An Exercise About Best Approximation in Hilbert Space

Question

Suppose $H$ is a Hilbert Space. Let $M \subset H$ be a closed linear subspace that is not reduced to $\{0\}$, $f \in H, f \not \in M^{\perp}$, prove that there exists a unique $u \in M$ such that \begin{equation*} u \in M, \|u\| = 1, (f,u) = \inf_{v \in M, \|v\| = 1}(f,v) \end{equation*}

Is there a version in a Banach space without inner product structure (we can assume $f \in H^*$)

Answer 1

There are counter-examples, that is, there are continuous linear functionals on Banach spaces such that the supremum in the definition of the norm of a functional is not attained.

For example, on the Banach space $c_0$, consider the linear functional $$ f(x) = \sum_{i=1}^\infty 2^{-i}x_i. $$ It is not hard to show that $\Vert f\Vert=1$, but there is no $x$ in $c_0$ such that $\Vert x\Vert=1$, and $|f(x)|=1$. The missing $x$ wants to be $(1,1,1,\ldots)$ but of course it cannot.

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The result you are looking for holds for reflexive Banach spaces (see MotylaNogaTomkaMazu's answer) and in fact, only for reflexive spaces. This is the content of James' Theorem [1] which says that if every continuous linear functional on $E$ attains its norm, then $E$ is reflexive.

After James' Theorem appeared, the notion of subreflexivity was proposed to refer to Banach spaces such that the set of norm-attaining linear functionals is dense in the dual. However Bishop and Phelps [2] proved that every Banach space is subreflexive so the term subreflexive fell from grace!

[1] James, Robert C., "Reflexivity and the supremum of linear functionals", Ann. of Math., 66 (1957), 159–169.

[2] Bishop, Errett and Phelps, R. R., "A proof that every Banach space is subreflexive", Bulletin of the American Mathematical Society, 67 (1961), 97–98.

Answer 2

First observe that $t(u) =(f,u)$ is continuous linear functional defined on $H.$ Then observe that $$\inf_{||v||=1 , v\in M} t(v) =\inf_{||v||\leq 1 , v\in M} t(v) . $$ Then observe that $(M , (,))$ is with the scalar product induced from $H$ is again a Hilbert space and hence reflexive with the unit ball $B=\{x\in M: ||x||\leq 1\}.$ But in the reflexive spaces the unit ball is compact in the weak topology and every linear continuous functional on reflexive space is weak continuous. So since $t$ is continuous function on compact set $B$ there is an $u\in B$ such that $t(u) =\inf_{v\in B} t(v).$ But since $t$ is linear therefore must $||u||=1.$