Infimum of inner product over a closed subspace Let $M\subset H$ a closed linear subspace of a Hilbert space H. Suppose $f \in H$ but $f \not\in M^{\perp}$. Show that
$$ \inf_{x\in M,\|x\|=1} (f,x)$$
is attained and the min is unique.
So what I've tried so far was to use extreme value theorem. I know that the closed unit circle is compact with respect to $\sigma(H,H^*)$ due to reflexivity and $M$ is weakly closed as well so the intersection would be weakly compact, EVT would come in here until I realized the infimum is only over the shell intersect M, so my idea stops there.
I also tried constructing a sequence $(f,x_n)$ converging to the infimum, since M is also reflexive and each $x_n$ has norm 1, there exists a weakly convergent subsequence. If $x$ is the weak limit and $\|x\|=1$, I would be done. But I don't know how to show $\|x\|=1$, I do know however $\|x\| \leq 1$ simply by weak convergence.
Just looking for a hint!!
Thank you!
 A: You are almost there. As you said we know that the unit ball is weak* compact, so by the EVT there exists $u$ in the closed unit ball so that $\inf_{\|x\|\leq 1}(f,x)=(f,u)$. We also note that $f\neq 0$ so $u\neq 0$. Set $\alpha=1/\|u\|$. We claim that $\inf_{\|x\|=1}(f,x)=(f,\alpha u)$. Assume not: Then there exists a $v$ on the unit sphere such that $(f,v)<(f,\alpha u$). Then $\frac{1}{\alpha}(f,v)<(f,u)$, meaning that $(f,\|u\|v)<(f,u)$, which contradicts the choice of $u$.
This approach, however, is not the best approach in order to see uniqueness. We need to exploit the geometry of the Hilbert space. Note that we can write $f=P_Mf+P_{M^\perp}f$. Thus for any $x\in M$ we have $(f,x)=(P_Mf,x)$. Combining this with Cauchy-Schwarz, we see that
$$\inf_{\|x\|=1}(f,x)\geq-\|P_Mf\|.$$
Clearly this infimum is achieved at $x=\frac{-P_Mf}{\|P_Mf\|}.$ Finally we remember that equality is achieved in Cauchy-Schwarz if and only if the two elements are linearly dependent. I.e. for any $x,y\in H$ $|(y,x)|=\|y\|\|x\|$ if and only if $x=\alpha y$. As any ray only intersects the unit sphere at a unique point, we are done.
