In a remark to the projection theorem for Hilbert spaces I read this conjecture of a more general projection theorem:

Let $X$ be a reflexive Banach space and $K\subset X$ nonempty, closed and convex. Then for every $x\in X$ there exists $y\in K$ such that $$\| x-y\|=d(x,K)=\inf_{z\in K} \|x-z\|$$

Now I tried showing this similarly to the projection theorem for Hilbert spaces, but this didn't get me far as the proof I used makes use of the parallelogram equation heavily, which does not hold in Banach spaces. So how can I show this?

  • $\begingroup$ Martin's answer here may help. $\endgroup$ Jun 26, 2014 at 19:12

2 Answers 2


Recall that:

  1. Since $X$ is reflexive, any closed ball is compact in the weak topology.
  2. Every convex closed subset of $X$ (in particular $K$) is closed in the weak topology

Fix any point $w \in K$. If $||x-w|| \leq d(x, K)$, we are ok.

Otherwise consider $K' = \{ y \in K : ||x-y|| \leq ||x-w|| \} = K \cap \bar B(x, ||x-w||)$.

$K'$ is convex and closed since it is the intersection of two convex and closed sets, so it is closed for the weak topology. Moreover it is compact in the weak topology, since it is a closed subset of $\bar B(x, ||x-w||)$ which is compact.

Now, if $X$ is endowed with the weak topology, the map $y \mapsto ||x-y||$ is lower semi-continuous, hence it has a minimum in $K'$, because $K'$ is compact. But clearly a minimum in $K'$ is a minimum in $K$.

So we showed that there exists $y \in K$ such that for every $z \in K$ holds $||x-y|| \leq ||x-z||$.

  • $\begingroup$ Why is the ball $\bar{B}(x, \| x- w\|) $ weakly compact? Don't we just have that $X$ is reflexive if and only if the closed unit ball is weakly compact? Thanks! $\endgroup$ Dec 6, 2020 at 19:20
  • 1
    $\begingroup$ In any Banach space all closed balls are homeomorphic. If the unit ball is weakly compact, then so are all other closed balls. $\endgroup$
    – Crostul
    Dec 6, 2020 at 23:39

Given $x\in X$, there exist $\{y_n\}\subset K$ such that $d(x,K)<||x-y_n||<d(x,K)+1/n$ by the definition of infimum. Since $X$ is reflexive Banach, by Banach-Eberlein-Smulian Theorem, there exist $\{y_{n_k}\}$ such that $y_{n_k}\rightharpoonup y$ for some $y\in K$. By Banach-Mazur Theorem, there is $\{p_n\}$ such that $p_n$ is a convex linear combination of $y_1,y_2,...,y_n$ and $p_n\rightarrow y$. Then $y\in K$ since K is convex and closed. Then $d(x,K)\leq d(x,y)=d(x,\lim_{n\rightarrow\infty}p_n)=\lim_{n\rightarrow\infty}d(x,p_n)\leq d(x,K)+1$, where the last inequality is from $p_n$ being convex linear combination of $y_1,y_2,...,y_n\ \forall n$. Doing the previous argument for $\{y_n\}_{n=2}^\infty,\{y_n\}_{n=3}^\infty,\{y_n\}_{n=3}^\infty,...$, we can get $d(x,K)\leq d(x,y)\leq d(x,K) + 1/n \ \forall n$, which implies $d(x,y)=d(x,K)$.

Banach-Eberlein-Smulian Theorem:Let $X$ be a Banach space, then it is reflexive iff every bounded sequence has a weakly convergent subsequence.

Banach-Mazur Theorem: If $x_n\rightharpoonup x$, then there exist $p_n$ such that $p_n\rightarrow x$ and $p_n$ is a convex linear combination of $x_1,x_2...x_n$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .