and this implies that $f(u_0)=\inf \{(f,u): u \in M , \|u\|= 1\}$.
I will leave the uniqueness part to you.
